Solution Manual For Finite Mathematics and Applied Calculus, 8th Edition by Stefan Waner,q Steven Co by digitaldownload87

Solutions Section 0.1 Section 0.1

1. 2(4 + (−1))(2 ⋅ −4) = 2(3)(−8) = (6)(−8) = −48

2. 3 + ([4 − 2] ⋅ 9) = 3 + (2 × 9) = 3 + 18 = 21

3. 20∕(3 * 4) − 1 = 20 − 1 = 53 − 1 = 23 12

4. 2 − (3 * 4)∕10 = 2 − 12 = 2 − 65 = 45 10

3 + ([3 + (−5)]) 3−2×2 3 + (−2) 1 = = = −1 3−4 −1

7. (2 − 5 * (−1))∕1 − 2 * (−1) 2 − 5 ⋅ (−1) = − 2 ⋅ (−1) 1 2+5 = +2=7+2=9 1

12 − (1 − 4) 2(5 − 1) ⋅ 2 − 1 12 − (−3) 15 = = =1 16 − 1 15

8. 2 − 5 * (−1)∕(1 − 2 * (−1)) 5 ⋅ (−1) =2− 1 − 2 ⋅ (−1) 5 11 −5 =2− +2=2+ = 1+2 3 5

2 × (−1) 2 2 2×1 2 = = =1 2 2

10. 2 + 4 ⋅ 3 2 = 2 + 4 × 9 = 2 + 36 = 38

11. 2 ⋅ 4 2 + 1 = 2 × 16 + 1 = 32 + 1 = 33

12. 1 − 3 ⋅ (−2) 2 × 2 = 1 − 3 × 4 × 2 = 1 − 24 = −23

13. 3^2+2^2+1 = 3 2 + 2 2 + 1 = 9 + 4 + 1 = 14

14. 2^(2^2-2)

9. 2 ⋅ (−1) 2∕2 =

15.

3 − 2(−3) 2 −6(4 − 1) 2 3 − 2 × 9 3 − 18 = = −6 × 9 −6(3) 2 5 −15 = = −54 18

17. 10*(1+1/10)^3 1 3 = 10(1 + ) = 10(1.1) 3 10 = 10 × 1.331 = 13.31

= 2 (2 −2) = 2 4−2 = 2 2 = 4

16.

1 − 2(1 − 4) 2 2(5 − 1) 2 ⋅ 2 1 − 2(−3) 2 1−2×9 = = 2 × 16 × 2 2(4) 2 ⋅ 2 1 − 18 17 = =− 64 64

18. 121/(1+1/10)^2 121 121 = = 2 1 1.1 2 !1 + " 10

121 = = 100 1.21

−2 ⋅ 3 2 [ −(4 − 1) 2 ] −2 × 9 ⎤ −18 = 3⎡ = 3[ 2 −9 ] ⎣⎢ −3 ⎦⎥ =3×2=6

19. 3

Solutions Section 0.1

1 2 2 21. 3⎡1 − (− ) ⎤ + 1 ⎣⎢ 2 ⎦⎥ 1 2 = 3[1 − ] + 1 4 3 2 9 = 3[ ] + 1 = 3[ ] + 1 4 16 27 43 = +1= 16 16 2 = # 12 $ − 12 = 14 − 14 = 0 2

8(1 − 4) 2 20. − [ −9(5 − 1) 2 ] 8×9 72 = −[ = −( ] −9 × 16 −144 ) 1 1 = −(− ) = 2 2 1 2 2 2 22. 3⎡ − ( ) ⎤ + 1 ⎣⎢ 9 3 ⎦⎥ 1 4 2 = 3[ − ] + 1 9 9 −3 2 −1 2 = 3[ ] + 1 = 3[ ] + 1 9 3 1 3 4 =3 +1= +1= 9 9 3 2

= 22 − # 21 $ = 21 − 41 = −2

23. (1/2)^2-1/2^2

24. 2/(1^2)-(2/1)^2

25. 3 × (2 − 5) = 3*(2-5)

26. 4 +

27.

3 = 3/(2-5) 2−5

3−1 = (3-1)/(8+6) 8+6 Note 3-1/8-6 is wrong, as it corresponds to 3 − 18 − 6. 29.

31. 3 −

33.

4+7 = 3-(4+7)/8 8

2 − 𝑥𝑦 2 = 2/(3+x)-x*y^2 3+𝑥

35. 3.1𝑥 3 − 4𝑥 −2 −

𝑥2 − 1

= 3.1x^3-4x^(-2)-60/(x^2-1)

28.

4−1 = (4-1)/3 3

30. 3 +

32.

5 = 4+5/9 or 4+(5/9) 9

3 = 3+3/(2-9) 2−9

4×2 2

!3"

34. 3 +

= 4*2/(2/3) or (4*2)/(2/3)

3+𝑥 = 3+(3+x)/(x*y) 𝑥𝑦

𝑥2 − 3 2 = 2.1x^(-3)-x^(-1)+(x^2-3)/2

36. 2.1𝑥 −3 − 𝑥 −1 +

37.

Solutions Section 0.1

!3"

5 = (2/3)/5

38.

!5"

= 2/(3/5)

39. 3 4−5 × 6 = 3^(4-5)*6 Note that the entire exponent is in parentheses.

40.

4 41. 3 1 + ( 100 )

42. 3

−3

= 2/(3+5^(7-9)) Note that the entire exponent is in parentheses. 1+4 ( 100 ) = 3((1+4)/100)^(-3)

= 3*(1+4/100)^(-3)

43. 3 2𝑥−1 + 4 𝑥 − 1 = 3^(2*x-1)+4^x-1 2

45. 2 2𝑥 −𝑥+1 = 2^(2x^2-x+1) Note that the entire exponent is in parentheses.

47.

4𝑒 −2𝑥 2 − 3𝑒 −2𝑥 = 4*e^(-2*x)/(2-3e^(-2*x)) or 4(*e^(-2*x))/(2-3e^(-2*x)) or (4*e^(-2*x))/(2-3e^(-2*x)) 2 2

49. 3(1 − !− 12 " ) + 1

= 3(1-(-1/2)^2)^2+1

2 3+5 7−9

−3

44. 2 𝑥 − (2 2𝑥) 2 = 2^(x^2)-(2^(2*x))^2 2

46. 2 2𝑥 −𝑥 + 1 = 2^(2x^2-x)+1 Note that the entire exponent is in parentheses. 𝑒 2𝑥 + 𝑒 −2𝑥 𝑒 2𝑥 − 𝑒 −2𝑥 = (e^(2*x)+e^(-2*x))/(e^(2*x)e^(-2*x)) 48.

1 2 2 2 50. 3⎛ − ( ) ⎞ + 1 ⎜⎝ 9 3 ⎟⎠ = 3(1/9-(2/3)^2)^2+1

Solutions Section 0.2 Section 0.2 1. 3 3 = 27

2. (−2) 3 = −8

3. −(2 ⋅ 3) 2 = −(2 2 ⋅ 3 2) = −(4 ⋅ 9) = −36 or −(2 ⋅ 3) 2 = −6 2 = −36

4. (4 ⋅ 2) 2 = 4 2 ⋅ 2 2 = 16 ⋅ 4 = 64 or (4 ⋅ 2) 2 = 8 2 = 64

5. (

−2 2 (−2) 2 4 = = 3 ) 9 32

7. (−2) −3 =

3 3 3 3 27 6. ( ) = 3 = 2 8 2

1 1 1 = =− 8 (−2) 3 −8

1 1 8. −2 −3 = − 3 = − 8 2

1 −2 1 1 9. ( ) = = = 16 2 4 1∕16 (1∕4)

10. (

11. 2 ⋅ 3 0 = 2 ⋅ 1 = 2

12. 3 ⋅ (−2) 0 = 3 ⋅ 1 = 3

13. 2 32 2 = 2 3+2 = 2 5 = 32 or 2 32 2 = 8 ⋅ 4 = 32

14. 3 23 = 3 23 1 = 3 2+1 = 3 3 = 27or 3 23 = 9 ⋅ 3 = 27

15. 2 22 −12 42 −4 = 2 2−1+4−4 = 2 1 = 2

16. 5 25 −35 25 −2 = 5 2−3+2−2 = 5 −1 =

17. 𝑥 3𝑥 2 = 𝑥 3+2 = 𝑥 5

18. 𝑥 4𝑥 −1 = 𝑥 4−1 = 𝑥 3

19. −𝑥 2𝑥 −3𝑦 = −𝑥 2−3𝑦 = −𝑥 −1𝑦 = −

𝑦 𝑥

1 1 9 −2 −2 = = = 2 3 ) 4∕9 4 (−2∕3)

1 5

20. −𝑥𝑦 −1𝑥 −1 = −𝑥 1−1𝑦 −1 = −𝑥 0𝑦 −1 1 =− 𝑦

21.

𝑥3 1 = 𝑥 3−4 = 𝑥 −1 = 𝑥 𝑥4

22.

𝑦5 = 𝑦 5−3 = 𝑦 2 3 𝑦

23.

𝑥 2𝑦 2 = 𝑥 2−(−1)𝑦 2−1 = 𝑥 3𝑦 𝑥 −1𝑦

24.

𝑥 −1𝑦 1 = 𝑥 −1−2𝑦 1−2 = 𝑥 −3𝑦 −1 = 3 𝑥 2𝑦 2 𝑥 𝑦

25.

(𝑥𝑦 −1𝑧 3) 2 𝑥 2(𝑦 −1) 2(𝑧 3) 2 = 𝑥 2𝑦𝑧 2 𝑥 2𝑦𝑧 2 2 −2 6 𝑥 𝑦 𝑧 = 2 2 = 𝑥 2−2𝑦 −2−1𝑧 6−2 𝑥 𝑦𝑧 𝑧4 = 𝑥 0𝑦 −3𝑧 4 = 3 𝑦

27. !

𝑥𝑦 −2𝑧 3 (𝑥𝑦 −2𝑧) 3 " = 𝑥 −1𝑧 (𝑥 −1𝑧) 3 3 −6 3 𝑥 𝑦 𝑧 = −3 3 = 𝑥 3−(−3)𝑦 −6𝑧 3−3 𝑥 𝑧 𝑥6 = 𝑥 6𝑦 −6𝑧 0 = 𝑦6 𝑥 −1𝑦 −2𝑧 2 −1−1 −2−1 2 −2 𝑦 𝑧 ) ) = (𝑥 𝑥𝑦 = (𝑥 −2𝑦 −3𝑧 2) −2 = 𝑥 4𝑦 6𝑧 −4 𝑥 4𝑦 6 = 4 𝑧 −2

29. (

31. √4 = 2

𝑥 2𝑦𝑧 2 𝑥 2𝑦𝑧 2 = (𝑥𝑦𝑧 −1) −1 𝑥 −1𝑦 −1(𝑧 −1) −1 𝑥 2𝑦𝑧 2 = −1 −1 = 𝑥 2−(−1)𝑦 1−(−1)𝑧 2−1 𝑥 𝑦 𝑧 = 𝑥 3𝑦 2𝑧 4

Solutions Section 0.2 26.

𝑥 2𝑦 −1𝑧 0 (𝑥 2𝑦 −1𝑧 0) 2 = 𝑥𝑦𝑧 ) (𝑥𝑦𝑧) 2 𝑥 4𝑦 −2 = 2 2 2 = 𝑥 4−2𝑦 −2−2𝑧 −2 𝑥 𝑦 𝑧 𝑥2 = 𝑥 2𝑦 −4𝑧 −2 = 4 2 𝑦 𝑧 2

28. (

30.

𝑥𝑦 −2 = (𝑥 1−2𝑦 −2+1𝑧 −1) −3 ( 𝑥 2𝑦 −1𝑧 ) = (𝑥 −1𝑦 −1𝑧 −1) −3 = 𝑥 3𝑦 3𝑧 3 −3

32. √5 ≈ 2.236

33.

1 √1 1 = = √ 4 √4 2

34.

1 √1 1 = = √ 9 √9 3

35.

16 √16 4 = = √9 3 √9

36.

9 √9 3 = = √ 4 √4 2

37.

√4

2 5

38.

√25

6 5

39. √9 + √16 = 3 + 4 = 7

40. √25 − √16 = 5 − 4 = 1

41. √9 + 16 = √25 = 5

42. √25 − 16 = √9 = 3

3 √27

43. √8 − 27 = √−19 ≈ −2.668 45. √27∕8 =

3 √8

3 2

44. √81 − 16 = √65 ≈ 2.839 3

46. √8 × 64 = √8 ⋅ √64 = 2 ⋅ 4 = 8

Solutions Section 0.2

47. √(−2) 2 = √4 = 2

49.

48. √(−1) 2 = √1 = 1

1 16 √16 4 (1 + 15) = = = =2 √4 √4 √4 2

50.

1 36 √36 6 (3 + 33) = = = =2 √9 √9 3 √9

51. √𝑎 2𝑏 2 = √𝑎 2√𝑏 2 = 𝑎𝑏

𝑎 2 √𝑎 2 𝑎 52. √ 2 = = 𝑏 𝑏 √𝑏 2

53. √(𝑥 + 9) 2 = 𝑥 + 9 (𝑥 + 9 > 0 because 𝑥 is positive.)

54. (√𝑥 + 9) = 𝑥 + 9

55. √𝑥 3(𝑎 3 + 𝑏 3) = √𝑥 3 √(𝑎 3 + 𝑏 3) 3

= 𝑥 √(𝑎 3 + 𝑏 3)

57.

(Not 𝑥(𝑎 + 𝑏))

4𝑥𝑦 3 4𝑦 2 √4√𝑦 2 2𝑦 = = = √ 2 √𝑥 √𝑥 𝑥 √𝑥 𝑦

56. √

58.

√𝑥 4 𝑥4 𝑥 = = 4 4 𝑎𝑏 𝑎 4𝑏 4 √𝑎 4 √𝑏 4

4(𝑥 2 + 𝑦 2) √4√𝑥 2 + 𝑦 2 = 𝑐2 √𝑐 2 √ =

2√𝑥 2 + 𝑦 2 𝑐

(Not 2(𝑥 + 𝑦)∕𝑐)

Solutions Section 0.3 Section 0.3

3 𝑥4

1 −4 1 𝑥 = 4 2 2𝑥

3 −2∕3 3 𝑥 = 4 4𝑥 2∕3

4 −3∕4 4 𝑦 = 5 5𝑦 3∕4

1 0.1𝑥 −2 𝑥 4 0.1 + = + 2 3 3 3𝑥 −4 3𝑥

1. 3𝑥 −4 =

6 −1 6 5. 1 − 0.3 = 1 − 0.3𝑥 2 − 5𝑥 −2 − 5 𝑥 𝑥

7. 2 2∕3 = √2 2

8. 3 4∕5 = √3 4

9. 𝑥 4∕3 = √𝑥 4 = √𝑥 3 √𝑥 = 𝑥 √𝑥 11. (𝑥 1∕2𝑦 1∕3)

1∕5

= √√ 𝑥 √ 𝑦 3

3 3 3 13. − 𝑥 −1∕4 = − =− 4 1∕4 2 2𝑥 2 √𝑥 15. 0.2𝑥 −2∕3 +

7𝑥 −1∕2 0.2 3√ 𝑥 = 3 + 7 √𝑥 2

17.

0.2 3𝑥 1∕2 + 7 𝑥 2∕3

3 3 = 5∕2 4(1 − 𝑥) 4√(1 − 𝑥) 5 3 = 4√(1 − 𝑥) 4√1 − 𝑥 3 = 4(1 − 𝑥) 2√1 − 𝑥

10. 𝑦 7∕4 = √𝑦 7 = √𝑦 4 √𝑦 3 = |𝑦| √𝑦 3 12. 𝑥 −1∕3𝑦 3∕2 =

14.

16.

18.

𝑦 3∕2 √𝑦 3 |𝑦|√𝑦 = 3 = 3 𝑥 1∕3 √𝑥 √𝑥

4 3∕2 4√𝑥 3 4|𝑥|√𝑥 𝑥 = = 5 5 5 3.1 11 −1∕7 11 𝑥 − = 3.1𝑥 4∕3 − −4∕3 7 𝑥 7𝑥 1∕7 3 3 11 11 = 3.1 √𝑥 4 − 7 = 3.1𝑥 √𝑥 − 7 7 √𝑥 7 √𝑥 9(1 − 𝑥) 7∕3 9 = 4 4(1 − 𝑥) −7∕3 3

9 √(1 − 𝑥) 7 9 √(1 − 𝑥) 6 √1 − 𝑥 = = 4 4 3 9(1 − 𝑥) 2 √1 − 𝑥 = 4

19. √3 = 3 1∕2

20. √8 = 8 1∕2

21. √𝑥 3 = 𝑥 3∕2

22. √𝑥 2 = 𝑥 2∕3

23. √𝑥𝑦 2 = (𝑥𝑦 2) 1∕3 25.

Solutions Section 0.3

𝑥2 𝑥2 = = 𝑥 2−1∕2 = 𝑥 3∕2 √𝑥 𝑥 1∕2

24. √𝑥 2𝑦 = (𝑥 2𝑦) 1∕2 26.

𝑥

√𝑥

𝑥

𝑥 1∕2

= 𝑥 1−1∕2 = 𝑥 1∕2

27.

3 3 = 𝑥 −2 2 5 5𝑥

28.

2 2 = 𝑥3 −3 5 5𝑥

29.

3𝑥 −1.2 1 3 1 − 2.1 = 𝑥 −1.2 − 𝑥 −2.1 2 2 3 3𝑥

30.

2 𝑥 2.1 2 1.2 1 2.1 − = 𝑥 − 𝑥 3 3 3 3𝑥 −1.2

31.

33.

2𝑥 𝑥 0.1 4 − + 1.1 3 2 3𝑥 2 1 4 = 𝑥 − 𝑥 0.1 + 𝑥 −1.1 3 2 3 3√𝑥 5 4 − + 4 3√𝑥 3𝑥√𝑥 3𝑥 1∕2 5 4 = − + 1∕2 4 3𝑥 3𝑥 ⋅ 𝑥 1∕2 3 5 4 = 𝑥 1∕2 − 𝑥 −1∕2 + 𝑥 −3∕2 4 3 3 5

3 √𝑥 2 7 3𝑥 2∕5 7 35. − = − 4 4 2𝑥 3∕2 2√𝑥 3 3 7 = 𝑥 2∕5 − 𝑥 −3∕2 4 2

34.

36.

38.

39. 4 −1∕24 7∕2 = 4 −1∕2+7∕2 = 4 6∕2 = 4 3 = 64

40.

(𝑥 2 + 1) 3

−

4 √𝑥 2 + 1 1 3 = − (𝑥 2 + 1) 3 (𝑥 2 + 1) 1∕3 3 = (𝑥 2 + 1) −3 − (𝑥 2 + 1) −1∕3 4

41. 3 2∕33 −1∕6 = 3 2∕3−1∕6 = 3 1∕2 = √3

4𝑥 2 𝑥 3∕2 2 + − 2 3 6 3𝑥 4 1 2 = 𝑥 2 + 𝑥 3∕2 − 𝑥 −2 3 6 3 3

5√𝑥

−

5√𝑥 7 + 3 8 2 √𝑥

3 5𝑥 1∕2 7 − + 1∕2 8 5𝑥 2𝑥 1∕3 3 −1∕2 5 1∕2 7 −1∕3 = 𝑥 − 𝑥 + 𝑥 5 8 2 =

37.

32.

1 2 1 2 − = − 5 3∕2 √ 3𝑥 3∕5 8𝑥 𝑥 3 √𝑥 3 8𝑥 1 2 = 𝑥 −3∕2 − 𝑥 −3∕5 8 3 3

3 √(𝑥 2 + 1) 7 2 − 4 3(𝑥 2 + 1) −3 2 2 3 = (𝑥 + 1) 3 − (𝑥 2 + 1) 7∕3 3 4 2 1∕𝑎 1 = 2 1∕𝑎−2∕𝑎 = 2 −1∕𝑎 = 2∕𝑎 1∕𝑎 2 2

42. 2 1∕32 −12 2∕32 −1∕3 = 2 1∕3−1+2∕3−1∕3 1 = 2 −1∕3 = 1∕3 2

43.

𝑥 3∕2 1 = 𝑥 3∕2−5∕2 = 𝑥 −1 = 5∕2 𝑥 𝑥

45.

𝑥 1∕2𝑦 2 = 𝑥 1∕2−(−1∕2)𝑦 2−1 = 𝑥𝑦 𝑥 −1∕2𝑦

47.

44.

𝑦 5∕4 = 𝑦 5∕4−3∕4 = 𝑦 1∕2 = √𝑦 𝑦 3∕4

46.

𝑥 −1∕2𝑦 = 𝑥 −1∕2−2𝑦 1−3∕2 = 𝑥 −5∕2𝑦 −1∕2 𝑥 2𝑦 3∕2

Solutions Section 0.3

𝑥 1∕3 𝑦 2∕3 𝑥 1∕3 𝑦 2∕3 ⋅ = ( 𝑦 ) (𝑥) 𝑦 1∕3 𝑥 2∕3 = 𝑥 1∕3−2∕3𝑦 −1∕3+2∕3 = 𝑥 −1∕3𝑦 1∕3 or ( 𝑥𝑦 ) 1∕3

48.

49. 𝑥 2 − 16 = 0 ⇒ 𝑥 2 = 16 ⇒ 𝑥 = ±√16 ⇒ 𝑥 = ±4

𝑥 −1∕3 𝑦 1∕3 𝑥 −1∕3 𝑦 1∕3 ⋅ = (𝑥) (𝑦) 𝑦 −1∕3 𝑥 1∕3 𝑥 −1∕3−1∕3𝑦 1∕3+1∕3 = 𝑥 −2∕3𝑦 2∕3 or ( 𝑥𝑦 ) 2∕3

50. 𝑥 2 − 1 = 0 ⇒ 𝑥 2 = 1 ⇒ 𝑥 = ±√1 ⇒ 𝑥 = ±1

4 4 = 0 ⇒ 𝑥2 = 9 9 4 2 ⇒𝑥=± ⇒𝑥=± √9 3

1 1 = 0 ⇒ 𝑥2 = 10 10 1 1 ⇒𝑥=± ⇒𝑥=± √ 10 √10

51. 𝑥 2 −

52. 𝑥 2 −

53. 𝑥 2 − (1 + 2𝑥) 2 = 0 ⇒ 𝑥 2 = (1 + 2𝑥) 2 ⇒ 𝑥 = ±1 + 2𝑥 If 𝑥 = 1 + 2𝑥, then −𝑥 = 1 ⇒ 𝑥 = −1. 1 If 𝑥 = −(1 + 2𝑥), then = −1 ⇒ 𝑥 = − . 3 1 So, 𝑥 = 1 or − . 3

54. 𝑥 2 − (2 − 3𝑥) 2 = 0 ⇒ 𝑥 2 = (2 − 3𝑥) 2 ⇒ 𝑥 = ±2 − 3𝑥 1 If 𝑥 = 2 − 3𝑥, then 4𝑥 = 2 ⇒ 𝑥 = . 2 If 𝑥 = −(2 − 3𝑥), then −2𝑥 = −2 ⇒ 𝑥 = 1. 1 So, 𝑥 = 1 or . 2

55. 𝑥 5 + 32 = 0 ⇒ 𝑥 5 = −32

56. 𝑥 4 − 81 = 0 ⇒ 𝑥 4 = 81 4 ⇒ 𝑥 = ± √81 = ±3

57. 𝑥 1∕2 − 4 = 0 ⇒ 𝑥 1∕2 = 4 ⇒ 𝑥 = 4 2 = 16

58. 𝑥 1∕3 − 2 = 0 ⇒ 𝑥 1∕3 = 2 ⇒ 𝑥 = 23 = 8

⇒𝑥=

5 √−32 = −2

1 1 =0⇒1= 2 2 𝑥 𝑥 ⇒ 𝑥 2 = 1 ⇒ 𝑥 = ±√1 = ±1

59. 1 −

61. (𝑥 − 4) −1∕3 = 2 ⇒ 𝑥 − 4 = 2 −3 = ⇒𝑥=4+

1 33 = 8 8

60.

1 8

2 6 2 6 − 4=0⇒ 3= 4 3 𝑥 𝑥 𝑥 𝑥 ⇒ 2𝑥 4 = 6𝑥 3 ⇒ 2𝑥 = 6 ⇒ 𝑥 = 3

62. (𝑥 − 4) 2∕3 + 1 = 5 ⇒ (𝑥 − 4) 2∕3 = 4 ⇒ 𝑥 − 4 = ±4 3∕2 = ±8 ⇒ 𝑥 = 4 ± 8 = −4 or 12

Solutions Section 0.4 Section 0.4

1. 𝑥(4𝑥 + 6) = 4𝑥 2 + 6𝑥

2. (4𝑦 − 2)𝑦 = 4𝑦 2 − 2𝑦

3. (2𝑥 − 𝑦)𝑦 = 2𝑥𝑦 − 𝑦 2

4. 𝑥(3𝑥 + 𝑦) = 3𝑥 2 + 𝑥𝑦

5. (𝑥 + 1)(𝑥 − 3) = 𝑥 2 + 𝑥 − 3𝑥 − 3 = 𝑥 2 − 2𝑥 − 3

6. (𝑦 + 3)(𝑦 + 4) = 𝑦 2 + 3𝑦 + 4𝑦 + 12 = 𝑦 2 + 7𝑦 + 12

7. (2𝑦 + 3)(𝑦 + 5) = 2𝑦 2 + 3𝑦 + 10𝑦 + 15 = 2𝑦 2 + 13𝑦 + 15

8. (2𝑥 − 2)(3𝑥 − 4) = 6𝑥 2 − 6𝑥 − 8𝑥 + 8 = 6𝑥 2 − 14𝑥 + 8

9. (2𝑥 − 3) 2 = 4𝑥 2 − 12𝑥 + 9

10. (3𝑥 + 1) 2 = 9𝑥 2 + 6𝑥 + 1

11. !𝑥 + 𝑥1 " = 𝑥 2 + 2 + 12

12. !𝑦 − 1𝑦 " = 𝑦 2 − 2 + 12

13. (2𝑥 − 3)(2𝑥 + 3) = (2𝑥) 2 − 3 2 = 4𝑥 2 − 9

14. (4 + 2𝑥)(4 − 2𝑥) = 4 2 − (2𝑥) 2 = 16 − 4𝑥 2

𝑥

𝑦

15. !𝑦 − 1𝑦 "!𝑦 + 1𝑦 " = 𝑦 2 − ! 1𝑦 " = 𝑦 2 − 12

16. (𝑥 − 𝑥 2)(𝑥 + 𝑥 2) = 𝑥 2 − (𝑥 2) 2 = 𝑥 2 − 𝑥 4

17. (𝑥 2 + 𝑥 − 1)(2𝑥 + 4) = 𝑥 2(2𝑥 + 4) + 𝑥(2𝑥 + 4) − 1(2𝑥 + 4) = 2𝑥 3 + 4𝑥 2 + 2𝑥 2 + 4𝑥 − 2𝑥 − 4 = 2𝑥 3 + 6𝑥 2 + 2𝑥 − 4

18. (3𝑥 + 1)(2𝑥 2 − 𝑥 + 1) = 3𝑥(2𝑥 2 − 𝑥 + 1) + 1(2𝑥 2 − 𝑥 + 1) = 6𝑥 3 − 3𝑥 2 + 3𝑥 + 2𝑥 2 − 𝑥 + 1 = 6𝑥 3 − 𝑥 2 + 2𝑥 + 1

𝑦

19. (𝑥 2 − 2𝑥 + 1) 2 = (𝑥 2 − 2𝑥 + 1)(𝑥 2 − 2𝑥 + 1) = 𝑥 2(𝑥 2 − 2𝑥 + 1) − 2𝑥(𝑥 2 − 2𝑥 + 1) + 1(𝑥 2 − 2𝑥 + 1) = 𝑥 4 − 2𝑥 3 + 𝑥 2 − 2𝑥 3 + 4𝑥 2 − 2𝑥 + 𝑥 2 − 2𝑥 + 1 = 𝑥 4 − 4𝑥 3 + 6𝑥 2 − 4𝑥 + 1 20. (𝑥 + 𝑦 − 𝑥𝑦) 2 = (𝑥 + 𝑦 − 𝑥𝑦)(𝑥 + 𝑦 − 𝑥𝑦) = 𝑥(𝑥 + 𝑦 − 𝑥𝑦) + 𝑦(𝑥 + 𝑦 − 𝑥𝑦) − 𝑥𝑦(𝑥 + 𝑦 − 𝑥𝑦) = 𝑥 2 + 𝑥𝑦 − 𝑥 2𝑦 + 𝑥𝑦 + 𝑦 2 − 𝑥𝑦 2 − 𝑥 2𝑦 − 𝑥𝑦 2 + 𝑥 2𝑦 2 = 𝑥 2 + 2𝑥𝑦 + 𝑦 2 − 2𝑥 2𝑦 − 2𝑥𝑦 2 + 𝑥 2𝑦 2

21. (𝑦 3 + 2𝑦 2 + 𝑦)(𝑦 2 + 2𝑦 − 1) = 𝑦 3(𝑦 2 + 2𝑦 − 1) + 2𝑦 2(𝑦 2 + 2𝑦 − 1) + 𝑦(𝑦 2 + 2𝑦 − 1) = 𝑦 5 + 2𝑦 4 − 𝑦 3 + 2𝑦 4 + 4𝑦 3 − 2𝑦 2 + 𝑦 3 + 2𝑦 2 − 𝑦 = 𝑦 5 + 4𝑦 4 + 4𝑦 3 − 𝑦

22. (𝑥 − 2𝑥 + 4)(3𝑥 − 𝑥 + 2) = 𝑥 3(3𝑥 2 − 𝑥 + 2) − 2𝑥 2(3𝑥 2 − 𝑥 + 2) + 4(3𝑥 2 − 𝑥 + 2) = 3𝑥 5 − 𝑥 4 + 2𝑥 3 − 6𝑥 4 + 2𝑥 3 − 4𝑥 2 + 12𝑥 2 − 4𝑥 + 8 = 3𝑥 5 − 7𝑥 4 + 4𝑥 3 + 8𝑥 2 − 4𝑥 + 8 3

Solutions Section 0.4

23. (𝑥 + 1)(𝑥 + 2) + (𝑥 + 1)(𝑥 + 3) = (𝑥 + 1)(𝑥 + 2 + 𝑥 + 3) = (𝑥 + 1)(2𝑥 + 5)

25. (𝑥 2 + 1) 5(𝑥 + 3) 4 + (𝑥 2 + 1) 6(𝑥 + 3) 3 = (𝑥 2 + 1) 5(𝑥 + 3) 3(𝑥 + 3 + 𝑥 2 + 1) = (𝑥 2 + 1) 5(𝑥 + 3) 3(𝑥 2 + 𝑥 + 4)

24. (𝑥 + 1)(𝑥 + 2) 2 + (𝑥 + 1) 2(𝑥 + 2) = (𝑥 + 1)(𝑥 + 2)(𝑥 + 2 + 𝑥 + 1) = (𝑥 + 1)(𝑥 + 2)(2𝑥 + 3)

26. 10𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 5 + 15𝑥 2(𝑥 2 + 1) 5(𝑥 3 + 1) 4 = 5𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 4[2(𝑥 3 + 1) + 3𝑥(𝑥 2 + 1)] = 5𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 4(5𝑥 3 + 3𝑥 + 2)

27. (𝑥 3 + 1)√𝑥 + 1 − (𝑥 3 + 1) 2√𝑥 + 1 = (𝑥 3 + 1)√𝑥 + 1 ⋅ [1 − (𝑥 3 + 1)] = −𝑥 3(𝑥 3 + 1)√𝑥 + 1

28. (𝑥 2 + 1)√𝑥 + 1 − √(𝑥 + 1) 3 = √𝑥 + 1 ⋅ [𝑥 2 + 1 − √(𝑥 + 1) 2] = √𝑥 + 1 ⋅ [𝑥 2 + 1 − (𝑥 + 1)] = (𝑥 2 − 𝑥)√𝑥 + 1 = 𝑥(𝑥 − 1)√𝑥 + 1

29. √(𝑥 + 1) 3 + √(𝑥 + 1) 5 = √(𝑥 + 1) 3 ⋅ [1 + √(𝑥 + 1) 2] = √(𝑥 + 1) 3 ⋅ (1 + 𝑥 + 1) = (𝑥 + 2)√(𝑥 + 1) 3

30. (𝑥 2 + 1) √(𝑥 + 1) 4 − √(𝑥 + 1) 7 3

= √(𝑥 + 1) 4 ⋅ [𝑥 2 + 1 − √(𝑥 + 1) 3] 3

= √(𝑥 + 1) 4 ⋅ [𝑥 2 + 1 − (𝑥 + 1)] 3

= (𝑥 2 − 𝑥) √(𝑥 + 1) 4 3

= 𝑥(𝑥 − 1) √(𝑥 + 1) 4

31. 𝑎 = 2, 𝑏 = 6, 𝑐 = 5, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = 6 2 − 4(2)(5) = 36 − 40 = −4 As the discriminant is negative, the expression does not factor at all.

32. 𝑎 = 4, 𝑏 = −6, 𝑐 = 2, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−6) 2 − 4(4)(2) = 36 − 32 = 4 = 2 2 As the discriminant is a perfect square, the expression factors over the integers.

33. 𝑎 = −3, 𝑏 = −2, 𝑐 = 3, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−2) 2 − 4(−3)(3) = 4 + 36 = 40 As the discriminant is positive but not a perfect square, the expression factors, but not over the integers. 34. 𝑎 = 1, 𝑏 = −4, 𝑐 = −7, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−4) 2 − 4(1)(−7) = 16 + 28 = 44 As the discriminant is positive but not a perfect square, the expression factors, but not over the integers. 35. 𝑎 = 8, 𝑏 = 12, 𝑐 = 4, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = 12 2 − 4(8)(4) = 144 − 128 = 16 = 4 2 As the discriminant is a perfect square, the expression factors over the integers.

Solutions Section 0.4 36. 𝑎 = 1, 𝑏 = 2, 𝑐 = 19, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = 2 2 − 4(1)(19) = 4 − 76 = −72 As the discriminant is negative, the expression does not factor at all.

37. 𝑎 = 40, 𝑏 = −64, 𝑐 = 24, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−64) 2 − 4(40)(24) = 4,096 − 3,840 = 256 = 16 2 As the discriminant is a perfect square, the expression factors over the integers. 38. 𝑎 = −10, 𝑏 = −32, 𝑐 = −32, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−32) 2 − 4(−10)(−32) = 1,024 − 1,280 = −256 As the discriminant is negative, the expression does not factor at all.

39. 𝑎 = 6, 𝑏 = −22, 𝑐 = 16, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = (−22) 2 − 4(6)(16) = 484 − 384 = 100 = 10 2 As the discriminant is a perfect square, the expression factors over the integers. 40. 𝑎 = 48, 𝑏 = 32, 𝑐 = 4, so that the discriminant is 𝑏 2 − 4𝑎𝑐 = 32 2 − 4(48)(4) = 1,024 − 768 = 256 = 16 2 As the discriminant is a perfect square, the expression factors over the integers. 41. a. 2𝑥 + 3𝑥 2 = 𝑥(2 + 3𝑥) b. 𝑥(2 + 3𝑥) = 0 𝑥 = 0 or 2 + 3𝑥 = 0 𝑥 = 0 or −2∕3

42. a. 𝑦 2 − 4𝑦 = 𝑦(𝑦 − 4) b. 𝑦(𝑦 − 4) = 0 𝑦 = 0 or 𝑦 − 4 = 0 𝑦 = 0 or 4

43. a. 6𝑥 3 − 2𝑥 2 = 2𝑥 2(3𝑥 − 1) b. 2𝑥 2(3𝑥 − 1) = 0 𝑥 2 = 0 or 3𝑥 − 1 = 0 𝑥 = 0 or 1∕3

44. a. 3𝑦 3 − 9𝑦 2 = 3𝑦 2(𝑦 − 3) b. 3𝑦 2(𝑦 − 3) = 0 𝑦 2 = 0 or 𝑦 − 3 = 0 𝑦 = 0 or 3

45. a. 𝑥 2 − 8𝑥 + 7 = (𝑥 − 1)(𝑥 − 7) b. (𝑥 − 1)(𝑥 − 7) = 0 𝑥 − 1 = 0 or 𝑥 − 7 = 0 𝑥 = 1 or 7

46. a. 𝑦 2 + 6𝑦 + 8 = (𝑦 + 2)(𝑦 + 4) b. (𝑦 + 2)(𝑦 + 4) = 0 𝑦 + 2 = 0 or 𝑦 + 4 = 0 𝑦 = −2 or −4

47. a. 𝑥 2 + 𝑥 − 12 = (𝑥 − 3)(𝑥 + 4) b. (𝑥 − 3)(𝑥 + 4) = 0 𝑥 − 3 = 0 or 𝑥 + 4 = 0 𝑥 = 3 or −4

48. a. 𝑦 2 + 𝑦 − 6 = (𝑦 − 2)(𝑦 + 3) b. (𝑦 − 2)(𝑦 + 3) = 0 𝑦 − 2 = 0 or 𝑦 + 3 = 0 𝑦 = 2 or −3

49. a. 2𝑥 2 − 3𝑥 − 2 = (2𝑥 + 1)(𝑥 − 2) b. (2𝑥 + 1)(𝑥 − 2) = 0 2𝑥 + 1 = 0 or 𝑥 − 2 = 0 𝑥 = −1∕2 or 2

50. a. 3𝑦 2 − 8𝑦 − 3 = (3𝑦 + 1)(𝑦 − 3) b. (3𝑦 + 1)(𝑦 − 3) = 0 3𝑦 + 1 = 0 or 𝑦 − 3 = 0 𝑦 = −1∕3 or 3

51. a. 6𝑥 2 + 13𝑥 + 6 = (2𝑥 + 3)(3𝑥 + 2) b. (2𝑥 + 3)(3𝑥 + 2) = 0 2𝑥 + 3 = 0 or 3𝑥 + 2 = 0 𝑥 = −3∕2 or −2∕3

52. a. 6𝑦 2 + 17𝑦 + 12 = (3𝑦 + 4)(2𝑦 + 3) b. (3𝑦 + 4)(2𝑦 + 3) = 0 3𝑦 + 4 = 0 or 2𝑦 + 3 = 0 𝑦 = −4∕3 or −3∕2

53. a. 12𝑥 2 + 𝑥 − 6 = (3𝑥 − 2)(4𝑥 + 3) b. (3𝑥 − 2)(4𝑥 + 3) = 0 3𝑥 − 2 = 0 or 4𝑥 + 3 = 0 𝑥 = 2∕3 or −3∕4

54. a. 20𝑦 2 + 7𝑦 − 3 = (4𝑦 − 1)(5𝑦 + 3) b. (4𝑦 − 1)(5𝑦 + 3) = 0 4𝑦 − 1 = 0 or 5𝑦 + 3 = 0 𝑦 = 1∕4 or −3∕5

55. a. 𝑥 2 + 4𝑥𝑦 + 4𝑦 2 = (𝑥 + 2𝑦) 2 b. (𝑥 + 2𝑦) 2 = 0 𝑥 + 2𝑦 = 0 𝑥 = −2𝑦

56. a. 4𝑦 2 − 4𝑥𝑦 + 𝑥 2 = (2𝑦 − 𝑥) 2 b. (2𝑦 − 𝑥) 2 = 0 2𝑦 − 𝑥 = 0 𝑦 = 𝑥∕2

57. a. 𝑥 4 − 5𝑥 2 + 4 = (𝑥 2 − 1)(𝑥 2 − 4) = (𝑥 + 1)(𝑥 − 1)(𝑥 + 2)(𝑥 − 2) b. (𝑥 + 1)(𝑥 − 1)(𝑥 + 2)(𝑥 − 2) = 0 𝑥 + 1 = 0 or 𝑥 − 1 = 0 or 𝑥 + 2 = 0 or 𝑥 − 2 = 0 𝑥 = ±1 or ±2

58. a. 𝑦 4 + 2𝑦 2 − 3 = (𝑦 2 − 1)(𝑦 2 + 3) = (𝑦 + 1)(𝑦 − 1)(𝑦 2 + 3) b. (𝑦 + 1)(𝑦 − 1)(𝑦 2 + 3) = 0 𝑦 + 1 = 0 or 𝑦 − 1 = 0 or 𝑦 2 + 3 = 0 𝑦 = ±1 (Notice that 𝑦 2 + 3 = 0 has no real solutions.)

59. a. 𝑥 2 − 3 = (𝑥 − √3)(𝑥 + √3) b. (𝑥 − √3)(𝑥 + √3) = 0 𝑥 − √3 = 0 or 𝑥 + √3 = 0 𝑥 = ±√3

60. a. 𝑦 2 − 7 = (𝑦 − √7)(𝑦 + √7) b. (𝑦 − √7)(𝑦 + √7) = 0 𝑦 − √7 = 0 or 𝑦 + √7 = 0 𝑦 = ±√7

Solutions Section 0.4

Solutions Section 0.5 Section 0.5 1.

𝑥 − 4 2𝑥 + 1 (𝑥 − 4)(2𝑥 + 1) 2𝑥 2 − 7𝑥 − 4 ⋅ = = 𝑥+1 𝑥−1 (𝑥 + 1)(𝑥 − 1) 𝑥2 − 1 2𝑥 − 3 𝑥 + 3 (2𝑥 − 3)(𝑥 + 3) 2𝑥 2 + 3𝑥 − 9 ⋅ = = 2 𝑥−2 𝑥+1 (𝑥 − 2)(𝑥 + 1) 𝑥 −𝑥−2

𝑥 − 4 2𝑥 + 1 (𝑥 − 4)(𝑥 − 1) + (𝑥 + 1)(2𝑥 + 1) 3𝑥 2 − 2𝑥 + 5 + = = 𝑥+1 𝑥−1 (𝑥 + 1)(𝑥 − 1) 𝑥2 − 1 2𝑥 − 3 𝑥 + 3 (2𝑥 − 3)(𝑥 + 1) + (𝑥 − 2)(𝑥 + 3) 3𝑥 2 − 9 + = = 2 𝑥−2 𝑥+1 (𝑥 − 2)(𝑥 + 1) 𝑥 −𝑥−2 𝑥2 𝑥 − 1 𝑥 2 − (𝑥 − 1) 𝑥 2 − 𝑥 + 1 − = = 𝑥+1 𝑥+1 𝑥+1 𝑥+1

(𝑥 2 − 1)(𝑥 − 1) − (𝑥 − 2) 𝑥 3 − 𝑥 2 − 2𝑥 + 3 𝑥2 − 1 1 − = = 𝑥−2 𝑥−1 (𝑥 − 2)(𝑥 − 1) 𝑥 2 − 3𝑥 + 2 1

+𝑥−1= 𝑥 ( 𝑥−1 ) 2

𝑥−2

! 𝑥2 "

−

𝑥 − 1 + 𝑥(𝑥 − 1) 𝑥 2 − 1 𝑥−1 +𝑥−1= = 𝑥 𝑥 𝑥

1 2𝑥 2 1 2𝑥 2 − 1 = − = 𝑥−2 𝑥−2 𝑥−2 𝑥−2

1 ⎡ 𝑥 − 3 1 ⎤ 1 ⎡ 𝑥 − 3 + 𝑥 ⎤ 2𝑥 − 3 + = = 𝑥 ⎢⎣ 𝑥𝑦 𝑦 ⎥⎦ 𝑥 ⎢⎣ 𝑥𝑦 ⎥⎦ 𝑥 2𝑦

10.

11.

12.

13.

14.

𝑦 2 ⎡ 2𝑥 − 3 𝑥 ⎤ 𝑦 2 ⎡ 2𝑥 − 3 + 𝑥 ⎤ 𝑦 2(3𝑥 − 3) 𝑦(3𝑥 − 3) 3𝑥𝑦 − 3𝑦 + = = = = 𝑥 ⎢⎣ 𝑦 𝑦 ⎥⎦ 𝑥 ⎢⎣ 𝑥𝑦 𝑥𝑦 𝑥 𝑥 ⎥⎦

(𝑥 + 1) 2(𝑥 + 2) 3 − (𝑥 + 1) 3(𝑥 + 2) 2 (𝑥 + 1) 2(𝑥 + 2) 2[(𝑥 + 2) − (𝑥 + 1)] (𝑥 + 1) 2 = = (𝑥 + 2) 4 (𝑥 + 2) 6 (𝑥 + 2) 6 6𝑥(𝑥 2 + 1) 2(𝑥 3 + 2) 3 − 9𝑥 2(𝑥 2 + 1) 3(𝑥 3 + 2) 2 (𝑥 3 + 2) 6 2 2 3 3𝑥(𝑥 + 1) (𝑥 + 2) 2[2(𝑥 3 + 2) − 3𝑥(𝑥 2 + 1)] 3𝑥(𝑥 2 + 1) 2(−𝑥 3 − 3𝑥 + 4) = = (𝑥 3 + 2) 4 (𝑥 3 + 2) 6 (𝑥 2 − 1)√𝑥 2 + 1 − 𝑥2 + 1

𝑥√𝑥 3 − 1 −

𝑥3 − 1

3𝑥 4

√𝑥 3−1

𝑥4 √𝑥 2+1

(𝑥 2 − 1)(𝑥 2 + 1) − 𝑥 4 (𝑥 2 + 1)√𝑥 2 + 1

𝑥(𝑥 3 − 1) − 3𝑥 4

(𝑥 3 − 1)√𝑥 3 − 1

−2𝑥 4 − 𝑥

−1

√(𝑥 2 + 1) 3

√(𝑥 3 − 1) 3

−𝑥(2𝑥 3 + 1) √(𝑥 3 − 1) 3

15.

1 − 12 𝑥 (𝑥+𝑦) 2

16.

1 − 13 𝑥 (𝑥+𝑦) 3

𝑦

Solutions Section 0.5

𝑥 2 − (𝑥 + 𝑦) 2 𝑥 2 − 𝑥 2 − 2𝑥𝑦 − 𝑦 2 −𝑦(2𝑥 + 𝑦) −(2𝑥 + 𝑦) = = = 2 2 2 2 2 2 𝑦𝑥 (𝑥 + 𝑦) 𝑦𝑥 (𝑥 + 𝑦) 𝑦𝑥 (𝑥 + 𝑦) 𝑥 2(𝑥 + 𝑦) 2

𝑥 2 − (𝑥 + 𝑦) 2 𝑥 3 − 𝑥 3 − 3𝑥 2𝑦 − 3𝑥𝑦 2 − 𝑦 3 = 𝑦 𝑦𝑥 2(𝑥 + 𝑦) 2 𝑦𝑥 3(𝑥 + 𝑦) 3 2 2 2 −𝑦(3𝑥 + 3𝑥𝑦 + 𝑦 ) −(3𝑥 + 3𝑥𝑦 + 𝑦 2) = = 𝑦𝑥 3(𝑥 + 𝑦) 3 𝑥 3(𝑥 + 𝑦) 3 =

Solutions Section 0.6 Section 0.6

1. 𝑥 + 1 = 0 ⇒ 𝑥 = 0 − 1 ⇒ 𝑥 = −1

2. 𝑥 − 3 = 1 ⇒ 𝑥 = 1 + 3 ⇒ 𝑥 = 4

3. −𝑥 + 5 = 0 ⇒ −𝑥 = −5 ⇒ 𝑥 = 5

4. 2𝑥 + 4 = 1 ⇒ 2𝑥 = −3 ⇒ 𝑥 = −

5. 4𝑥 − 5 = 8 ⇒ 4𝑥 = 13 ⇒ 𝑥 =

13 4 43 7

8. 3𝑥 + 1 = 𝑥 ⇒ 2𝑥 = −1 ⇒ 𝑥 = −

9. 𝑥 + 1 = 2𝑥 + 2 ⇒ −𝑥 = 1 ⇒ 𝑥 = −1 11. 𝑎𝑥 + 𝑏 = 𝑐 ⇒ 𝑎𝑥 = 𝑐 − 𝑏 ⇒ 𝑥 =

3 3 4 𝑥 + 1 = 0 ⇒ 𝑥 = −1 ⇒ 𝑥 = − 4 4 3

7. 7𝑥 + 55 = 98 ⇒ 7𝑥 = 43 ⇒ 𝑥 = −

𝑐−𝑏 𝑎

12. 𝑥 − 1 = 𝑐𝑥 + 𝑑 ⇒ (1 − 𝑐)𝑥 = 𝑑 + 1 ⇒ 𝑥 =

𝑑+1 1−𝑐

1 2

14. 𝑥 2 + 𝑥 + 1 = 0 ⇒ Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so this equation has no real solutions. 15. 𝑥 2 − 𝑥 + 1 = 0 ⇒ Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so this equation has no real solutions. 16. 2𝑥 2 − 4𝑥 + 3 ⇒ Δ = 𝑏 2 − 4𝑎𝑐 = −8 < 0, so this equation has no real solutions.

18. 3𝑥 2 − 1 = 0 ⇒ 𝑥 2 =

5 5 ⇒𝑥=± √2 2 1 1 ⇒𝑥=± √3 3

19. −𝑥 2 − 2𝑥 − 1 = 0 ⇒ −(𝑥 + 1) 2 = 0 ⇒ 𝑥 = −1

3 20. 2𝑥 2 − 𝑥 − 3 = 0 ⇒ (2𝑥 − 3)(𝑥 + 1) = 0 ⇒ 𝑥 = , −1 2 21.

1 2

10. 𝑥 + 1 = 3𝑥 + 1 ⇒ −2𝑥 = 0 ⇒ 𝑥 = 0

13. 2𝑥 2 + 7𝑥 − 4 = 0 ⇒ (2𝑥 − 1)(𝑥 + 4) = 0 ⇒ 𝑥 = −4,

17. 2𝑥 2 − 5 = 0 ⇒ 𝑥 2 =

3 2

1 2 3 𝑥 − 𝑥 − = 0 ⇒ 𝑥 2 − 2𝑥 − 3 = 0 ⇒ (𝑥 + 1)(𝑥 − 3) = 0 ⇒ 𝑥 = −1, 3 2 2

1 1 22. − 𝑥 2 − 𝑥 + 1 = 0 ⇒ 𝑥 2 + 𝑥 − 2 = 0 ⇒ (𝑥 + 2)(𝑥 − 1) = 0 ⇒ 𝑥 = −2, 1 2 2

−𝑏 ± √𝑏 2 − 4𝑎𝑐 1 ± √5 = 2𝑎 2

Solutions Section 0.6

23. 𝑥 2 − 𝑥 = 1 ⇒ 𝑥 2 − 𝑥 − 1 = 0 ⇒ 𝑥 =

24. 16𝑥 2 = −24𝑥 − 9 ⇒ 16𝑥 2 + 24𝑥 + 9 = 0(4𝑥 + 3) 2 = 0 ⇒ 𝑥 = − 25. 𝑥 = 2 −

3 4

1 ⇒ 𝑥 2 = 2𝑥 − 1 ⇒ 𝑥 2 − 2𝑥 + 1 = 0 ⇒ (𝑥 − 1) 2 = 0 ⇒ 𝑥 = 1 𝑥

1 ⇒ (𝑥 + 4)(𝑥 − 2) = 1 ⇒ 𝑥 2 + 2𝑥 − 8 = 1 ⇒ 𝑥 2 + 2𝑥 − 9 − 0 𝑥−2 −𝑏 ± √𝑏 2 − 4𝑎𝑐 −2 ± √40 ⇒𝑥= = = −1 ± √10 2𝑎 2

26. 𝑥 + 4 =

27. 𝑥 4 − 10𝑥 2 + 9 = 0 ⇒ (𝑥 2 − 1)(𝑥 2 − 9) = 0 ⇒ 𝑥 2 = 1 or 𝑥 2 = 0 ⇒ 𝑥 = ±1, ±3 28. 𝑥 4 − 2𝑥 2 + 1 = 0 ⇒ (𝑥 2 − 1) 2 = 0 ⇒ 𝑥 = ±1 29. 𝑥 4 + 𝑥 2 − 1 = 0 ⇒ 𝑥 2 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐 −1 ± √5 −1 ± √5 ⇒ 𝑥2 = ⇒ 𝑥 = ±√ 2𝑎 2 2

30. 𝑥 3 + 2𝑥 2 + 𝑥 = 0 ⇒ 𝑥(𝑥 2 + 2𝑥 + 1) = 0 ⇒ 𝑥(𝑥 + 1) 2 = 0 ⇒ 𝑥 = 0, −1

31. 𝑥 3 + 16𝑥 2 + 11𝑥 + 6 = 0 ⇒ (𝑥 + 1)(𝑥 + 2)(𝑥 + 3) = 0 ⇒ 𝑥 = −1, −2, −3 32. 𝑥 3 − 6𝑥 2 + 12𝑥 − 8 = 0 ⇒ (𝑥 − 2) 3 = 0 ⇒ 𝑥 = 2

33. 𝑥 3 + 4𝑥 2 + 4𝑥 + 3 = 0 ⇒ (𝑥 + 3)(𝑥 2 + 𝑥 + 1) = 0 ⇒ 𝑥 = −3 (For 𝑥 2 + 𝑥 + 1 = 0, Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so there are no real solutions to this quadratic equation.) 3

34. 𝑦 3 + 64 = 0 ⇒ 𝑦 3 = −64 ⇒ 𝑦 = √−64 = −4 3

35. 𝑥 3 − 1 = 0 ⇒ 𝑥 3 = 1 ⇒ 𝑥 = √1 = 1 36. 𝑥 3 − 27 = 0 3 ⇒ 𝑥 3 = 27 ⇒ 𝑥 = √27 = 3

37. 𝑦 3 + 3𝑦 2 + 3𝑦 + 2 = 0 ⇒ (𝑦 + 2)(𝑦 2 + 𝑦 + 1) = 0 ⇒ 𝑦 = −2 (For 𝑦 2 + 𝑦 + 1 = 0, Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so there are no real solutions to this quadratic equation.) 38. 𝑦 3 − 2𝑦 2 − 2𝑦 − 3 = 0 ⇒ (𝑦 − 3)(𝑦 2 + 𝑦 + 1) = 0 ⇒ 𝑦 = 3 (For 𝑦 2 + 𝑦 + 1 = 0, Δ = 𝑏 2 − 4𝑎𝑐 = −3 < 0, so there are no real solutions to this quadratic equation.) 39. 𝑥 3 − 𝑥 2 − 5𝑥 + 5 = 0 ⇒ (𝑥 − 1)(𝑥 2 − 5) = 0 ⇒ 𝑥 = 1, ± √5

40. 𝑥 3 − 𝑥 2 − 3𝑥 + 3 = 0 ⇒ (𝑥 − 1)(𝑥 2 − 3) = 0 ⇒ 𝑥 = 1, ± √3

Solutions Section 0.6 41. 2𝑥 − 𝑥 − 2𝑥 + 1 = 0 ⇒ (2𝑥 − 1)(𝑥 4 − 1) = 0 ⇒ (2𝑥 2 − 1)(𝑥 2 − 1)(𝑥 2 + 1) = 0 (Think of the cubic you get by substituting 𝑦 for 𝑥 2) 1 ⇒ 𝑥 = ±1, ± √2 6

42. 3𝑥 6 − 𝑥 4 − 12𝑥 2 + 4 = 0 ⇒ (3𝑥 2 − 1)(𝑥 4 − 4) = 0 ⇒ (3𝑥 2 − 1)(𝑥 2 − 2)(𝑥 2 + 2) = 0 (Think of the cubic you get by substituting 𝑦 for 𝑥 2) 1 ⇒ 𝑥 = ±√2, ± √3

43. (𝑥 2 + 3𝑥 + 2)(𝑥 2 − 5𝑥 + 6) = 0 ⇒ (𝑥 + 2)(𝑥 + 1)(𝑥 − 2)(𝑥 − 3) = 0 ⇒ 𝑥 = −2, −1, 2, 3 44. (𝑥 2 − 4𝑥 + 4) 2(𝑥 2 + 6𝑥 + 5) 3 = 0 ⇒ (𝑥 − 2) 4(𝑥 + 1) 3(𝑥 + 5) 3 = 0 ⇒ 𝑥 = −5, −1, 2

Solutions Section 0.7 Section 0.7

1. 𝑥 4 − 3𝑥 3 = 0, 𝑥 3(𝑥 − 3) = 0, 𝑥 = 0, 3

2. 𝑥 6 − 9𝑥 4 = 0, 𝑥 4(𝑥 2 − 9) = 0, 𝑥 = 0, ±3

3. 𝑥 4 − 4𝑥 2 = −4, 𝑥 4 − 4𝑥 2 + 4 = 0, (𝑥 2 − 2) 2 = 0, 𝑥 = ±√2

4. 𝑥 4 − 𝑥 2 = 6, 𝑥 4 − 𝑥 2 − 6 = 0, (𝑥 2 − 3)(𝑥 2 + 2) = 0, 𝑥 = ±√3

5. (𝑥 + 1)(𝑥 + 2) + (𝑥 + 1)(𝑥 + 3) = 0, (𝑥 + 1)(𝑥 + 2 + 𝑥 + 3) = 0, (𝑥 + 1)(2𝑥 + 5) = 0, 𝑥 = −1, −5∕2

6. (𝑥 + 1)(𝑥 + 2) 2 + (𝑥 + 1) 2(𝑥 + 2) = 0, (𝑥 + 1)(𝑥 + 2)(𝑥 + 2 + 𝑥 + 1) = 0, (𝑥 + 1)(𝑥 + 2)(2𝑥 + 3) = 0, 𝑥 = −1, −2, −3∕2 7. (𝑥 2 + 1) 5(𝑥 + 3) 4 + (𝑥 2 + 1) 6(𝑥 + 3) 3 = 0, (𝑥 2 + 1) 5(𝑥 + 3) 3(𝑥 + 3 + 𝑥 2 + 1) = 0, (𝑥 2 + 1) 5(𝑥 + 3) 3(𝑥 2 + 𝑥 + 4) = 0, 𝑥 = −3 (Neither 𝑥 2 + 1 = 0 nor 𝑥 2 + 𝑥 + 4 = 0 has a real solution.) 8. 10𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 5 − 10𝑥 2(𝑥 2 + 1) 5(𝑥 3 + 1) 4 = 0, 10𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 4[𝑥 3 + 1 − 𝑥(𝑥 2 + 1)] = 0, 10𝑥(𝑥 2 + 1) 4(𝑥 3 + 1) 4(1 − 𝑥) = 0, 𝑥 = −1, 0, 1 9. (𝑥 3 + 1)√𝑥 + 1 − (𝑥 3 + 1) 2√𝑥 + 1 = 0, (𝑥 3 + 1)√𝑥 + 1 [1 − (𝑥 3 + 1)] = 0, −𝑥 3(𝑥 3 + 1)√𝑥 + 1 = 0, 𝑥 = 0, −1

10. (𝑥 2 + 1)√𝑥 + 1 − √(𝑥 + 1) 3 = 0, √𝑥 + 1 [𝑥 2 + 1 − (𝑥 + 1)] = 0, (𝑥 2 − 𝑥)√𝑥 + 1 = 0, 𝑥(𝑥 − 1)√𝑥 + 1 = 0, 𝑥 = −1, 0, 1

11. √(𝑥 + 1) 3 + √(𝑥 + 1) 5 = 0, √(𝑥 + 1) 3 (1 + 𝑥 + 1) = 0, (𝑥 + 2)√(𝑥 + 1) 3 = 0, 𝑥 = −1 (𝑥 = −2 is not a solution because √(𝑥 + 1) 3 is not defined for 𝑥 = −2.) 3

12. (𝑥 2 + 1) √(𝑥 + 1) 4 − √(𝑥 + 1) 7 = 0, 3

√(𝑥 + 1) 4 [𝑥 2 + 1 − (𝑥 + 1)] = 0, 3

(𝑥 2 − 𝑥) √(𝑥 + 1) 4 = 0, 𝑥(𝑥 − 1) √(𝑥 + 1) 4 = 0, 𝑥 = −1, 0, 1

13. (𝑥 + 1) 2(2𝑥 + 3) − (𝑥 + 1)(2𝑥 + 3) 2 = 0, (𝑥 + 1)(2𝑥 + 3)(𝑥 + 1 − 2𝑥 − 3) = 0, (𝑥 + 1)(2𝑥 + 3)(−𝑥 − 2) = 0, 𝑥 = −2, −3∕2, −1

Solutions Section 0.7 14. (𝑥 − 1) (𝑥 + 2) − (𝑥 − 1) (𝑥 + 2) 2 = 0, (𝑥 2 − 1) 2(𝑥 + 2) 2(𝑥 + 2 − 𝑥 2 + 1) = 0, −(𝑥 2 − 1) 2(𝑥 + 2) 2(𝑥 2 − 𝑥 − 3) = 0, 𝑥 = −2, −1, 1, (1 ± √13)∕2 2

15.

16.

17.

6𝑥(𝑥 2 + 1) 2(𝑥 2 + 2) 4 − 8𝑥(𝑥 2 + 1) 3(𝑥 2 + 2) 3 = 0, (𝑥 2 + 2) 8 2𝑥(𝑥 2 + 1) 2(𝑥 2 + 2) 3[3(𝑥 2 + 2) − 4(𝑥 2 + 1)] = 0, (𝑥 2 + 2) 8 −2𝑥(𝑥 2 + 1) 2(𝑥 2 − 2) = 0, (𝑥 2 + 2) 5 −2𝑥(𝑥 2 + 1) 2(𝑥 2 − 2) = 0, 𝑥 = 0, ±√2 2(𝑥 2 − 1)√𝑥 2 + 1 − 𝑥2 + 1

𝑥4 − 2

4𝑥√𝑥 3 − 1 − 𝑥3 − 1

𝑥 4 − 4𝑥

19. 𝑥 − 20. 1 −

22.

23.

𝑥4 √𝑥 2+1

= 0,

2(𝑥 2 − 1)(𝑥 2 + 1) − 𝑥 4 (𝑥 2 + 1)√𝑥 2 + 1 4

= 0, 𝑥 4 − 2 = 0, 𝑥 = ± √2

3𝑥 4 √𝑥 3−1

(𝑥 3 − 1)√𝑥 3 − 1

21.

(𝑥 + 1) 2(𝑥 + 2) 3 − (𝑥 + 1) 3(𝑥 + 2) 2 = 0, (𝑥 + 2) 6 (𝑥 + 1) 2(𝑥 + 2) 2[(𝑥 + 2) − (𝑥 + 1)] = 0, (𝑥 + 2) 6 (𝑥 + 1) 2 = 0, (𝑥 + 1) 2 = 0, 𝑥 = −1 (𝑥 + 2) 4

(𝑥 2 + 1)√𝑥 2 + 1

18.

= 0,

4𝑥(𝑥 3 − 1) − 3𝑥 4 (𝑥 3 − 1)√𝑥 3 − 1

= 0,

= 0, 𝑥 4 − 4𝑥 = 0, 𝑥(𝑥 3 − 4) = 0, 𝑥 = 0, ± √4

1 = 0, 𝑥 2 − 1 = 0, 𝑥 = ±1 𝑥

4 = 0, 𝑥 2 − 4 = 0, 𝑥 = ±2 𝑥2

1 9 − 3 = 0, 𝑥 2 − 9 = 0, 𝑥 = ±3 𝑥 𝑥

1 1 − = 0, 𝑥 + 1 − 𝑥 2 = 0, 𝑥 2 − 𝑥 − 1 = 0, 𝑥 = (1 ± √5)∕2 2 𝑥+1 𝑥 (𝑥 − 4)(𝑥 − 1) − 𝑥(𝑥 + 1) 𝑥−4 𝑥 − = 0, = 0, 𝑥+1 𝑥−1 (𝑥 + 1)(𝑥 − 1) −6𝑥 + 4 = 0, −6𝑥 + 4 = 0, 𝑥 = 2∕3 (𝑥 + 1)(𝑥 − 1)

Solutions Section 0.7 (2𝑥 − 3)(𝑥 + 1) − (2𝑥 + 3)(𝑥 − 1) 2𝑥 − 3 2𝑥 + 3 24. − = 0, = 0, 𝑥−1 𝑥+1 (𝑥 − 1)(𝑥 + 1) −2𝑥 = 0, −2𝑥 = 0, 𝑥 = 0 (𝑥 − 1)(𝑥 + 1) 25.

26.

3𝑥(𝑥 + 4) + (𝑥 + 1)(𝑥 + 4) 𝑥+4 𝑥+4 + = 0, = 0, 𝑥+1 3𝑥 3𝑥(𝑥 + 1) (𝑥 + 4)(3𝑥 + 𝑥 + 1) (𝑥 + 4)(4𝑥 + 1) = 0, = 0, 3𝑥(𝑥 + 1) 3𝑥(𝑥 + 1) (𝑥 + 4)(4𝑥 + 1) = 0, 𝑥 = −4, −1∕4

(2𝑥 − 3)(𝑥 + 1) − 𝑥(2𝑥 − 3) 2𝑥 − 3 2𝑥 − 3 − = 0, = 0, 𝑥 𝑥+1 𝑥(𝑥 + 1) (2𝑥 − 3)(𝑥 + 1 − 𝑥) (2𝑥 − 3) = 0, = 0, 2𝑥 − 3 = 0, 𝑥 = 3∕2 𝑥(𝑥 + 1) 𝑥(𝑥 + 1)

Solutions Section 0.8 Section 0.8

1. 𝑃 (0, 2), 𝑄(4, −2), 𝑅(−2, 3), 𝑆(−3.5, −1.5), 𝑇 (−2.5, 0), 𝑈(2, 2.5)

2. 𝑃 (−2, 2), 𝑄(3.5, 2), 𝑅(0, −3), 𝑆(−3.5, −1.5), 𝑇 (2.5, 0), 𝑈(−2, 2.5) 3.

5. Solve the equation 𝑥 + 𝑦 = 1 for 𝑦 to get 𝑦 = 1 − 𝑥. Then plot some points: 𝑥

−2

𝑦=1−𝑥

−1 2

−1

Graph:

6. Solve the equation 𝑦 − 𝑥 = −1 for 𝑦 to get 𝑦 = −1 + 𝑥. Then plot some points: 𝑥

𝑦 = −1 + 𝑥

−2 −3

−1 −2

−1

1 0

2 1

Solutions Section 0.8 Graph:

7. Solve the equation 2𝑦 − 𝑥 2 = 1 for 𝑦 to get 𝑦 = 𝑦=

𝑥

−2

1 + 𝑥2 2

2.5

−1 1

0.5

1 + 𝑥2 . Then plot some points: 2

2.5

Graph:

8. Solve the equation 2𝑦 + √𝑥 = 1 for 𝑦 to get 𝑦 = 𝑦=

𝑥

1 − √𝑥 2

Graph:

0.5

1 0

−0.5

−1

−1.5

1 − √𝑥 . Then plot some points: 2

Solutions Section 0.8 4 9. Solve the equation 𝑥𝑦 = 4 for 𝑦 to get 𝑦 = . Then plot some points: 𝑥 𝑦=

𝑥

−3

4 𝑥

−2

− 43

−1

−2

−4

3 4 3

Graph:

1 10. Solve the equation 𝑥 2𝑦 = −1 for 𝑦 to get 𝑦 = − 2 . Then plot some points: 𝑥 𝑥

1 𝑦=− 2 𝑥

−3

− 19

−2

− 14

−1 −1

−1

− 14

− 19

Graph:

11. Solve the equation 𝑥𝑦 = 𝑥 2 + 1 for 𝑦 to get 𝑦 = 𝑥 + 𝑦=𝑥+

𝑥

1 𝑥

−2

− 52

−1 −2

− 12 − 52

1 2 5 2

1 2

2 5 2

1 . Then plot some points: 𝑥

Solutions Section 0.8 Graph:

12. Solve the equation 𝑥𝑦 = 2𝑥 3 + 1 for 𝑦 to get 𝑦 = 2𝑥 2 + 𝑦 = 2𝑥 2 +

𝑥

1 𝑥

−2 15 2

−1 1

− 12 − 32

1 2 5 2

1 3

17 2

Graph:

13. √(2 − 1) 2 + (−2 + 1) 2 = √2 14. √(6 − 1) 2 + (1 − 0) 2 = √26

15. √(0 − 𝑎) 2 + (𝑏 − 0) 2 = √𝑎 2 + 𝑏 2

16. √(𝑏 − 𝑎) 2 + (𝑏 − 0) 2 = √2(𝑏 − 𝑎) 2 or √2|𝑏 − 𝑎| 17. Set the two distances equal and solve: √(1 − 0) * 2 + (𝑘 − 0) 2 = √(1 − 2) 2 + (𝑘 − 1) 2 ⇒ √1 + 𝑘 2 = √2 − 2𝑘 + 𝑘 2 ⇒ 1 + 𝑘 2 = 2 − 2𝑘 + 𝑘 2 1 ⇒ 2𝑘 = 1 ⇒ 𝑘 = 2 18. Set the two distances equal and solve: √(𝑘 + 1) * 2 + (𝑘 − 0) 2 = √(𝑘 − 0) 2 + (𝑘 − 2) 2 ⇒ √2𝑘 2 + 2𝑘 + 1 = √2𝑘 2 − 4𝑘 + 4 ⇒ 2𝑘 2 + 2𝑘 + 1 = 2𝑘 2 − 4𝑘 + 4 1 ⇒ 6𝑘 = 3 ⇒ 𝑘 = 2

1 . Then plot some points: 𝑥

Solutions Section 0.8 19. Circle with center (0, 0) and radius 3. 20. The single point (0, 0).

Solutions Section 0.9 Section 0.9 1. Exponential Form

10 2 = 100

4 3 = 64

4 4 = 256

0.45 0 = 1

8 1∕2 = 2√2

1 Logarithmic Form log10 100 = 2 log4 64 = 3 log4 256 = 4 log0.45 1 = 0 log8 2√2 = 2

2. Exponential Form Logarithmic Form

10 1 = 10

5 5 = 3,125

log10 10 = 1

Exponential Form

0.5 3 = 0.125

Exponential Form

0.3 2 = 0.09

Logarithmic Form log0.5 0.125 = 3 3.

Logarithmic Form log0.3 0.09 = 2 Exponential Form Logarithmic Form 4. Exponential Form Logarithmic Form Exponential Form

log6 216 = 3

log4 √2 = 14

1 log3 ! 81 " = −4

10 −3 = 0.001

1 !2" = 1

1 3 −4 = 81

log1∕2 1 = 0

log10 0.001 = −3

1 log9 81 = −2

log2 1,024 = 10

log64 14 = − 13

2 −1 = 12

10 5 = 100,000

10 −5 = 0.00001

16 −1∕4 = 12

0.25 2 = 0.0625

1 9 −2 = 81

log2 12 = −1 2 11 = 2,048

Logarithmic Form log2 2,048 = 11

2 10 = 1,024

1 log4 ! 64 " = −3

6 3 = 216

log5 3,125 = 5 4 1∕4 = √2

1 4 −3 = 64

64 −1∕3 = 14

log10 100,000 = 5 log10 0.00001 = −5 log16 12 = − 14

log0.25 0.0625 = 2

5. log4 16 = the power to which we need to raise 4 in order to get 16. Because 4 boxed 2 = 16, this power is 2, so log4 16 = 2.

6. log5 125 = the power to which we need to raise 5 in order to get 125. Because 5 boxed 3 = 125, this power is 3, so log5 125 = 3.

1 1 1 7. log5 25 = the power to which we need to raise 5 in order to get 25 . Because 5 boxed −2 = 25 , this power is −2, so 1 log5 25 = −2.

1 1 1 8. log4 64 = the power to which we need to raise 4 in order to get 64 . Because 4 boxed −3 = 64 , this power is −3, so 1 log4 64 = −3.

Solutions Section 0.9 9. log 100,000 = log10 100,000 = the power to which we need to raise 10 in order to get 100,000. Because 10 boxed 5 = 100,000, this power is 5, so log 100,000 = 5. 10. log 1,000 = log10 1,000 = the power to which we need to raise 10 in order to get 1,000. Because 10 boxed 3 = 1,000, this power is 3, so log 1,000 = 3.

11. log16 16 = the power to which we need to raise 16 in order to get 16. Because 16 boxed 1 = 16, this power is 1, so log16 16 = 1. 1 1 1 1 boxed 1 1 = the power to which we need to raise in order to get . Because ! " = , this power is 1, 2 2 2 2 2 1 so log1∕2 = 1. 2 12. log1∕2

1 1 1 = the power to which we need to raise 4 in order to get . Because 4 boxed −2 = , this power is −2, 16 16 16 1 so log4 = −2. 16 13. log4

14. log2 log2

1 1 1 = the power to which we need to raise 2 in order to get . Because 2 boxed −3 = , this power is −3, so 8 8 8

1 = −3. 8

15. log2 √2 = the power to which we need to raise 2 in order to get √2. Because 2 boxed 1∕2 = √2, this power is 1 1 , so log2 √2 = . 2 2 16. log4 √2 = the power to which we need to raise 4 in order to get √2. Because 4 boxed 1∕4 = √2, this power is 1 1 , so log4 √2 = . 4 4 17. By Identity (1), log𝑏 3 + log𝑏 4 = log𝑏 (3 × 4) = log𝑏 boxed 12. 3 18. By Identity (2), log𝑏 3 − log𝑏 4 = log𝑏 boxed . 4

19. log𝑏 2 − log𝑏 5 − log𝑏 4 = log𝑏 2 − (log𝑏 5 + log𝑏 4) = log𝑏 2 − (log𝑏 (5 × 4)) (Identity 1) 2 1 = log𝑏 ! " (Identity 2) = log𝑏 boxed 5×4 10 20. log𝑏 3 + log𝑏 2 − log𝑏 7 = log𝑏 (3 × 2) − log𝑏 7 (Identity 1) = log𝑏 !

3×2 6 " (Identity 2) = log𝑏 boxed 7 7

3 3 21. log𝑏 3 − 3 log𝑏 2 = log𝑏 3 − log𝑏 2 3 (Identity 3) = log𝑏 ! 3 " (Identity 2) = log𝑏 boxed 8 2

22. 3 log𝑏 2 + 2 log𝑏 3 = log𝑏 2 3 + log𝑏 3 2 (Identity 3) = log𝑏 (2 3 × 3 2) (Identity 1) = log𝑏 boxed 72 23. 4 log𝑏 𝑥 + 5 log𝑏 𝑦 = log𝑏 𝑥 4 + log𝑏 𝑦 5 (Identity 3) = log𝑏 boxed 𝑥 4𝑦 5 (Identity 1)

Solutions Section 0.9 𝑝3 24. 3 log𝑏 𝑝 − 2 log𝑏 𝑞 = log𝑏 𝑝 3 − log𝑏 𝑞 2 (Identity 3) = log𝑏 boxed 2 (Identity 2) 𝑞 25. 2 log𝑏 𝑥 + 3 log𝑏 𝑦 − 4 log𝑏 𝑧 = log𝑏 𝑥 2 + log𝑏 𝑦 3 − log𝑏 𝑧 4 (Identity 3) 𝑥 2𝑦 3 = log𝑏 (𝑥 2𝑦 3) − log𝑏 𝑧 4 (Identity 1) = log𝑏 boxed 4 (Identity 2) 𝑧 26. 4 log𝑏 𝑝 − 3 log𝑏 𝑞 − 2 log𝑏 𝑟 = log𝑏 𝑝 4 − log𝑏 𝑞 3 − log𝑏 𝑟 2 (Identity 3)

= log𝑏 𝑝 4 − (log𝑏 𝑞 3 + log𝑏 𝑟 2) = log𝑏 𝑝 4 − log𝑏 (𝑞 3𝑟 2) (Identity 1) = log𝑏 boxed

27. 𝑥 log𝑏 2 − 2 log𝑏 𝑥 = log𝑏 2 𝑥 − log𝑏 𝑥 2 (Identity 3) = log𝑏 boxed

2𝑥 (Identity 2) 𝑥2

28. 𝑝 log𝑏 𝑞 + 𝑞 log𝑏 𝑝 = log𝑏 𝑞 𝑝 + log𝑏 𝑝 𝑞 (Identity 3) = log𝑏 boxed 𝑞 𝑝𝑝 𝑞 (Identity 1) 29. log 21 = log(3 × 7) = log 3 + log 7 = 𝑏 + 𝑐 (Identity 1)

30. log 14 = log(2 × 7) = log 2 + log 7 = 𝑎 + 𝑐 (Identity 1)

31. log 42 = log(2 × 3 × 7) = log 2 + log 3 + log 7 = 𝑎 + 𝑏 + 𝑐 (Identity 1) 32. log 28 = log(2 × 2 × 7) = log 2 + log 2 + log 7 = 2𝑎 + 𝑐 (Identity 1) 1 33. log! " = − log 7 = −𝑐 (Identity 5) 7

1 34. log! " = − log 3 = −𝑏 (Identity 5) 3

2 35. log! " = log 2 − log 3 = 𝑎 − 𝑏 (Identity 2) 3 7 36. log! " = log 7 − log 2 = 𝑐 − 𝑎 (Identity 2) 2

4 37. log! " = log 4 − log 7 (Identity 2) = log 2 + log 2 − log 7 (Identity 1) = 2𝑎 − 𝑐 7

2 38. log! " = log 2 − log 9 (Identity 2) = log 2 − (log 3 + log 3) (Identity 1) = 𝑎 − 2𝑏 9 39. log 16 = log 2 4 = 4 log 2 = 4𝑎 (Identity 3) 40. log 81 = log 3 4 = 4 log 3 = 4𝑏 (Identity 3) 41. log 0.03 = log!

3 " = log 3 − log 10 2 = log 3 − 2 log 10 = 𝑏 − 2 10 2

42. log 7,000 = log(7 × 10 3) = log 7 + log 10 3 = log 7 + 3 log 10 = 𝑐 + 3

𝑝4 (Identity 2) 𝑞 3𝑟 2

43. log 5 = log!

10 " = log 10 − log 2 = 1 − 𝑎 (Identity 2) 2

44. log 25 = log!

Solutions Section 0.9

100 " = log 100 − log 4 = log 10 2 − log 2 2 = 2 log 10 − 2 log 2 = 2 − 2𝑎 4

45. log √7 = log(7 1∕2) = 46. log!

1 𝑐 log 7 = 2 2

2 1 𝑏 " = log(2) − log √3 = log(2) − log(3 1∕2) = log 2 − log 3 = 𝑎 − √3 2 2

47. 4 = 2 𝑥 is exactly the equation we solve in order to calculate log2 4; answer: 2. Alternatively, write the equation 4 = 2 𝑥 in logarithmic form to get 𝑥 = log2 4 = 2.

48. 81 = 3 𝑥 is exactly the equation we solve in order to calculate log3 81; answer: 4. Alternatively, write the equation 81 = 3 𝑥 in logarithmic form to get 𝑥 = log3 81 = 4. 49. Take the base 3 logarithm of both sides of the given equation 27 = 3 2𝑥−1 to get log3 27 = log3 3 2𝑥−1 3 = (2𝑥 − 1) log3 3 3 = 2𝑥 − 1 3+1 𝑥= =2 2

By Identity 3 Solve for 𝑥.

By Identity 4

50. Take the base 4 logarithm of both sides of the given equation 4 2−3𝑥 = 256 to get log4 4 2−3𝑥 = log4 256 (2 − 3𝑥) log4 4 = 4 2 − 3𝑥 = 4 2−4 2 𝑥= =− 3 3

By Identity 3 Solve for 𝑥.

By Identity 4

1 51. Take the base 5 logarithm of both sides of the given equation 5 −𝑥+1 = to get 125 1 −𝑥+1 log5 5 = log5 ! " 125 (−𝑥 + 1) log5 5 = − log5 125 By Identities 3 and 5 −𝑥 + 1 = −3 𝑥=1+3=4

By Identity 4

𝑥 2 − 12 = −3 𝑥 = ±√9 = ±3

By Identity 4

Solve for 𝑥.

2 1 52. Take the base 3 logarithm of both sides of the given equation 3 𝑥 −12 = to get 27 2 1 log3 3 𝑥 −12 = log3 ! " 27 2 (𝑥 − 12) log3 3 = − log3 27 By Identities 3 and 5

Solve for 𝑥.

53. First divide both sides of the given equation by 50 to get 2.4 = 2 3𝑡. Take the common logarithm of both sides: log 2.4 = log(2 3𝑡)

log 2.4 = 3𝑡 log 2 log 2.4 3𝑡 = log 2 log 2.4 𝑡= ≈ 0.4210 3 log 2

Solutions Section 0.9 By Identity 3 Divide.

Solve for 𝑡.

If instead you take the base-2 logarithm, the answer is represented as

log2 (2.4) . 3

54. First divide both sides of the given equation by 500 to get 2 = 1.1 2𝑡. Take the common logarithm of both sides: log 2 = log(1.1 2𝑡) log 2 = 2𝑡 log 1.1 By Identity 3 log 2 2𝑡 = Divide. log 1.1 log 2 𝑡= ≈ 3.6363 Solve for 𝑡. 2 log 1.1 55. First divide both sides of the given equation by 300 to get sides:

10 " = log(1.3 4𝑡−1) 3 log 10 − log 3 = (4𝑡 − 1) log 1.3 log 10 − log 3 4𝑡 − 1 = log 1.3 log 10 − log 3 1 𝑡= ! + 1" ≈ 1.3972 4 log 1.3 log!

By Identities 2 and 3 Divide.

Solve for 𝑡.

56. First divide both sides of the given equation by 700 to get both sides:100 log! " = log(1.04 3𝑡+1) 7 log 100 − log 7 = (3𝑡 + 1) log 1.04 log 100 − log 7 3𝑡 + 1 = log 1.04 1 log 100 − log 7 𝑡= ! − 1" ≈ 22.2675 3 log 1.04

10 = 1.3 4𝑡−1. Take the common logarithm of both 3

100 = 1.04 3𝑡+1. Take the common logarithm of 7

By Identities 2 and 3 Divide.

Solve for 𝑡.

Solutions Section 1.1 Section 1.1 1. Using the table: 2. Using the table:

a. 𝑓(0) = 2

a. 𝑓(−1) = 4

b. 𝑓(2) = −0.5 b. 𝑓(1) = −1

3. Using the table: a. 𝑓(2) − 𝑓(−2) = −0.5 − 2 = −2.5 c. −2𝑓(−1) = −2(4) = −8 4. Using the table: a. 𝑓(1) − 𝑓(−1) = −1 − 4 = −5 c. 3𝑓(−2) = 3(2) = 6 5. From the graph, we estimate:

a. 𝑓(1) = 20

In a similar way, we find: c. 𝑓(3) = 30 f. 𝑓(3 − 2) = 𝑓(1) = 20 6. From the graph, we estimate:

7. From the graph, we estimate:

b. 𝑓(1)𝑓(−2) = (−1)(2) = −2

b. 𝑓(2) = 30

d. 𝑓(5) = 20\\e. 𝑓(3) − 𝑓(2) = 30 − 30 = 0

a. 𝑓(1) = 20

In a similar way, we find: c. 𝑓(3) = 10 f. 𝑓(3 − 2) = 𝑓(1) = 20

b. 𝑓(−1)𝑓(−2) = (4)(2) = 8

b. 𝑓(2) = 10

d. 𝑓(5) = 20 \\e. 𝑓(3) − 𝑓(2) = 10 − 10 = 0

a. 𝑓(−1) = 0

b. 𝑓(1) = −3 since the solid dot is on (1, −3).

In a similar way, we estimate c. 𝑓(3) = 3 d. Since 𝑓(3) = 3 and 𝑓(1) = −3,

𝑓(3) − 𝑓(1) 3 − (−3) = = 3. 3−1 3−1

8. From the graph, we estimate:

Solutions Section 1.1 a. 𝑓(−3) = 3 b. 𝑓(−1) = −2 since the solid dot is on (−1, −2).

In a similar way, we estimate c. 𝑓(1) = 0 𝑓(3) − 𝑓(1) 2 − 0 d. Since 𝑓(3) = 2 and 𝑓(1) = 0, = = 1. 3−1 3−1

1 , with its natural domain. 𝑥2 The natural domain consists of all 𝑥 for which 𝑓(𝑥) makes sense: all real numbers other than 0. 1 1 63 a. Since 4 is in the natural domain, 𝑓(4) is defined, and 𝑓(4) = 4 − 2 = 4 − . = 16 16 4 b. Since 0 is not in the natural domain, 𝑓(0) is not defined. 1 1 c. Since −1 is in the natural domain, 𝑓(−1) = −1 − = −1 − = −2. 2 1 (−1) 9. 𝑓(𝑥) = 𝑥 −

10. 𝑓(𝑥) =

2 − 𝑥 2, with domain [2, ∞) 𝑥

a. Since 4 is in [2, ∞), 𝑓(4) is defined, and 𝑓(4) = b. Since 0 is not in [2, ∞), 𝑓(0) is not defined.

2 1 31 − 4 2 = − 16 = − . 4 2 2 c. Since 1 is not in [2, ∞), 𝑓(1) is not defined.

11. 𝑓(𝑥) = √𝑥 + 10, with domain [−10, 0) a. Since 0 is not in [−10, 0), 𝑓(0) is not defined. b. Since 9 is not in [−10, 0), 𝑓(9) is not defined. c. Since −10 is in [−10, 0), 𝑓(−10) is defined, and 𝑓(−10) = √−10 + 10 = √0 = 0

12. 𝑓(𝑥) = √9 − 𝑥 2, with domain (−3, 3) a. Since 0 is in (−3, 3), 𝑓(0) is defined, and 𝑓(0) = √9 − 0 = 3. b. Since 3 is not in (−3, 3), 𝑓(3) is not defined. c. Since −3 is not in (−3, 3), 𝑓(−3) is not defined. 13. 𝑓(𝑥) = 4𝑥 − 3 a. 𝑓(−1) = 4(−1) − 3 = −4 − 3 = −7 b. 𝑓(0) = 4(0) − 3 = 0 − 3 = −3 c. 𝑓(1) = 4(1) − 3 = 4 − 3 = 1 d. Substitute 𝑦 for 𝑥 to obtain 𝑓(𝑦) = 4𝑦 − 3 e. Substitute (𝑎 + 𝑏) for 𝑥 to obtain 𝑓(𝑎 + 𝑏) = 4(𝑎 + 𝑏) − 3.

14. 𝑓(𝑥) = −3𝑥 + 4 a. 𝑓(−1) = −3(−1) + 4 = 3 + 4 = 7 b. 𝑓(0) = −3(0) + 4 = 0 + 4 = 4 c. 𝑓(1) = −3(1) + 4 = −3 + 4 = 1 d. Substitute 𝑦 for 𝑥 to obtain 𝑓(𝑦) = −3𝑦 + 4 e. Substitute (𝑎 + 𝑏) for 𝑥 to obtain 𝑓(𝑎 + 𝑏) = −3(𝑎 + 𝑏) + 4. 15. 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 3 a. 𝑓(0) = (0) 2 + 2(0) + 3 = 0 + 0 + 3 = 3

b. 𝑓(1) = 1 2 + 2(1) + 3 = 1 + 2 + 3 = 6

Solutions Section 1.1 c. 𝑓(−1) = (−1) + 2(−1) + 3 = 1 − 2 + 3 = 2 d. 𝑓(−3) = (−3) 2 + 2(−3) + 3 = 9 − 6 + 3 = 6 e. Substitute 𝑎 for 𝑥 to obtain 𝑓(𝑎) = 𝑎 2 + 2𝑎 + 3. f. Substitute (𝑥 + ℎ) for 𝑥 to obtain 𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2 + 2(𝑥 + ℎ) + 3. 2

16. 𝑔(𝑥) = 2𝑥 2 − 𝑥 + 1 a. 𝑔(0) = 2(0) 2 − 0 + 1 = 0 − 0 + 1 = 1 b. 𝑔(−1) = 2(−1) 2 − (−1) + 1 = 2 + 1 + 1 = 4 c. Substitute 𝑟 for 𝑥 to obtain 𝑔(𝑟) = 2𝑟 2 − 𝑟 + 1. d. Substitute (𝑥 + ℎ) for 𝑥 to obtain 𝑔(𝑥 + ℎ) = 2(𝑥 + ℎ) 2 − (𝑥 + ℎ) + 1.

1 𝑠 1 1 a. 𝑔(1) = 1 2 + = 1 + 1 = 2 b. 𝑔(−1) = (−1) 2 + =1−1=0 1 −1 1 1 65 1 c. 𝑔(4) = 4 2 + = 16 + = or 16.25 d. Substitute 𝑥 for 𝑠 to obtain 𝑔(𝑥) = 𝑥 2 + 4 4 4 𝑥 1 2 e. Substitute (𝑠 + ℎ) for 𝑠 to obtain 𝑔(𝑠 + ℎ) = (𝑠 + ℎ) + 𝑠+ℎ 1 1 f. 𝑔(𝑠 + ℎ) − 𝑔(𝑠) = Answer to part (e) − Original function = !(𝑠 + ℎ) 2 + " − !𝑠 2 + " 𝑠+ℎ 𝑠 17. 𝑔(𝑠) = 𝑠 2 +

1 𝑟+4 1 1 1 1 a. ℎ(0) = b. ℎ(−3) = = = =1 0+4 4 (−3) + 4 1 1 1 1 c. ℎ(−5) = d. Substitute 𝑥 2 for 𝑟 to obtain ℎ(𝑥 2) = 2 . = = −1 (−5) + 4 (−1) 𝑥 +4 1 1 e. Substitute (𝑥 2 + 1) for 𝑟 to obtain ℎ(𝑥 2 + 1) = 2 . = 2 (𝑥 + 1) + 4 𝑥 + 5 1 f. ℎ(𝑥 2) + 1 = Answer to part (d) + 1 = 2 + 1\\ 𝑥 +4 18. ℎ(𝑟) =

19. 𝑓(𝑥) = −𝑥 3 (domain (−∞, ∞))

20. 𝑓(𝑥) = 𝑥 3 (domain [0, ∞))

Technology formula: -(x^3)\\

Technology formula: x^3

21. 𝑓(𝑥) = 𝑥 (domain (−∞, ∞)) 4

22. 𝑓(𝑥) = √𝑥 (domain (−∞, ∞))

23. 𝑓(𝑥) =

1 (𝑥 ≠ 0) 𝑥2

24. 𝑓(𝑥) = 𝑥 +

1 (𝑥 ≠ 0) 𝑥

Solutions Section 1.1 Technology formula: x^4

Technology formula: x^(1/3)

Technology formula: 1/(x^2)

Technology formula: x+1/x

25. a. 𝑓(𝑥) = 𝑥 (−1 ≤ 𝑥 ≤ 1) Since the graph of 𝑓(𝑥) = 𝑥 is a diagonal 45 ∘ line through the origin inclining up from left to right, the correct graph is (A). b. 𝑓(𝑥) = −𝑥 (−1 ≤ 𝑥 ≤ 1) Since the graph of 𝑓(𝑥) = −𝑥 is a diagonal 45 ∘ line through the origin inclining down from left to right, the correct graph is (D). c. 𝑓(𝑥) = √𝑥 (0 < 𝑥 < 4) Since the graph of 𝑓(𝑥) = √𝑥 is the top half of a sideways parabola, the correct graph is (E). 1 d. 𝑓(𝑥) = 𝑥 + − 2 (0 < 𝑥 < 4) 𝑥 If we plot a few points like 𝑥 = 1∕2, 1, 2, and 3, we find that the correct graph is (F). e. 𝑓(𝑥) = |𝑥| (−1 ≤ 𝑥 ≤ 1) Since the graph of 𝑓(𝑥) = |𝑥|is a "V"-shape with its vertex at the origin, the correct graph is (C). f. 𝑓(𝑥) = 𝑥 − 1 (−1 ≤ 𝑥 ≤ 1)

Solutions Section 1.1 Since the graph of 𝑓(𝑥) = 𝑥 − 1 is a straight line through (0, −1) and (1, 0), the correct graph is (B).

26. a. 𝑓(𝑥) = −𝑥 + 3 (0 < 𝑥 ≤ 3) Since the graph of 𝑓(𝑥) = −𝑥 + 3 is a straight line inclining down from left to right, the correct graph must be (D). b. 𝑓(𝑥) = 2 − |𝑥| (−2 < 𝑥 ≤ 2) Since 𝑓(𝑥) = 2 − |𝑥| is obtained from the graph of 𝑦 = |𝑥| by flipping it vertically (the minus sign in front of |𝑥|) and then moving it 2 units vertically up (adding 2 to all the values), the correct graph is (F). c. 𝑓(𝑥) = √𝑥 + 2 (−2 < 𝑥 ≤ 2) The graph of𝑓(𝑥) = √𝑥 + 2 is similar to that of 𝑦 = √𝑥, which is half a parabola on its side, and the correct graph is (A). d. 𝑓(𝑥) = −𝑥 2 + 2 (−2 < 𝑥 ≤ 2) The graph of 𝑓(𝑥) = −𝑥 2 + 2 is a parabola opening down, so the correct graph is (C). 1 e. 𝑓(𝑥) = − 1 𝑥 1 The graph of 𝑓(𝑥) = − 1 (0 < 𝑥 ≤ 3) is part of a hyperbola, and the correct graph is (E). 𝑥 f. 𝑓(𝑥) = 𝑥 2 − 1 (−2 < 𝑥 ≤ 2) The graph of 𝑓(𝑥) = 𝑥 2 − 1 is a parabola opening up, so the correct graph is (B). 27. Technology formula: 0.1*x^2 - 4*x+5 Table of values: 𝑥

𝑓(𝑥)

0 5

1.1

−5

𝑔(𝑥) 39.9

−4

−3

30.3

21.5

−2

13.5

−1

6.3

0.5

1.5

2.5

3.5

−0.1 −5.7 −10.5 −14.5 −17.7 −20.1

29. Technology formula: (x^2-1)/(x^2+1) Table of values: 𝑥

−2.6 −6.1 −9.4 −12.5 −15.4 −18.1 −20.6 −22.9 −25

28. Technology formula: 0.4*x^2-6*x-0.1 Table of values: 𝑥

4.5

5.5

6.5

7.5

8.5

9.5

10.5

ℎ(𝑥) −0.6000 0.3846 0.7241 0.8491 0.9059 0.9360 0.9538 0.9651 0.9727 0.9781 0.9820 30. Technology formula: (2*x^2+1)/(2*x^2-1) Table of values: 𝑥

−1

𝑟(𝑥) 3.0000 −1.0000 3.0000 1.2857 1.1176 1.0645 1.0408 1.0282 1.0206 1.0157 1.0124

𝑥 if − 4 ≤ 𝑥 < 0 {2 if 0 ≤ 𝑥 ≤ 4 Technology formula: x*(x\lt 0)+2*(x\gt =0) (For a graphing calculator, use ≥ instead of >=.) 31. 𝑓(𝑥) =

Solutions Section 1.1

a. 𝑓(−1) = −1. We used the first formula, since −1 is in [−4, 0). b. 𝑓(0) = 2. We used the second formula, since 0 is in [0, 4]. c. 𝑓(1) = 2. We used the second formula, since 1 is in [0, 4].

−1 if − 4 ≤ 𝑥 ≤ 0 {𝑥 if 0 < 𝑥 ≤ 4 Technology formula: (-1)*(x\lt =0)+x*(x\gt 0) (For a graphing calculator, use ≤ instead of <=.) 32. 𝑓(𝑥) =

a. 𝑓(−1) = −1. We used the first formula, since −1 is in [−4, 0]. b. 𝑓(0) = −1. We used the first formula, since 0 is in [−4, 0]. c. 𝑓(1) = 1. We used the second formula, since 1 is in (0, 4].

⎧𝑥 2 if − 2 < 𝑥 ≤ 0 33. 𝑓(𝑥) = ⎪ ⎨ 1 if 0 < 𝑥 ≤ 4 ⎪𝑥 ⎩ Technology formula: (x^2)*(x<=0)+(1/x)*(0<x) (For a graphing calculator, use ≤ instead of <=.)

a. 𝑓(−1) = 1 2 = 1. We used the first formula, since −1 is in (−2, 0]. b. 𝑓(0) = 0 2 = 0. We used the first formula, since 0 is in (−2, 0]. c. 𝑓(1) = 1∕1 = 1. We used the second formula, since 1 is in (0, 4].

if − 2 < 𝑥 ≤ 0 −𝑥 {√𝑥 if 0 < 𝑥 < 4 Technology formula: Excel: (-1\*x^2)\*(x\lt =0)+SQRT(ABS(x))\*(x\gt 0) TI-83/84 Plus: (-1\*x^2)\*(x≤0)+ √(x)\*(x\gt 0) 2

34. 𝑓(𝑥) =

Solutions Section 1.1

a. 𝑓(−1) = −(−1) 2 = −1. We used the first formula, since −1 is in (−2, 0]. b. 𝑓(0) = −0 2 = 0. We used the first formula, since 0 is in (−2, 0]. c. 𝑓(1) = √1 = 1. We used the second formula, since 1 is in (0, 4).

if − 1 < 𝑥 ≤ 0 ⎧𝑥 35. 𝑓(𝑥) = ⎪ 𝑥 1 if 0 < 𝑥 ≤ 2 + ⎨ ⎪ 𝑥 if 2 < 𝑥 ≤ 4 ⎩ Technology formula: x*(x<=0)+(x+1)*(0<x)*(x<=2)+x*(2<x) (For a graphing calculator, use ≤ instead of <=.)

a. 𝑓(0) = 0. We used the first formula, since 0 is in (−1, 0]. b. 𝑓(1) = 1 + 1 = 2. We used the second formula, since 1 is in (0, 2]. c. 𝑓(2) = 2 + 1 = 3. We used the second formula, since 2 is in (0, 2]. d. 𝑓(3) = 3. We used the third formula, since 3 is in (2, 4].

if − 1 < 𝑥 < 0 ⎧−𝑥 𝑥 2 if 0 ≤ 𝑥 ≤ 2 − 36. 𝑓(𝑥) = ⎪ ⎨ ⎪ if 2 < 𝑥 ≤ 4 ⎩−𝑥 Technology formula: x*(x\lt 0)+(x-2)*(0\lt =x)*(x\lt =2)+(-x)*(2\lt x) (For a graphing calculator, use ≤ instead of <=.) y

1 -1

2 -2 -4

Solutions Section 1.1

a. 𝑓(0) = 0 − 2 = −2. We used the second formula, since 0 is in [0, 2]. b. 𝑓(1) = 1 − 2 = −1. We used the second formula, since 1 is in [0, 2]. c. 𝑓(2) = 2 − 2 = 0. We used the second formula, since 2 is in [0, 2]. d. 𝑓(3) = −3. We used the third formula, since 3 is in (2, 4]. 37. 𝑓(𝑥) = 𝑥 2 a. 𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2 Therefore,

𝑓(𝑥 + ℎ) − 𝑓(𝑥) = (𝑥 + ℎ) 2 − 𝑥 2 = 𝑥 2 + 2𝑥ℎ + ℎ 2 − 𝑥 2 = 2𝑥ℎ + ℎ 2 = ℎ(2𝑥 + ℎ)

b. Using the answer to part (a), 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ(2𝑥 + ℎ) = = 2𝑥 + ℎ ℎ ℎ

38. 𝑓(𝑥) = 3𝑥 − 1 a. 𝑓(𝑥 + ℎ) = 3(𝑥 + ℎ) − 1 = 3𝑥 + 3ℎ − 1 Therefore, 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 3𝑥 + 3ℎ − 1 − (3𝑥 − 1) = 3𝑥 + 3ℎ − 1 − 3𝑥 + 1 = 3ℎ

b. Using the answer to part (a), 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 3ℎ = =3 ℎ ℎ

39. 𝑓(𝑥) = 2 − 𝑥 2 a. 𝑓(𝑥 + ℎ) = 2 − (𝑥 + ℎ) 2 Therefore,

𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 2 − (𝑥 + ℎ) 2 − (2 − 𝑥 2) = 2 − 𝑥 2 − 2𝑥ℎ − ℎ 2 − 2 + 𝑥 2 = −2𝑥ℎ − ℎ 2 = −ℎ(2𝑥 + ℎ)

b. Using the answer to part (a), 𝑓(𝑥 + ℎ) − 𝑓(𝑥) −ℎ(2𝑥 + ℎ) = = −(2𝑥 + ℎ) ℎ ℎ 40. 𝑓(𝑥) = 𝑥 2 + 𝑥 a. 𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2 + (𝑥 + ℎ)Therefore,

𝑓(𝑥 + ℎ) − 𝑓(𝑥) = (𝑥 + ℎ) 2 + (𝑥 + ℎ) − (𝑥 2 + 𝑥) = 𝑥 2 + 2𝑥ℎ + ℎ 2 + 𝑥 + ℎ − 𝑥 2 − 𝑥 = 2𝑥ℎ + ℎ 2 + ℎ = ℎ(2𝑥 + ℎ + 1)

b. Using the answer to part (a), 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ(2𝑥 + ℎ + 1) = = 2𝑥 + ℎ + 1 ℎ ℎ

41. From the table, a. 𝑝(2) = 0.67; Pemex produced 0.67 billion barrels of crude oil in 2017 (𝑡 = 2). 𝑝(3) = 0.61; Pemex produced 0.61 billion barrels of crude oil in 2018 (𝑡 = 3). 𝑝(6) = 0.62; Pemex produced 0.62 billion barrels of crude oil in 2021 (𝑡 = 6).

Solutions Section 1.1 b. 𝑝(6) − 𝑝(3) = 0.62 − 0.61 = 0.01; Annual crude oil production by Pemex increased by 0.01 billion barrels from 2018 (𝑡 = 3) to 2021 (𝑡 = 6). 42. From the table, a. 𝑠(0) = 0.69; Pemex produced 0.69 billion barrels of offshore crude oil in 2015 (𝑡 = 0). 𝑠(2) = 0.56; Pemex produced 0.56 billion barrels of offshore crude oil in 2017 (𝑡 = 2). 𝑠(4) = 0.51; Pemex produced 0.51 billion barrels of offshore crude oil in 2019 (𝑡 = 4). b. 𝑠(4) − 𝑠(0) = 0.51 − 0.69 = −0.18; Annual offshore crude oil production by Pemex decreased by 0.18 billion barrels from 2015 (𝑡 = 0) to 2019 (𝑡 = 4). 43. a. Graph of 𝑝:

The graph suggests a curve, so we exclude the linear models (A) and (C). Model (B) gives a parabola that curves up whereas the curve suggested by the graph curves down, leaving us with Model (D). Also, model (D) gives almost the exact values shown in the chart. (Use the technology formula -0.25x^2+3.5x+57.) b. Using Model (D), 𝑝(5) = −0.25(5) 2 + 3.5(5) + 57 = 68.25 Interpretation: 𝑡 = 5 represents 5 years since the start of 2013, or the start of 2018. Thus, we interpret the answer as follows: Approximately 68.25% of U.S. adults used Facebook at the start of 2018 c. The graph becomes less steep as time increases, indicating decelerating Facebook membership over the period 2013–2021. 44. a. Graph of 𝑝:

The graph suggests a curve, so we exclude the linear model (B). Model (D) gives a parabola that curves up whereas the curve suggested by the graph curves down, leaving us with Models (A) and (C). Model (A) gives values that round to the values shown in the chart wheeras Model (C) is way off. (Use the technology formula -0.32x^2+3.6x+13. b. Using Model (A), 𝑝(7.5) = −0.32(7.5) 2 + 3.6(7.5) + 13 = 22 Interpretation: 𝑡 = 7.5 represents 7.5 years since the start of 2013, or midway through 2020. Thus, we interpret the answer as follows: Approximately 22% of U.S. adults used Twitter midway through 2020. c. The graph becomes less steep as time increases, indicating decelerating Twitter membership over the period 2013–2021.

45. From the graph, 𝑓(11) ≈ 800. Because 𝑓 is the number of thousands of housing starts in year 11, we interpret the result as follows: Approximately 800,000 homes were started in 2016. Similarly, 𝑓(15) ≈ 1,000. Approximately 1,000,000 homes were started in 2020. Also, we estimate 𝑓(2.5) ≈ 800. Because 𝑡 = 2.5 is midway between 2007 and 2008, we interpret the result as follows: approximately 800,000 homes were started in the year beginning midway through 2007.

Solutions Section 1.1 46. From the graph, 𝑓(3) ≈ 600. Because 𝑓 is the number of thousands of housing starts in year 3, we interpret the result as follows: Approximately 600,000 homes were started in 2008. 𝑓(12) ≈ 900: Approximately 900,000 homes were started in 2016. 𝑓(14.5) ≈ 950: Approximately 950,000 homes were started in the year beginning midway through 2019.

47. 𝑓(14 − 10) = 𝑓(4) ≈ 400. Interpretation: Approximately 400,000 homes were started in 2009 (𝑡 = 4). 𝑓(14) − 𝑓(10) ≈ 900 − 700 = 200. Interpretation: 𝑓(14) − 𝑓(10) is the change in the number of housing starts (in thousands) from 2015 to 2019; there were approximately 200,000 more housing starts in 2019 than in 2015. 48. 𝑓(15 − 1) = 𝑓(14) ≈ 900. Interpretation: Approximately 900,000 homes were started in 2019 (𝑡 = 14). 𝑓(15) − 𝑓(1) ≈ 1,000 − 1,500 = −500. Interpretation: 𝑓(15) − 𝑓(1) is the change in the number of housing starts (in thousands) from 2006 to 2020; there were approximately 500,000 fewer housing starts in 2020 than in 2006.

49. 𝑓(𝑡 + 5) − 𝑓(𝑡) measures the change from year 𝑡 to the year five years later. It is greatest when the line segment from year 𝑡 to year 𝑡 + 5 is steepest upward-sloping. From the graph, or by computing differences of values estimated from the graph, this occurs when 𝑡 = 6 or 7: 𝑓(11) − 𝑓(6) = 800 − 400 = 400; 𝑓(12) − 𝑓(7) = 900 − 500 = 400. Interpretation: The greatest five-year increase in the number of housing starts occurred in 2011–2016 and again in 2012–2017. 50. 𝑓(𝑡) − 𝑓(𝑡 − 1) measures the change from year 𝑡 − 1 to the following year. It is least when the the line segment from year 𝑡 − 1 to year 𝑡 is steepest downward-sloping. From the graph, this occurs when 𝑡 = 2, for a change of 𝑓(2) − 𝑓(1) = 1,000 − 1,500 = −500. Interpretation: The greatest annual decrease in the number of housing starts occurred in 2006–2007.

51. a. From the graph, 𝑏(3) ≈ 50 and 𝑏(5) ≈ 35. The (50-day average) price of Bitcoin was approximately $50,000 on June 1, 2021 (𝑡 = 3) and $35,000 on August 1, 2021 (𝑡 = 5). 𝑏(7) − 𝑏(5) ≈ 50 − 35 = 15. The Bitcoin price increased by around $15,000 from August 1 to October 1, 2021. b. Increasing most rapidly when the graph is steepest upward from left to right on the interal [1, 3], which occurs at 𝑡 = 1 (integer answer required). Thus, during the period April 1–June 1, 2021, the price of Bitcoin was increasing most rapidly around April 1. c. Decreasing most rapidly when the graph is steepest down from left to right, which occurs at 𝑡 = 3 (integer answer required). Thus, during the period April 1–June 1, 2021, the price of Butcoin was decreasing most rapidly around June 1. 52. a. From the graph, 𝑒(2) ≈ 2.5 and 𝑒(4) ≈ 2.5. The price of Etherium was approximately $2,500 on May 1, 2021 (𝑡 = 2) and again on July 1, 2021 (𝑡 = 4) . 𝑒(3) − 𝑒(1) ≈ 2 − 3 = −1. The price of Etherium decreased by around $1,000 from April 1 to June 1, 2021. b. Increasing most rapidly when the graph is steepest upward from left to right on the interal [1, 5], which occurs at 𝑡 = 4 (integer answer required). Thus, during the period April 1–August 1, 2021, the price of Etherium was increasing most rapidly around July 1. c. Decreasing most rapidly when the graph is steepest down from left to right, which occurs at 𝑡 = 2 (integer answer required). Thus, during the period April 1–August 1, 2021, the price of Etherium was decreasing most rapidly around May 1. 53. 𝑟(𝑥) = 0.4𝑥 2 + 𝑥 + 26.5 (0 ≤ 𝑥 ≤ 5) a. The domain of 𝑟 is [0, 5].

𝑟(0) = 0.4(0) + 0 + 26.5 = 26.5; 𝑟(1) = 0.4(1) 2 + 1 + 26.5 = 27.9 𝑟(3) = 0.4(3) 2 + 3 + 26.5 = 33.1 𝑟(5) = 0.4(5) 2 + 5 + 26.5 = 41.5 Food delivery app revenues in the U.S. were projected to be $26.5 billion in 2020, $27.9 billion in 2021, $33.1 billion in 2023, and $41.5 billion in 2025. b. Graph: 2

Solutions Section 1.1

(As the curve is concave up the projected revenue was accelerating.)

54. 𝑟(𝑥) = −0.2𝑥 2 + 3𝑥 + 18 (0 ≤ 𝑥 ≤ 5) a. The domain of 𝑟 is [0, 5]. 𝑟(0) = −0.2(0) 2 + 3(0) + 18 = 18; 𝑟(1) = −0.2(1) 2 + 3(1) + 18 = 20.8 𝑟(3) = −0.2(3) 2 + 3(3) + 18 = 25.2 𝑟(5) = −0.2(5) 2 + 3(5) + 18 = 28 Food delivery app revenues in Europe were projected to be $18 billion in 2020, $20.8 billion in 2021, $25.2 billion in 2023, and $28 billion in 2025. b. Graph

(As the graph is concave down the projected revenue was decelerating). 55. a. The model is valid for the range 1958 (𝑡 = 0) through 1966 (𝑡 = 8). Thus, an appropriate domain is [0, 8]. 𝑡 ≥ 0 is not an appropriate domain because it would predict federal funding of NASA beyond 1966, whereas the model is based only on data up to 1966. 4.5 b. 𝑝(𝑡) = 2 1.07 (𝑡−8) 4.5 ⇒ 𝑝(5) = ≈ 2.4 Technology formula: 4.5/(1.07^((t-8)^2)) 2 1.07 (5−8) 𝑡 = 5 represents 1958 + 5 = 1963, and therefore we interpret the result as follows: In 1963, 2.4% of the U.S. federal budget was allocated to NASA. c. 𝑝(𝑡) is increasing most rapidly when the graph is steepest upward-sloping from left to right, and, among the given values of 𝑡, this occurs when 𝑡 = 5. Thus, the percentage of the budget allocated to NASA was

Solutions Section 1.1 increasing most rapidly in 1963.

56. a. [1, 55]; [0, 55] is not an appropriate domain because 𝑝 is undefined at 0. 5 b. 𝑝(𝑡) = 0.03 + 0.6 𝑡 5 ⇒ 𝑝(40) = 0.03 + ≈ 0.58 Technology formula: 0.03+5/t^0.6 40 0.6 𝑡 = 40 represents 1965 + 40 = 2005, and therefore we interpret the result as follows: In 2005, 0.58% of the US federal budget was allocated to NASA. c. If we evaluate 𝑝(𝑡) for 𝑡 = 100, 1,000, 100,000, 1,000,000 . . . , we find values of 𝑝(𝑡) decreasing toward 0.03. Thus, in the (very) long term, the percentage of the budget allocated to NASA is predicted to approach 0.03% 12, 200 57. a. 𝑝(𝑡) = 100⎛1 − 4.48 ⎞ (𝑡 ≥ 8.5) ⎝⎜ ⎠⎟ 𝑡 Technology formula: 100*(1-12200/t^4.48) b. Graph:

c. Table of values: 𝑡

𝑝(𝑡) 35.2

59.6

73.6

82.2

87.5

91.1

93.4

95.1

96.3

97.1

97.7

98.2

d. From the table, 𝑝(12) = 82.2, so that 82.2% of children are able to speak in at least single words by the age of 12 months. e. We seek the first value of 𝑡 such that 𝑝(𝑡) is at least 90. Since 𝑡 = 14 has this property (𝑝(14) = 91.1) we conclude that, at 14 months, 90% or more children are able to speak in at least single words. 5.27 × 10 17 (𝑡 ≥ 30) " 𝑡 12 Technology formula: 100*(1-5.27*10^17/t^12) b.Graph: 58. a. 𝑝(𝑡) = 100!1 −

c. Table of values: 𝑡

𝑝(𝑡)

0.8

33.1

54.3

68.4

77.9

84.4

88.9

92.0

94.2

95.7

96.9

d. From the table, 𝑝(36) = 88.9, so that 88.9% of children are able to speak in sentences of five or more words by the age of 36 months.

Solutions Section 1.1 e. We seek the first value of 𝑡 such that 𝑝(𝑡) is at least 75. Since 𝑡 = 34 has this property (𝑝(34) = 77.9) we conclude that, at 34 months, 75% or more children are able to speak in sentences of five or more words. if 0 ≤ 𝑡 < 16 ⎧8(1.22) 𝑡 ⎪ 59. 𝑣(𝑡) = ⎨400𝑡 − 6,200 if 16 ≤ 𝑡 < 25 ⎪0.45(𝑡 − 25) 3 + 3,800 if 25 ≤ 𝑡 ≤ 40 ⎩

a. 𝑣(10) = 8(1.22) 10 ≈ 58. We used the first formula, since 10 is in [0, 16). 𝑣(16) = 400(16) − 6,200 = 200. We used the second formula, since 16 is in [16, 25). 𝑣(40) = 0.45(40 − 25) 3 + 3,800 ≈ 5,319 We used the third formula, since 40 is in [25, 40]. Interpretation: Processor speeds were about 58 MHz in 1990, 200 MHz in 1996, and 5,319 MHz in 2020. b. Technology formula (using 𝑥 as the independent variable): (8*(1.22)^x)*(x<16)+(400*x-6200)*(x>=16)*(x<25) +(0.45*(x-25)^3+3800)*(x>=25) (For a graphing calculator, use ≤ instead of <=.) c. Using the above technology formula (for instance, on the Function Evaluator and Grapher on the Web site) we obtain the graph and table of values. Graph:

Table of values: 𝑡

𝑣(𝑡)

200 1,800 3,400 3,800 4,000 4,400 5,300

d. From either the graph or the table, we see that the speed reached 3,000 MHz between 𝑡 = 20 and 𝑡 = 24. We can obtain a more precise answer algebraically by using the formula for the corresponding portion of the graph: 3,000 = 400𝑡 − 6,200 giving 𝑡=

9,200 = 23 400

Since 𝑡 is time since 1980, 𝑡 = 23 corresponds to 2003.

⎧0.12𝑡 2 + 0.04𝑡 + 0.2 if 0 ≤ 𝑡 < 12 60. 𝑣(𝑡) = ⎪ if 12 ≤ 𝑡 < 26 ⎨1.1(1.22) 𝑡 ⎪400𝑡 − 10,200 if 26 ≤ 𝑡 ≤ 30 ⎩ 2 a. 𝑣(2) = 0.12(2) + 0.04(2) + 0.2 = 0.76. We used the first formula, since 2 is in [0, 12). 𝑣(12) = 1.1(1.22) 12 ≈ 12. We used the second formula, since 12 is in [12, 26). 𝑣(28) = 400(28) − 10, 200 = 1, 000. We used the third formula, since 28 is in [26, 30]. Interpretation: Processor speeds were about 0.76 MHz in 1972, 12 MHz in 1982, and 1,000 MHz in 1998.

Solutions Section 1.1

b. Technology formula (using 𝑥 as the independent variable): (0.12*x^2+0.04*x+0.2)*(x<12)+(1.1*(1.22)^x)*(x>=12)*(x<26) +(400*x-10200)*(x>=26) (For a graphing calculator, use ≤ instead of <=.) c. Using the above technology formula (for instance, on the Function Evaluator and Grapher on the Web site) we obtain the graph and table of values. Graph:

Table of values: 𝑡

𝑣(𝑡) 0.20 0.76

2.3

4.8

8.2

130

200 1,000 1,800

d. From either the graph or the table, we see that the speed reached 500 MHz around 𝑡 = 27. We can obtain a more precise answer algebraically by using the formula for the corresponding portion of the graph: 500 = 400𝑡 − 10,200

giving 𝑡=

10,700 = 26.75 ≈ 27 to the nearest year 400

Since 𝑡 is time since 1970, 𝑡 = 27 corresponds to 1997. 61. a. Each row of the table gives us a formula with a condition: First row in words: 10% of the amount over $0 if your income is over $0 and not over $9,950. Translation to formula: 0.10𝑥 if 0 < 𝑥 ≤ 9,950. Second row in words: $995.00 + 12% of the amount over $9,950 if your income is over $9,950 and not over $40,525. Translation to formula: 995.00 + 0.12(𝑥 − 9, 950) if 9,950 < 𝑥 ≤ 40,525. Continuing in this way leads to the following piecewise-defined function: ⎧0.10𝑥 ⎪995.00 + 0.12(𝑥 − 9,950) ⎪ ⎪4,664 + 0.22(𝑥 − 40,525) 𝑇 (𝑥) = ⎪ ⎨14,751 + 0.24(𝑥 − 86,375) ⎪33,603 + 0.32(𝑥 − 164,925) ⎪ ⎪47,843 + 0.35(𝑥 − 209,425) ⎪157,804.25 + 0.37(𝑥 − 523,600) ⎩

if 0 < 𝑥 ≤ 9,950 if 9,950 < 𝑥 ≤ 40,525 if 40,525 < 𝑥 ≤ 86,375 if 86,375 < 𝑥 ≤ 164,925 if 164,925 < 𝑥 ≤ 209,425 if 209,425 < 𝑥 ≤ 523,600 if 523,600 < 𝑥

Solutions Section 1.1

b. A taxable income of $45,000 falls in the bracket 40,525 < 𝑥 ≤ 86,375 and so we use the formula 4,664 + 0.22(𝑥 − 40,525) : 4,664 + 0.22(45,000 − 40,525) = 4,664 + 0.22(4,475) = $5,648.50 62. a. Each row of the table gives us a formula with a condition: First row in words: 10% of the amount over $0 if your income is over $0 and not over $8,700. Translation to formula: 0.10𝑥 if 0 < 𝑥 ≤ 8,700. Second row in words: $870.00 + 15% of the amount over $8,700 if your income is over $8,700 and not over $35,350. Translation to formula: 870.00 + 0.15(𝑥 − 8,700) if 8,700 < 𝑥 ≤ 35,350. Continuing in this way leads to the following piecewise-defined function: ⎧0.10𝑥 ⎪870.00 + 0.15(𝑥 − 8,700) ⎪ ⎪4,867.50 + 0.22(𝑥 − 35,350) 𝑇 (𝑥) = ⎨ 17,442.50 + 0.28(𝑥 − 85,650) ⎪ ⎪43,482.50 + 0.33(𝑥 − 178,650) ⎪112,683.50 + 0.35(𝑥 − 388,350) ⎩

if 0 < 𝑥 ≤ 8,700 if 8,700 < 𝑥 ≤ 35,350 if 35,350 < 𝑥 ≤ 85,650 if 85,650 < 𝑥 ≤ 178,650 if 178,650 < 𝑥 ≤ 388,350 if 388,350 < 𝑥

b. A taxable income of $45,000 falls in the bracket 35,350 < 𝑥 ≤ 85,650 and so we use the formula 4,867.50 + 0.22(𝑥 − 35,350) : 4,867.50 + 0.22(45,000 − 35,350) = 4,867.50 + 0.22(9,650) = $9,990.50.

63. The dependent variable is a function of the independent variable. Here, the market price of gold 𝑚 is a function of time 𝑡. Thus, the independent variable is 𝑡 and the dependent variable is 𝑚. 64. The dependent variable is a function of the independent variable. Here, the weekly profit 𝑃 is a function of the selling price 𝑠. Thus, the independent variable is 𝑠 and the dependent variable is 𝑃 .

65. To obtain the function notation, write the dependent variable as a function of the independent variable. Thus 𝑦 = 4𝑥 2 − 2 can be written as 𝑓(𝑥) = 4𝑥 2 − 2 or 𝑦(𝑥) = 4𝑥 2 − 2 66. To obtain the equation notation, introduce a dependent variable instead of the function notation. Thus 𝐶(𝑡) = −0.34𝑡 2 + 0.1𝑡 can be written as 𝑐 = −0.34𝑡 2 + 0.1𝑡 or 𝑦 = −0.34𝑡 2 + 0.1𝑡 67. False. A graph usually gives infinitely many values of the function while a numerical table will give only a finite number of values.

68. True. An algebraically specified function 𝑓 is specified by algebraic formulas for 𝑓(𝑥). Given such formulas, we can construct the graph of 𝑓 by plotting the points (𝑥, 𝑓(𝑥)) for values of 𝑥 in the domain of 𝑓. 69. False. In a numerically specified function, only certain values of the function are specified so we cannot know its value on every real number in [0, 10], whereas an algebraically specified function would give values for every real number in [0, 10].

Solutions Section 1.1 70. False. A graphically specified function is specified by a graph. However, we cannot always expect to find an algebraic formula whose graph is exactly the graph that is given. 71. Functions with infinitely many points in their domain (such as 𝑓(𝑥) = 𝑥 2) cannot be specified numerically. So, the assertion is false.

72. A numerical model supplies only the values of a function at specific values of the independent variable, whereas an algebraic model supplies the value of a function at every point in its domain. Thus, an algebraic model supplies more information.

73. (Answers may vary.) Take 𝑓(𝑥) = 𝑥 2. Then a. 𝑓(3 + 2) = 𝑓(5) = 5 2 = 25, whereas 𝑓(3) + 𝑓(2) = 3 2 + 2 2 = 9 + 4 = 13. Thus, 𝑓(3 + 2) ≠ 𝑓(3) + 𝑓(2). b. 𝑓(3 − 2) = 𝑓(1) = 1 2 = 1, whereas 𝑓(3) − 𝑓(2) = 3 2 − 2 2 = 9 − 4 = 5. Thus, 𝑓(3 − 2) ≠ 𝑓(3) − 𝑓(2). 74. (Answers may vary.) Take 𝑓(𝑥) = 𝑥 + 1. Then a. 𝑓(3 × 2) = 𝑓(6) = 6 + 1 = 7, whereas 3𝑓(2) = 3(2 + 1) = 3 × 3 = 9. Thus, 𝑓(3 × 2) ≠ 3𝑓(2). b. 𝑓(3 × 2) = 𝑓(6) = 6 + 1 = 7, whereas 𝑓(3)𝑓(2) = (3 + 1)(2 + 1) = 4 × 3 = 12. Thus, 𝑓(3 × 2) ≠ 𝑓(3)𝑓(2).

75. As the text reminds us: to evaluate 𝑓 of a quantity (such as 𝑥 + ℎ) replace 𝑥 everywhere by the whole quantity 𝑥 + ℎ : 𝑓(𝑥) = 𝑥2 − 1 𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2 − 1.

76. Knowing 𝑓(𝑥) for two values of 𝑥 does not convey any information about 𝑓(𝑥) at any other value of 𝑥. Interpolation is only a way of estimating 𝑓(𝑥) at values of 𝑥 not given.

77. If two functions are specified by the same formula 𝑓(𝑥), say, their graphs must follow the same curve 𝑦 = 𝑓(𝑥). However, it is the domain of the function that specifies what portion of the curve appears on the graph. Thus, if the functions have different domains, their graphs will be different portions of the curve 𝑦 = 𝑓(𝑥). 78. If we plot points of the graphs 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑔(𝑥), we see that, since 𝑔(𝑥) = 𝑓(𝑥) + 10, we must add 10 to the 𝑦-coordinate of each point in the graph of 𝑓 to get a point on the graph of 𝑔. Thus, the graph of 𝑔 is 10 units higher up than the graph of 𝑓. 79. Suppose we already have the graph of 𝑓 and want to construct the graph of 𝑔. We can plot a point of the graph of 𝑔 as follows: Choose a value for 𝑥 (𝑥 = 7, say) and then "look back" 5 units to read off 𝑓(𝑥 − 5) (𝑓(2) in this instance). This value gives the 𝑦-coordinate we want. In other words, points on the graph of 𝑔 are obtained by "looking back 5 units" to the graph of 𝑓 and then copying that portion of the curve. Put another way, the graph of 𝑔 is the same as the graph of 𝑓, but shifted 5 units to the right:

80. Suppose we already have the graph of 𝑓 and want to construct the graph of 𝑔. We can plot a point of

Solutions Section 1.1 the graph of 𝑔 as follows: Choose a value for 𝑥 (𝑥 = 7, say) and then look on the other side of the 𝑦-axis to read off 𝑓(−𝑥) (𝑓(−7) in this instance). This value gives the 𝑦-coordinate we want. In other words, points on the graph of 𝑔 are obtained by "looking back" to the graph of 𝑓 on the opposite side of the 𝑦-axis and then copying that portion of the curve. Put another way, the graph of 𝑔(𝑥) is the mirror image of the graph of 𝑓(𝑥) in the 𝑦-axis:

Solutions Section 1.2 Section 1.2

1. 𝑓(𝑥) = 𝑥 2 + 1 with domain (−∞, ∞) 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) a. 𝑠(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) = (𝑥 2 + 1) + (𝑥 − 1) = 𝑥 2 + 𝑥 b. Since both functions are defined for every real number 𝑥, the domain of 𝑠 is the set of all real numbers: (−∞, ∞). c. 𝑠(−3) = (−3) 2 + (−3) = 9 − 3 = 6

2. 𝑓(𝑥) = 𝑥 2 + 1 with domain (−∞, ∞) 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) a. 𝑑(𝑥) = 𝑔(𝑥) − 𝑓(𝑥) = (𝑥 − 1) − (𝑥 2 + 1) = −𝑥 2 + 𝑥 − 2 b. Since both functions are defined for every real number 𝑥, the domain of 𝑑 is the set of all real numbers: (−∞, ∞). c. 𝑑(−1) = −(−1) 2 + (−1) − 2 = −4 3. 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) 𝑢(𝑥) = √𝑥 + 10 with domain [−10, 0) a. 𝑝(𝑥) = 𝑔(𝑥)𝑢(𝑥) = (𝑥 − 1)√𝑥 + 10 b. The domain of 𝑝 consists of all real numbers 𝑥 simultaneously in the domains of 𝑔 and 𝑢; that is, [−10, 0). c. 𝑝(−6) = (−6 − 1)√−6 + 10 = (−7)(2) = −14

4. ℎ(𝑥) = 𝑥 + 4 with domain [10, ∞) 𝑣(𝑥) = √10 − 𝑥 with domain [0, 10] a. 𝑝(𝑥) = ℎ(𝑥)𝑣(𝑥) = (𝑥 + 4)√10 − 𝑥 b. The domain of 𝑝 consists of all real numbers 𝑥 simultaneously in the domains of ℎ and 𝑣; that is, the single point 𝑥 = 10. c. As 1 is not in the domain of 𝑝, 𝑝(1) is not defined.

5. 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) 𝑣(𝑥) = √10 − 𝑥 with domain [0, 10] 𝑣(𝑥) √10 − 𝑥 a. 𝑞(𝑥) = = 𝑔(𝑥) 𝑥−1 b. The domain of 𝑞 consists of all real numbers 𝑥 simultaneously in the domains of 𝑣 and 𝑔 such that 𝑔(𝑥) ≠ 0. Since 𝑔(𝑥) = 0 when 𝑥 − 1 = 0, or 𝑥 = 1 we exclude 𝑥 = 1 from the domain of the quotient. Thus, the domain consists of all 𝑥 in [0, 10] excluding 𝑥 = 1 (since 𝑔(1) = 0), or 0 ≤ 𝑥 ≤ 10; 𝑥 ≠ 1. c. As 1 is not in the domain of 𝑞, 𝑞(1) is not defined.

6. 𝑔(𝑥) = 𝑥 − 1 with domain (−∞, ∞) 𝑣(𝑥) = √10 − 𝑥 with domain [0, 10] 𝑔(𝑥) 𝑥−1 a. 𝑞(𝑥) = = 𝑣(𝑥) √10 − 𝑥 b. The domain of 𝑞 consists of all real numbers 𝑥 simultaneously in the domains of 𝑣 and 𝑔 such that 𝑣(𝑥) ≠ 0. Since 𝑣(𝑥) = 0 when √10 − 𝑥 = 0, or 𝑥 = 10 we exclude 𝑥 = 10 from the domain of the quotient. Thus, the domain consists of all 𝑥 in [0, 10] excluding 𝑥 = 10; that is, [0, 10).

Solutions Section 1.2 1−1 c. 1 is in the domain of 𝑞, and 𝑞(1) = =0 √10 − 1 7. 𝑓(𝑥) = 𝑥 2 + 1 with domain (−∞, ∞) a. 𝑚(𝑥) = 5𝑓(𝑥) = 5(𝑥 2 + 1) b. The domain of 𝑚 is the same as the domain of 𝑓 : (−∞, ∞). c. 𝑚(1) = 5𝑓(1) = 5(1 2 + 1) = 10 8. 𝑢(𝑥) = √𝑥 + 10 with domain [−10, 0) a. 𝑚(𝑥) = 3𝑢(𝑥) = 3√𝑥 + 10 b. The domain of 𝑚 is the same as the domain of 𝑢 : [−10, 0) c. 𝑚(−1) = 3𝑢(−1) = 3√−1 + 10 = 9

9. Number of music files = Starting number + New files = 200 + 10 × Number of days So, 𝑁(𝑡) = 200 + 10𝑡 (𝑁 = number of music files, 𝑡 = time in days) 10. Free space left = Current amount − Decrease = 50 − 5× Number of months So, 𝑆(𝑡) = 50 − 5𝑡 (𝑆 = space on your HD, 𝑡 = time in months)

11. The number of hours you study, ℎ(𝑛), equals 4 on Sunday through Thursday and equals 0 on the remaining days. Since Sunday corresponds to 𝑛 = 1 and Thursday to 𝑛 = 5, we get 4 if 1 ≤ 𝑛 ≤ 5 . {0 if 𝑛 > 5

12. The number of hours you watch movies, ℎ(𝑛), equals 5 on Saturday (𝑛 = 7) and Sunday (𝑛 = 1) and equals 2 on the remaining days. 5 if 𝑛 = 1or 𝑛 = 7 . {2 otherwise

13. For a linear cost function, 𝐶(𝑥) = 𝑚𝑥 + 𝑏. Here, 𝑚 = marginal cost = $1,500 per piano, 𝑏 = fixed cost = $1,000. Thus, the daily cost function is 𝐶(𝑥) = 1,500𝑥 + 1,000. a. The cost of manufacturing 3 pianos is 𝐶(3) = 1,500(3) + 1,000 = 4,500 + 1,000 = $5,500. b. The cost of manufacturing each additional piano (such as the third one or the 11th one) is the marginal cost, 𝑚 = $1,500. c. Same answer as (b). d. Variable cost = part of the cost function that depends on 𝑥 = $1,500𝑥 Fixed cost = constant summand of the cost function = $1,000 Marginal cost = slope of the cost function = $1,500 per piano

Solutions Section 1.2 e. Graph: C 8,000 7,000 6,000 5,000 4,000 3,000 2,000 x

1,000

14. For a linear cost function, 𝐶(𝑥) = 𝑚𝑥 + 𝑏. Here, 𝑚 = marginal cost = $88 per tuxedo, 𝑏 = fixed cost = $20. Thus, the cost function is 𝐶(𝑥) = 88𝑥 + 20. a. The cost of renting 2 tuxes is 𝐶(2) = 88(2) + 20 = $196 b. The cost of each additional tux is the marginal cost 𝑚 = $88. c. Same answer as (b). d. Variable cost = part of the cost function that depends on 𝑥 = $88𝑥 Fixed cost = constant summand of the cost function = $20 Marginal cost = slope of the cost function = $88 per tuxedo e. Graph: C 400 350 300 250 200 150 100 50 0

15. a. For a linear cost function, 𝐶(𝑥) = 𝑚𝑥 + 𝑏. Here, 𝑚 = marginal cost = $0.40 per copy, 𝑏 = fixed cost = $70. Thus, the cost function is 𝐶(𝑥) = 0.4𝑥 + 70. The revenue function is 𝑅(𝑥) = 0.50𝑥. (𝑥 copies at 50¢ per copy) The profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0.5𝑥 − (0.4𝑥 + 70) = 0.5𝑥 − 0.4𝑥 − 70 = 0.1𝑥 − 70

b. 𝑃 (500) = 0.1(500) − 70 = 50 − 70 = −20 Since 𝑃 is negative, this represents a loss of $20. c. For breakeven, 𝑃 (𝑥) = 0 :

0.1𝑥 − 70 = 0 0.1𝑥 = 70 70 𝑥= = 700 copies 0.1

Solutions Section 1.2

16. a. For a linear cost function, 𝐶(𝑥) = 𝑚𝑥 + 𝑏. Here, 𝑚 = marginal cost = $0.15 per serving, 𝑏 = fixed cost = $350. Thus, the cost function is 𝐶(𝑥) = 0.15𝑥 + 350. The revenue function is 𝑅(𝑥) = 0.50𝑥. The profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0.50𝑥 − (0.15𝑥 + 350) = 0.35𝑥 − 350

b. For break-even, 𝑃 (𝑥) = 0 : 0.35𝑥 − 350 = 0 0.35𝑥 = 350 𝑥 = 1,000 servings c. 𝑃 (1,500) = 0.35(1,500) − 350 = 525 − 350 = $175, representing a profit of $175. 17. The revenue per jersey is $100. Therefore, Revenue 𝑅(𝑥) = $100𝑥. Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 100𝑥 − (2,000 + 10𝑥 + 0.2𝑥 2) = −2,000 + 90𝑥 − 0.2𝑥 2

To break even, 𝑃 (𝑥) = 0, so −2,000 + 90𝑥 − 0.2𝑥 2 = 0. This is a quadratic equation with 𝑎 = −0.2, 𝑏 = 90, 𝑐 = −2,000 and solution −𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 −90 ± √(90) 2 − 4(−2,000)(−0.2) = ≈ 23.44 or 426.56 jerseys. 2(−0.2)

𝑥=

Since the second value is outside the domain, we use the first: 𝑥 = 23.44 jerseys. To make a profit, 𝑥 should be larger than this value: at least 24 jerseys. 18. The revenue per pair is $120. Therefore, Revenue 𝑅(𝑥) = $120𝑥. Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 120𝑥 − (3,000 + 8𝑥 + 0.1𝑥 2) = −3,000 + 112 − 0.1𝑥 2

To break even, 𝑃 (𝑥) = 0, so −3,000 + 112 − 0.1𝑥 2 = 0. This is a quadratic equation with 𝑎 = −0.1, 𝑏 = 112, 𝑐 = −3,000 and solution 𝑥=

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 −90 ± √(112) 2 − 4(−3,000)(−0.1) = ≈ 27.46 or 1092.54 jerseys. 2(−0.1)

Solutions Section 1.2 Since the second value is outside the domain, we use the first: 𝑥 = 27.46 jerseys. To make a profit, 𝑥 should be larger than this value: at least 28 pairs of cleats.

19. The revenue from one thousand square feet (𝑥 = 1) is $0.1 million. Therefore, Revenue 𝑅(𝑥) = $0.1𝑥. Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0.1𝑥 − (1.7 + 0.12𝑥 − 0.0001𝑥 2) = −1.7 − 0.02𝑥 + 0.0001𝑥 2

To break even, 𝑃 (𝑥) = 0, so −1.7 − 0.02𝑥 + 0.0001𝑥 2 = 0. This is a quadratic equation with 𝑎 = 0.0001, 𝑏 = −0.02, 𝑐 = −1.7 and solution\\

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 0.02 ± √(−0.02) 2 − 4(0.0001)(−1.7) 0.02 ± 0.03286 = ≈ 264 thousand square feet = 2(0.0001) 0.0002

𝑥=

20. The revenue from one thousand square feet (𝑥 = 1) is $0.2 million. Therefore, Revenue 𝑅(𝑥) = $0.2𝑥. Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0.2𝑥 − (1.7 + 0.14𝑥 − 0.0001𝑥 2) = −1.7 + 0.06𝑥 + 0.0001𝑥 2

To break even, 𝑃 (𝑥) = 0, so −1.7 + 0.06𝑥 + 0.0001𝑥 2 = 0. This is a quadratic equation with 𝑎 = 0.0001, 𝑏 = 0.06, 𝑐 = −1.7 and solution\\

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎 −0.06 ± √(−0.06) 2 − 4(0.0001)(−1.7) −0.06 ± 0.0654 = ≈ 27 thousand square feet = 2(0.0001) 0.0002

𝑥=

21. The hourly profit function is given by Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) (Hourly) cost function: This is a fixed cost of $10,200 only: 𝐶(𝑥) = 10,200 (Hourly) revenue function: This is a variable of $200 per passenger cost only: 𝑅(𝑥) = 200𝑥 Thus, the profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 200𝑥 − 10,200 For the domain of 𝑃 (𝑥), the number of passengers 𝑥 cannot exceed the capacity: 381. Also, 𝑥 cannot be negative. Thus, the domain is given by 0 ≤ 𝑥 ≤ 381, or [0, 381]. For breakeven, 𝑃 (𝑥) = 0 200𝑥 − 10,200 = 0 10,200 200𝑥 = 10,200, or 𝑥 = = 51 200 If 𝑥 is larger than this, then the profit function is positive, and so there should be at least 52 passengers; 𝑥 ≥ 52, for a profit.

Solutions Section 1.2 22. The hourly profit function is given by Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) (Hourly) cost function: This is a fixed cost of $6,500 only: 𝐶(𝑥) = 6,500 (Hourly) revenue function: This is a variable of $100 per passenger cost only: 𝑅(𝑥) = 200𝑥 Thus, the profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 200𝑥 − 6,500 For the domain of 𝑃 (𝑥), the number of passengers 𝑥 cannot exceed the capacity: 150 Also, 𝑥 cannot be negative. Thus, the domain is given by 0 ≤ 𝑥 ≤ 150, or [0, 150]. For breakeven, 𝑃 (𝑥) = 0 200𝑥 − 5,600 = 0 5,600 200𝑥 = 5,600, or 𝑥 = = 28 200 If 𝑥 is larger than this, then the profit function is positive, and so there should be at least 29 passengers; 𝑥 ≥ 29, for a profit. 23. To compute the break-even point, we use the profit function: Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑅(𝑥) = 2𝑥 $2 per unit 𝐶(𝑥) = Variable Cost + Fixed Cost = 40% of Revenue + 6,000 = 0.4(2𝑥) + 6,000 = 0.8𝑥 + 6,000 Thus, 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 2𝑥 − (0.8𝑥 + 6,000) = 1.2𝑥 − 6,000 For breakeven, 𝑃 (𝑥) = 0 1.2𝑥 − 6,000 = 0 6,000 1.2𝑥 = 6,000, so 𝑥 = = 5,000 1.2𝑥 Therefore, 5,000 units should be made to break even. 24. To compute the break-even point, we use the profit function: Profit = Revenue − Cost 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑅(𝑥) = 5𝑥 $5 per unit 𝐶(𝑥) = Variable Cost + Fixed Cost = 30% of Revenue + 7,000 = 0.3(5𝑥) + 7,000 = 1.5𝑥 + 7,000 Thus, 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 5𝑥 − (1.5𝑥 + 7,000) = 3.5𝑥 − 7,000 For breakeven, 𝑃 (𝑥) = 0 3.5𝑥 = 7,000, so 𝑥 = 2,000 units. 25. To compute the break-even point, we use the revenue and cost functions: 𝑅(𝑥) = Selling price × Number of units = 𝑆𝑃 𝑥 𝐶(𝑥) = Variable Cost + Fixed Cost = 𝑉 𝐶𝑥 + 𝐹 𝐶 (Note that "variable cost per unit" is marginal cost.) For breakeven 𝑅(𝑥) = 𝐶(𝑥) 𝑆𝑃 𝑥 = 𝑉 𝐶𝑥 + 𝐹 𝐶 Solve for 𝑥 : 𝑆𝑃 𝑥 − 𝑉 𝐶𝑥 = 𝐹 𝐶 𝐹𝐶 𝑥(𝑆𝑃 − 𝑉 𝐶) = 𝐹 𝐶, so 𝑥 = . 𝑆𝑃 − 𝑉 𝐶

Solutions Section 1.2 26. To compute the break-even point, we use the revenue and cost functions: 𝑅(𝑥) = 𝑆𝑃 𝑥; 𝐶(𝑥) = 𝑉 𝐶𝑥 + 𝐹 𝐶 At breakeven 𝑅(𝐵𝐸) = 𝐶(𝐵𝐸) 𝑆𝑃 (𝐵𝐸) = 𝑉 𝐶(𝐵𝐸) + 𝐹 𝐶 Thus, 𝐹 𝐶 = 𝑆𝑃 (𝐵𝐸) − 𝑉 𝐶(𝐵𝐸) = 𝐵𝐸(𝑆𝑃 − 𝑉 𝐶).

27. Take 𝑥 to be the number of grams of perfume he buys and sells. The profit function is given by Profit = Revenue − Cost: that is, 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) Cost function 𝐶(𝑥) : Fixed costs: 20,000 Cheap perfume @ $20 per g: 20𝑥 Transportation @ $30 per 100 g: 0.3𝑥 Thus the cost function is 𝐶(𝑥) = 20𝑥 + 0.3𝑥 + 20,000 = 20.3𝑥 + 20,000 Revenue function R(x) 𝑅(𝑥) = 600𝑥 $600 per gram Thus, the profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) 𝑃 (𝑥) = 600𝑥 − (20.3𝑥 + 20,000) = 579.7𝑥 − 20,000, with domain 𝑥 ≥ 0. For breakeven, 𝑃 (𝑥) = 0 579.7𝑥 − 20,000 = 0 20,000 579.7𝑥 = 20,000, so 𝑥 = ≈ 34.50 579.7 Thus, he should buy and sell 34.50 grams of perfume per day to break even.

28. Take 𝑥 to be the number of grams of perfume he buys and sells. The profit function is given by Profit = Revenue − Cost Cost function: 𝐶(𝑥) = 400𝑥 + 30𝑥 = 430𝑥 Revenue: 𝑅(𝑥) = 420𝑥 Thus, the profit function is 𝑃 (𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 420𝑥 − 430𝑥 = −10𝑥, with domain 𝑥 ≥ 0. For breakeven, −10𝑥 = 0, so 𝑥 = 0 grams per day; he should shut down the operation. 29. a. To graph the demand function we use technology with the formula 760/x-1 and with xMin = 400 and xMax = 700. Graph:

760 760 − 1 = 0.9, 𝑞(500) = − 1 = 0.52 400 500 So, the change in demand is 0.52 − 0.9 = −0.38 billion units, which means that demand decreases by about 380 million units sold per year. b. 𝑞(400) =

Solutions Section 1.2 c. The value of 𝑞 on the graph decreases by smaller and smaller amounts as we move to the right on the graph, indicating that the demand decreases at a smaller and smaller rate (Choice (D)). 30. a. To graph the demand function we use technology with the formula 350/x+0.5 and with xMin = 100 and xMax = 300. Graph:

350 350 + 0.5 = 4, 𝑞(200) = + 0.5 = 2.25 100 200 So, the change in demand is 2.25 − 4 = −1.75 billion units, which means that demand decreases by about 1.75 billion units sold per year. c. The value of 𝑞 on the graph increases by larger and larger amounts as we move to the left on the graph, indicating that the demand increases at a greater and greater rate (Choice (A)). b. 𝑞(100) =

31. The price at which there is neither a shortage nor surplus is the equilibrium price, which occurs when demand = supply: −3𝑝 + 700 = 2𝑝 − 500 5𝑝 = 1200 𝑝 = $240 per skateboard 32. The price at which there is neither a shortage nor surplus is the equilibrium price, which occurs when demand = supply: −5𝑝 + 50 = 3𝑝 − 30 8𝑝 = 80 𝑝 = $10 per skateboard 33. a. The equilibrium price occurs when demand = supply: −𝑝 + 57 = 4𝑝 − 136 5𝑝 = 193 𝑝 = 38.6, or $38,600 per vehicle. b. Since $38,400 is below the equilibrium price, there would be a shortage at that price. To calculate it, compute demand and supply: Demand: 𝑞 = −38.4 + 57 = 18.6 million vehicles Supply: 𝑞 = 4(38.4) − 136 = 17.6 million vehicles Shortage = Demand − Supply = 18.6 − 17.6 = 1 million vehicles 34. a. The equilibrium price occurs when demand = supply: −0.15𝑝 + 23 = 3.85𝑝 − 131 4𝑝 = 154 𝑝 = 38.5, or $38,500 per vehicle. b. Since $38,600 is above the equilibrium price, there would be a surplus at that price. To calculate it, compute demand and supply: Demand: 𝑞 = −0.15(38.6) + 23 = 17.21 million vehicles

Solutions Section 1.2 Supply: 𝑞 = 3.85(38.6) − 131 = 17.61 million vehicles Surplus = Supply − Demand = 17.61 − 17.21 = 0.4 million vehicles 35. a. For equilibrium, Demand = Supply: 760 − 1 = 0.00304𝑝 − 1 𝑝 760 = 0.0.00304𝑝 𝑝 Cross-multiply: 760 0.00304𝑝 2 = 760 ⇒ 𝑝 2 = = 250,000 0.00304 So 𝑝 = √250,000 = $500. Thus, the equilibrium price is $500, and the equilibrium demand (or supply) is 760∕500 − 1 = 0.52 billion phones b. Graph:

Technology formulas: Demand: y = 760/x-1 Supply: y = 0.00304x-1 The graphs cross at (500, 0.52) confirming the calclation in part (a). c. Since $400 is below the equilibrium price, there would be a shortage at that price. To calculate it, compute demand and supply: 760 Demand: 𝑞 = − 1 = 0.9 billion phones 400 Supply: 0.00304(400) − 1 = 0.216 billion phones Shortage = Demand − Supply ≈ 0.9 − 0.216 = 0.684 billion phones, or 684 million phones. 36. a. For equilibrium, Demand = Supply: 350 + 0.5 = 0.0056𝑝 + 0.5 𝑝 350 = 0.0056𝑝 𝑝 Cross-multiply: 350 0.0056𝑝 2 = 350 ⇒ 𝑝 2 = = 62,500 0.0056 So 𝑝 = √62,500 = $250. Thus, the equilibrium price is $250, and the equilibrium demand (or supply) is 350∕250 + 0.5 = 1.9 billion phones

Solutions Section 1.2 b. Graph:

Technology formulas: Demand: y = 350/x+0.5 Supply: y = 0.0056x+0.5 The graphs cross at (250, 1.9) confirming the calclation in part (a). c. Since $200 is below the equilibrium price, there would be a shortage at that price. To calculate it, compute demand and supply: 350 Demand: 𝑞 = + 0.5 = 2.25 billion phones 200 Supply: 0.0056(200) + 0.5 = 1.62 billion phones Shortage = Demand − Supply ≈ 2.25 − 1.62 = 0.63 billion phones, or 630 million phones. 37. 𝐶(𝑞) = 2,000 + 100𝑞 2 a. 𝐶(10) = 2,000 + 100(10) 2 = 2,000 + 10,000 = $12,000 b. 𝑁 = 𝐶 − 𝑆, so 𝑁(𝑞) = 𝐶(𝑞) − 𝑆(𝑞) = 2,000 + 100𝑞 2 − 500𝑞

This is the cost of removing 𝑞 lb of PCPs per day after the subsidy is taken into account. c. 𝑁(20) = 2,000 + 100(20) 2 − 500(20) = 2,000 + 40,000 − 10,000 = $32,000

38. 𝐶(𝑞) = 1,000 + 100√𝑞 a. 𝐶(100) = 1,000 + 100√100 = 1,000 + 100(10) = $2,000 b. 𝑁 = 𝐶 − 𝑆, so 𝑁(𝑞) = 𝐶(𝑞) − 𝑆(𝑞) = 1,000 + 100√𝑞 − 200𝑞 This is the cost for dental coverage to the company if it has 𝑞 employees, after the subsidy is taken into account. c. 𝑁(100) = 1,000 + 100√100 − 200(100) = 1000 + 100(10) − 20,000 = −$18,000 The company makes $18,000 from the government for dental coverage if it employs 100 people. 39. The technology formulas are: (A) -0.2*t^2+t+16 (B) 0.2*t^2+t+16 (C) t+16 The following table shows the values predicted by the three models:

𝑡

Solutions Section 1.2 0

(A)

17.2

16.8

14.8

13.2

(B)

18.8

23.2

29.2

32.8

(C)

𝑆(𝑡)

As shown in the table, the values predicted by model (B) are much closer to the observed values 𝑆(𝑡) than those predicted by the other models. b. Since 1998 corresponds to 𝑡 = 8, 𝑆(𝑡) = 0.2𝑡 2 + 𝑡 + 16 𝑆(8) = 0.2(8) 2 + 8 + 16 = 36.8 So the spending on corrections in 1998 was predicted to be approximately $37 billion. 40. The technology formulas are: (A) 16+2*t (B) 16+t+0.5*t^2 (C) 16+t-0.5*t^2 The following table shows the values predicted by the three models: 𝑡

(A)

(B)

47.5

(C)

-1.5

𝑆(𝑡)

As shown in the table, the values predicted by model (A) are much closer to the observed values 𝑆(𝑡) than those predicted by the other models. b. Since 1998 corresponds to 𝑡 = 8, 𝑆(𝑡) = 16 + 2𝑡 𝑆(8) = 16 + 2(8) = 32 So the spending on corrections in 1998 was predicted to be $32 billion. 41. The technology formulas are: (A) 0.005*x+20.75 (B) 0.01*x+20+25/x (C) 0.0005*x^2-0.07*x+23.25 (The stars are optional for some technologies like graphing calculators and the Evaluator and Grapher App) Here is result from the Evaluator and Grapher App: Functions box (Cartesian mode):

Evaluator box (Set to 2 decimal places):

Model (C) fits the data perfectly to two decimal places—more closely than any of the other models.

Solutions Section 1.2 b. Graph of model (C):

0.0005*x^2-0.07*x+23.25

The lowest point on the graph occurs at 𝑥 = 70 with a 𝑦-coordinate of 20.8. Thus, the lowest cost per shirt is $20.80, which the team can obtain by buying 70 shirts. 42. The technology formulas are: (A) 0.05*x+20.75 (B) 0.1*x+20+25/x (C) 0.0008*x^2-0.07*x+23.25 (The stars are optional for some technologies like graphing calculators and the Evaluator and Grapher App) Here is result from the Evaluator and Grapher App: Functions box (Cartesian mode):

Evaluator box (Set to 2 decimal places):

Model (B) fits the data perfectly to two decimal places—more closely than any of the other models. b. Graph of model (B):

0.1*x+20+25/x

The lowest point on the graph occurs at 𝑥 = 16 with a y-coordinate of 23.1625. Thus, the lowest cost per hat is $23.16, which the team can obtain by buying 16 hats. 43. Here are the technology formulas as entered in the online Function Evaluator and Grapher at the Web Site: (A) 0.075*0.83^x (B) 0.24/(x+3) (C) 0.00054*x^2-0.012*x+0.075 (D) -0.006*x+0.065 The following graph shows all four curves together with the plotted points (entered as shown in the margin technology note with Example 5).

Solutions Section 1.2

As shown in the graph, the values predicted by models (A) and (C) are much closer to the observed values than those predicted by the other models. b. Since 2030 corresponds to 𝑡 = 20, Model (A): 𝑐(20) = 0.075 * 0.83 20 ≈ $0.00181 Model (C): 𝑐(20) = 0.00054(20 2) − 0.012(20) + 0.075 ≈ $0.0510 So model (A) gives the lower price: approximately $0.0018 per GB. 44. Here are the technology formulas as entered in the online Function Evaluator and Grapher at the Web Site: (A) -0.038+0.8/(x+7) (B) 0.075*1.21^(-x) (C) 0.00034*(x-14.5)^2 (D) -0.022+0.8/(x+9.32) The following graph shows all three curves together with the plotted points (entered as shown in the margin technology note with Example 5).

As shown in the graph, the values predicted by models (A), (B), and (C) are much closer to the observed values than those predicted by (D). b. Since 2030 corresponds to 𝑡 = 20, 0.8 Model (A): 𝑐(20) = −0.038 + ≈ −$0.084 (unreasonable as prices cannot be negative) 20 + 7 Model (B): 𝑐(20) = 0.075(1.21 −20) ≈ $0.0017 (reasonable given the current trends) Model (C): 𝑐(20) = 0.00034(20 − 14.5) 2 ≈ $0.010 (unreasonable that the price would be the same again in 10 years) 45. A plot of the given points gives a straight line (Option (A)). Options (B) and (C) give curves, so (A) is the best choice.

Solutions Section 1.2 46. Plotting the three data points suggests a concave down curve, suggesting that a linear model may not be the best fit. An exponential model 𝑝(𝑡) = 𝐴𝑏 𝑡 would result in a concave up curve, regardless of whether 𝑏 is larger than 1 or less than 1. This leaves a quadratic model as the only possible choice. In fact, a quadratic can always be found that passes through any three points not on the same straight line with different 𝑥-coordinates. Therefore, a quadratic model would give an exact fit. 47. A plot of the given data suggests a concave-down curve that becomes steeper downward as the price 𝑝 increases, suggesting Model (D). Model (A) would predict increasing demand with increasing price, Model (B) would correspond to a descending curve that becomes less steep as 𝑝 increases (a concave up curve), and Model (C) would give a concave-up parabola. 48. Model (B) is the best choice; Model (A) would predict increasing demand with increasing price, Model (D) would correspond to a concave down parabola, and Model (C), would predict demand that, rather than flattening out as the price increases, would begin to climb again. 49. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = $1,000, 𝑟 = 6∕100 = 0.06, 𝑛 = 4, 𝑡 = 4. 0.06 4×4 4 ) = 1,000(1.015) 16 ≈ $1,268.99

𝐴(4) = 1,000(1 +

50. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = $10,000, 𝑟 = 2∕100 = 0.02, 𝑛 = 4, 𝑡 = 5 0.02 4×5 4 ) = 1,000(1.005) 20 ≈ $11,048.96

𝐴(5) = 1,000(1 +

51. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = 5,000, 𝑟 = 0.15∕100 = 0.0015, and 𝑛 = 12. We get the model 𝐴(𝑡) = 5,000(1 + 0.0015∕12) 12𝑡 In June 2028 (𝑡 = 7), the deposit would be worth 5,000(1 + 0.0015∕12) 12(7) ≈ $5,053. 52. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = 4,000, 𝑟 = 0.0120, and 𝑛 = 365. We get the model 𝐴(𝑡) = 4,000(1 + 0.0120∕365) 365𝑡 In March 2029 (𝑡 = 8), the deposit would be worth 4,000(1 + 0.0120∕365) 365(8) ≈ $4,403.

53. 𝑃 (0) = 200, 𝑃 (1) = 230, 𝑃 (2) = 260, ... and so on. Thus, the population is increasing by 30 per year. 54. 𝐵(0) = 5000, 𝐵(1) = 4800, 𝐵(2) = 4600, ... and so on. Thus, the balance is decreasing by $200 per day.

Solutions Section 1.2 55. Curve fitting. The model is based on fitting a curve to a given set of observed data. 56. Analytical. The model is obtained by analyzing the situation being modeled.

57. The given model is 𝑐(𝑡) = 4 − 0.2𝑡. This tells us that 𝑐 is $4 at time 𝑡 = 0 (January) and is decreasing by $0.20 per month. So, the cost of downloading a movie was $4 in January and is decreasing by 20¢ per month. 58. The given model is 𝑐(𝑡) = 4 − 0.2𝑡 and therefore passes through the points (𝑡, 𝑐) = (0, 4) and (1, 3.8). So, the cost of downloading a movie was $4 in January and $3.80 in February. 59. In a linear cost function, the variable cost is 𝑥 times the marginal cost.

60. In a linear cost function, the marginal cost is the additional (or incremental) cost per item. 61. Yes, as long as the supply is going up at a faster rate, as illustrated by the following graph:

62. There would be a shortage at any given price. Therefore, consumers would be willing to pay more for a scarce commodity and sellers would naturally oblige by charging more, resulting in an upward spiral of prices for the commodity.

63. Extrapolate both models and choose the one that gives the most reasonable predictions.

64. No; as long as 𝑎 is negative, the value of 𝑠(𝑡) for large 𝑡 will be negative, making the model unreasonable for large value of 𝑡.

65. The value of 𝑓 − 𝑔 at 𝑥 is 𝑓(𝑥) − 𝑔(𝑥). Since 𝑓(𝑥) ≥ 𝑔(𝑥) for every 𝑥, it follows that 𝑓(𝑥) − 𝑔(𝑥) ≥ 0 for every 𝑥. 66. The value of

𝑓 𝑓(𝑥) 𝑓(𝑥) at 𝑥 is . Since 𝑓(𝑥) > 𝑔(𝑥) > 0 for every 𝑥, it follows that > 1 for every 𝑥. 𝑔 𝑔(𝑥) 𝑔(𝑥)

𝑓 𝑓(𝑥) 𝑓 at 𝑥 are ratios , it follows that the units of measurement of are units of 𝑓 𝑔 𝑔(𝑥) 𝑔 per unit of 𝑔; that is, books per person. 67. Since the values of

68. Write 𝑓(𝑥) = 𝑚𝑥 + 𝑏 and 𝑔(𝑥) = 𝑛𝑥 + 𝑐. Then 𝑓(𝑥) − 𝑔(𝑥) = 𝑚𝑥 + 𝑏 − (𝑛𝑥 + 𝑐) = (𝑚 − 𝑛)𝑥 + (𝑏 − 𝑐), also a linear function.

Solutions Section 1.3 Section 1.3 1.

𝑥 𝑦

−1 5

0 8

We calculate the slope 𝑚 first. The first two points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 0 − (−1) = 1 Δ𝑦 = 8 − 5 = 3 This gives a slope of Δ𝑦 3 𝑚= = = 3. Δ𝑥 1 Now look at the second and third points: The change in 𝑥 is again Δ𝑥 = 1 − 0 = 1 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 Δ𝑦 = 3(1) = 3 This means that the missing value of 𝑦 is 8 + Δ𝑦 = 8 + 3 = 11. 2.

𝑥 𝑦

−1

−3

We calculate the slope 𝑚 first. The first two points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 0 − (−1) = 1 Δ𝑦 = −3 − (−1) = −2 This gives a slope of Δ𝑦 −2 𝑚= = = −2. Δ𝑥 1 Now look at the second and third points: The change in 𝑥 is again Δ𝑥 = 1 − 0 = 1 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 Δ𝑦 = −2(1) = −2 This means that the missing value of 𝑦 is −3 + Δ𝑦 = −3 + (−2) = −5. 3.

𝑥 𝑦

−1

−2

We calculate the slope 𝑚 first. The first two points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 3 − 2 = 1 Δ𝑦 = −2 − (−1) = −1 This gives a slope of Δ𝑦 −1 𝑚= = = −1. Δ𝑥 1 Now look at the second and third points: The change in 𝑥 is Δ𝑥 = 5 − 3 = 2 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 Δ𝑦 = (−1)(2) = −2 This means that the missing value of 𝑦 is −2 + Δ𝑦 = −2 + (−2) = −4.

𝑥

𝑦

−1

−2

𝑥

−2

Solutions Section 1.3

We calculate the slope 𝑚 first. The first two points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 4 − 2 = 2 Δ𝑦 = −2 − (−1) = −1 This gives a slope of Δ𝑦 −1 1 𝑚= = =− . Δ𝑥 2 2 Now look at the second and third points: The change in 𝑥 is Δ𝑥 = 5 − 4 = 1 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 1 1 Δ𝑦 = !− "(1) = − 2 2 This means that the missing value of 𝑦 is 1 5 −2 + Δ𝑦 = −2 + !− " = − or −2.5. 2 2 5.

𝑦

We calculate the slope 𝑚 first. The first and third points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 2 − (−2) = 4 Δ𝑦 = 10 − 4 = 6 This gives a slope of Δ𝑦 6 3 𝑚= = = . Δ𝑥 4 2 Now look at the first and second points: The change in 𝑥 is Δ𝑥 = 0 − (−2) = 2 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 3 Δ𝑦 = !− "(2) = 3 2 This means that the missing value of 𝑦 is 4 + Δ𝑦 = 4 + 3 = 7. 6.

𝑥 𝑦

−1

−5

We calculate the slope 𝑚 first. The first and third points shown give changes in 𝑥 and 𝑦 of Δ𝑥 = 6 − 0 = 6 Δ𝑦 = −5 − (−1) = −4 This gives a slope of Δ𝑦 −4 2 𝑚= = =− . Δ𝑥 6 3 Now look at the first and second points: The change in 𝑥 is Δ𝑥 = 3 − 0 = 3 and so Δ𝑦 must be given by the formula Δ𝑦 = 𝑚Δ𝑥 2 Δ𝑦 = !− "(3) = −2 3

Solutions Section 1.3 This means that the missing value of 𝑦 is −1 + Δ𝑦 = −1 + (−2) = −3 7. From the table, 𝑏 = 𝑓(0) = −2. The slope (using the first two points) is 𝑦 − 𝑦1 −2 − (−1) −1 1 𝑚= 2 = = =− . 𝑥2 − 𝑥1 0 − (−2) 2 2 Thus, the linear equation is 1 𝑥 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = − 𝑥 − 2, or 𝑓(𝑥) = − − 2. 2 2 8. From the table, 𝑏 = 𝑓(0) = 3. The slope (using the first two points) is 𝑦 − 𝑦1 2−1 1 𝑚= 2 = = 𝑥2 − 𝑥1 (−3) − (−6) 3 Thus, the linear equation is 1 𝑥 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = 𝑥 + 3, or 𝑓(𝑥) = + 3. 3 3

9. The slope (using the first two points) is 𝑦 − 𝑦1 −2 − (−1) −1 𝑚= 2 = = = −1. 𝑥2 − 𝑥1 −3 − (−4) 1 To obtain 𝑓(0) = 𝑏, use the formula for 𝑏 : 𝑓(0) = 𝑏 = 𝑦1 − 𝑚𝑥1 = −1 − (−1)(−4) = −5 This gives 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = −𝑥 − 5. 10. The slope (using the first two points) is 𝑦 − 𝑦1 6−4 2 𝑚= 2 = = =2 𝑥2 − 𝑥1 2 − 1 1 To obtain 𝑓(0) = 𝑏, use the formula for 𝑏 : 𝑓(0) = 𝑏 = 𝑦1 − 𝑚𝑥1 = 4 − (2)(1) = 2 This gives 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = 2𝑥 + 2.

Using the point (𝑥1 , 𝑦1 ) = (−4, −1)

Using the point (𝑥1 , 𝑦1 ) = (1, 4)

11. In the table, 𝑥 increases in steps of 1 and 𝑓 increases in steps of 4, showing that 𝑓 is linear with slope Δ𝑦 4 𝑚= = =4 Δ𝑥 1 and intercept 𝑏 = 𝑓(0) = 6 giving 𝑓(𝑥) = 𝑚𝑥 + 𝑏 = 4𝑥 + 6. The function 𝑔 does not increase in equal steps, so 𝑔 is not linear.

12. In the table, 𝑥 increases in steps of 10 and 𝑔 increases in steps of 5, showing that 𝑔 is linear with slope Δ𝑦 5 1 𝑚= = = Δ𝑥 10 2 and intercept 𝑏 = 𝑔(0) = −4 giving 1 𝑔(𝑥) = 𝑚𝑥 + 𝑏 = 𝑥 − 4. 2

Solutions Section 1.3 The function 𝑓 does not increase in equal steps, so 𝑓 is not linear.

13. In the first three points listed in the table, 𝑥 increases in steps of 3, but 𝑓 does not increase in equal steps, whereas 𝑔 increases in steps of 6. Thus, based on the first three points, only 𝑔 could possibly be linear, with slope Δ𝑦 6 𝑚= = =2 Δ𝑥 3 and intercept 𝑏 = 𝑔(0) = −1 giving 𝑔(𝑥) = 𝑚𝑥 + 𝑏 = 2𝑥 − 1. We can now check that the remaining points in the table fit the formula 𝑔(𝑥) = 2𝑥 − 1, showing that 𝑔 is indeed linear.

14. In the first and last pairs of points listed in the table, 𝑥 increases in steps of 3, but 𝑓 does not increase in equal steps, whereas 𝑔 increases in steps of 9. Thus, based on those points, only 𝑔 could possibly be linear, with slope Δ𝑦 9 𝑚= = =3 Δ𝑥 3 and intercept 𝑏 = 𝑔(0) = −1 giving 𝑔(𝑥) = 𝑚𝑥 + 𝑏 = 3𝑥 − 1. We can now check that the remaining points in the table fit the formula 𝑔(𝑥) = 3𝑥 − 1, showing that 𝑔 is indeed linear. 15. Slope = coefficient of 𝑥 = − 16. Slope = coefficient of 𝑥 = 17. Slope = coefficient of 𝑥 =

2 3 1 6

18. Write the equation as 𝑦 = − Slope = coefficient of 𝑥 = −

2 3

3 2

2𝑥 1 + 3 3

19. If we solve for 𝑥 we find that the given equation represents the vertical line 𝑥 = −1∕3, and so its slope is infinite (undefined). 20. 8𝑥 − 2𝑦 = 1. Solving for 𝑦 : 2𝑦 = 8𝑥 − 1 1 𝑦 = 4𝑥 − 2 Slope = coefficient of 𝑥 = 4 21. 3𝑦 + 1 = 0. Solving for 𝑦 : 3𝑦 = −1 1 𝑦=− 3

Slope = coefficient of 𝑥 = 0

Solutions Section 1.3

22. If we solve for 𝑥 we find that the given equation represents the vertical line 𝑥 = −3∕2, and so its slope is infinite (undefined). 23. 4𝑥 + 3𝑦 = 7. Solve for 𝑦 : 3𝑦 = −4𝑥 + 7 4 7 𝑦=− 𝑥+ 3 3 4 Slope = coefficient of 𝑥 = − 3 24. 2𝑦 + 3 = 0. Solve for 𝑦 : 2𝑦 = −3 3 𝑦=− 2 Slope = coefficient of 𝑥 = 0 25. 𝑦 = 2𝑥 − 1 𝑦-intercept = −1, slope = 2

26. 𝑦 = 𝑥 − 3 𝑦-intercept = −3, slope = 1

27. 𝑦-intercept = 2, slope = −

1 28. 𝑦 = − 𝑥 + 3 2

𝑦-intercept = 3, slope = −

29. 𝑦 +

2 3

1 2

1 1 𝑥 = −4. Solve for 𝑦 to obtain 𝑦 = − 𝑥 − 4 4 4

y-intercept = −4, slope = −

1 4

Solutions Section 1.3

1 1 𝑥 = −2. Solve for 𝑦 to obtain 𝑦 = 𝑥 − 2 4 4 1 𝑦-intercept = −2, slope = 4 30. 𝑦 −

31. 7𝑥 − 2𝑦 = 7. Solve for 𝑦 : 7 7 −2𝑦 = −7𝑥 + 7, so 𝑦 = 𝑥 − 2 2 7 7 𝑦-intercept = − = −3.5, slope = = 3.5 2 2

32. 2𝑥 − 3𝑦 = 1. Solve for 𝑦 : 2 1 −3𝑦 = −2𝑥 + 1, so 𝑦 = 𝑥 − 3 3 1 2 𝑦-intercept = − , slope = 3 3

33. 3𝑥 = 8. Solve for 𝑥 to obtain 𝑥 = The graph is a vertical line:

8 . 3

Solutions Section 1.3 7 34. 2𝑥 = −7. Solve for 𝑥 to obtain 𝑥 = − = −3.5. 2 The graph is a vertical line:

35. 6𝑦 = 9. Solve for 𝑦 to obtain 𝑦 =

9 3 = = 1.5 6 2

36. 3𝑦 = 4. Solve for 𝑦 to obtain 𝑦 =

4 3

𝑦-intercept =

3 = 1.5, slope = 0. The graph is a horizontal line: 2

4 𝑦-intercept = , slope = 0. The graph is a horizontal line: 3

37. 2𝑥 = 3𝑦. Solve for 𝑦 to obtain 𝑦 = 𝑦-intercept = 0, slope =

2 3

2 𝑥 3

3 38. 3𝑥 = −2𝑦. Solve for 𝑦 to obtain 𝑦 = − 𝑥 2 3 𝑦-intercept = 0, slope = − 2

39. (0, 0) and (1, 2) 𝑦 − 𝑦1 2−0 𝑚= 2 = =2 𝑥2 − 𝑥1 1 − 0

40. (0, 0) and (−1, 2) 𝑦 − 𝑦1 2−0 𝑚= 2 = = −2 𝑥2 − 𝑥1 −1 − 0

41. (−1, −2) and (0, 0) 𝑦 − 𝑦1 0 − (−2) 𝑚= 2 = =2 𝑥2 − 𝑥1 0 − (−1)

42. (2, 1) and (0, 0) 𝑦 − 𝑦1 0−1 1 𝑚= 2 = = 𝑥2 − 𝑥1 0 − 2 2

Solutions Section 1.3

43. (4, 3) and (5, 1) 𝑦 − 𝑦1 1 − 3 −2 𝑚= 2 = = = −2 𝑥2 − 𝑥1 5 − 4 1

44. (4, 3) and (4, 1) 𝑦 − 𝑦1 1−3 𝑚= 2 Undefined = 𝑥2 − 𝑥1 4 − 4

45. (1, −1) and (1, −2) 𝑦 − 𝑦1 −2 − (−1) 𝑚= 2 Undefined = 𝑥2 − 𝑥1 1−1

46. (−2, 2) and (−1, −1) 𝑦 − 𝑦1 −1 − 2 −3 𝑚= 2 = = = −3 𝑥2 − 𝑥1 −1 − (−2) 1

47. (2, 3.5) and (4, 6.5) 𝑦 − 𝑦1 6.5 − 3.5 3 𝑚= 2 = = = 1.5 𝑥2 − 𝑥1 4−2 2

48. (10, −3.5) and (0, −1.5) 𝑦 − 𝑦1 −1.5 − (−3.5) 2 𝑚= 2 = = = −0.2 𝑥2 − 𝑥1 0 − 10 −10

49. (300, 20.2) and (400, 11.2) 𝑦 − 𝑦1 11.2 − 20.2 −9 𝑚= 2 = = = −0.09 𝑥2 − 𝑥1 400 − 300 100

50. (1, −20.2) and (2, 3.2) 𝑦 − 𝑦1 3.2 − (−20.2) 23.4 𝑚= 2 = = = 23.4 𝑥2 − 𝑥1 2−1 1

1 3 51. (0, 1) and !− , " 2 4 𝑦2 − 𝑦1 3∕4 − 1 −1∕4 2 1 𝑚= = = = = 𝑥2 − 𝑥1 −1∕2 − 0 −1∕2 4 2

1 1 3 52. ! , 1" and !− , " 2 2 4 𝑦2 − 𝑦1 3∕4 − 1 −1∕4 1 𝑚= = = = 𝑥2 − 𝑥1 −1∕2 − 1∕2 4 −1

53. (𝑎, 𝑏) and (𝑐, 𝑑) (𝑎 ≠ 𝑐) 𝑦 − 𝑦1 𝑑−𝑏 𝑚= 2 = 𝑥2 − 𝑥1 𝑐 − 𝑎

54. (𝑎, 𝑏) and (𝑐, 𝑏) (𝑎 ≠ 𝑐) 𝑦 − 𝑦1 𝑏−𝑏 𝑚= 2 = =0 𝑥2 − 𝑥1 𝑐 − 𝑎

55. (𝑎, 𝑏) and (𝑎, 𝑑) (𝑏 ≠ 𝑑) 𝑦 − 𝑦1 𝑑−𝑏 𝑚= 2 \ \ Undefined = 𝑥2 − 𝑥1 𝑎 − 𝑎

56. (𝑎, 𝑏) and (−𝑎, −𝑏) (𝑎 ≠ 0) 𝑦 − 𝑦1 −𝑏 − 𝑏 −2𝑏 𝑏 𝑚= 2 = = = 𝑥2 − 𝑥1 −𝑎 − 𝑎 −2𝑎 𝑎

57. (−𝑎, 𝑏) and (𝑎, −𝑏) (𝑎 ≠ 0) 𝑦 − 𝑦1 𝑏 −𝑏 − 𝑏 −2𝑏 𝑚= 2 = = =− 𝑥2 − 𝑥1 𝑎 − (−𝑎) 2𝑎 𝑎

58. (𝑎, 𝑏) and (𝑏, 𝑎) (𝑎 ≠ 𝑏) 𝑦 − 𝑦1 𝑎−𝑏 𝑚= 2 = = −1 𝑥2 − 𝑥1 𝑏 − 𝑎

59. a. 𝑚 =

Δ𝑦 1 = =1 Δ𝑥 1

Solutions Section 1.3

60. a. 𝑚 =

Δ𝑦 −1 = = −1 Δ𝑥 1

b. 𝑚 =

Δ𝑦 1 = Δ𝑥 2

b. 𝑚 =

Δ𝑦 2 = =2 Δ𝑥 1

c. 𝑚 =

Δ𝑦 0 = =0 Δ𝑥 1

c. 𝑚 =

Δ𝑦 0 = =0 Δ𝑥 1

d. 𝑚 =

Δ𝑦 3 = =3 Δ𝑥 1

d. 𝑚 =

Δ𝑦 1 = =1 Δ𝑥 1

e. 𝑚 =

Δ𝑦 −2 1 = =− Δ𝑥 4 2

e. 𝑚 =

f. 𝑚 =

Δ𝑦 −1 1 = =− Δ𝑥 3 3 Δ𝑦 −1 = = −1 Δ𝑥 1

f. Vertical line; undefined slope

g. 𝑚 =

Δ𝑦 3 = =3 Δ𝑥 1

g. Vertical line; undefined slope

h. 𝑚 =

Δ𝑦 −1 1 = =− Δ𝑥 4 4

h. 𝑚 =

Δ𝑦 −1 1 = =− Δ𝑥 3 3

i. 𝑚 =

Δ𝑦 −2 = = −2 Δ𝑥 1

i. 𝑚 =

Δ𝑦 −2 = = −2 Δ𝑥 1

61. Through (1, 3) with slope 3 Point: (1, 3) Slope: 𝑚 = 3 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 3 − 3(1) = 0 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 3𝑥 + 0, or 𝑦 = 3𝑥 Solutions Section 1.3

62. Through (2, 1) with slope 2 Point: (2, 1) Slope: 𝑚 = 2 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 1 − 2(2) = −3 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 2𝑥 − 3 3 1 63. Through !1, − " with slope 4 4 3 1 Point: !1, − " Slope: 𝑚 = 4 4

Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 1 1 64. Through !0, − " with slope 3 3 1 1 Point: !0, − " Slope: 𝑚 = 3 3

3 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = − − (1) = −1 4 4

1 𝑥−1 4

1 1 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = − − (0) = − 3 3 3 1 1 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 − 3 3 65. Through (20, −3.5) and increasing at a rate of 10 units of 𝑦 per unit of 𝑥 Δ𝑦 10 Point: (20, −3.5) Slope: 𝑚 = = = 10 Δ𝑥 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −3.5 − (10)(20) = −3.5 − 200 = −203.5 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 10𝑥 − 203.5

66. Through (3.5, −10) and increasing at a rate of 1 unit of 𝑦 per 2 units of 𝑥 Δ𝑦 1 Point: (3.5, −10) Slope: 𝑚 = = = 0.5 Δ𝑥 2 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −10 − (0.5)(3.5) = −11.75 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 0.5𝑥 − 11.75 67. Through (2, −4) and (1, 1)

𝑦2 − 𝑦1 5 = = −5 𝑥2 − 𝑥1 −1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −4 − (−5)(2) = 6 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −5𝑥 + 6 Point:(2, −4)

Slope: 𝑚 =

68. Through (1, −4) and (−1, −1) 𝑦 − 𝑦1 3 Point: (1, −4) Slope: 𝑚 = 2 = = −1.5 𝑥2 − 𝑥1 −2 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −4 − (−1.5)(1) = −2.5 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −1.5𝑥 − 2.5

69. Through (1, −0.75) and (0.5, 0.75) 𝑦 − 𝑦1 0.75 − (−0.75) 1.5 Point: (1, −0.75) Slope: 𝑚 = 2 = = = −3 𝑥2 − 𝑥1 0.5 − 1 −0.5 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −0.75 − (−3)(1) = −0.75 + 3 = 2.25 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −3𝑥 + 2.25

70. Through (0.5, −0.75) and (1, −3.75) 𝑦 − 𝑦1 −3.75 − (−0.75) −3 Point: (0.5, −0.75) Slope: 𝑚 = 2 = = = −6 𝑥2 − 𝑥1 1 − 0.5 0.5 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −0.75 − (−6)(0.5) = −0.75 + 3 = 2.25 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −6𝑥 + 2.25 Solutions Section 1.3

71. Through (6, 6) and parallel to the line 𝑥 + 𝑦 = 4 Point: (6, 6) Slope: Same as slope of 𝑥 + 𝑦 = 4. To find the slope, solve for 𝑦, getting 𝑦 = −𝑥 + 4. Thus, 𝑚 = −1. Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 6 − (−1)(6) = 6 + 6 = 12 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −𝑥 + 12 72. Through (1∕3, −1) and parallel to the line 3𝑥 − 4𝑦 = 8 Point: (1∕3, −1)

Slope: Same as slope of 3𝑥 − 4𝑦 = 8. To find the slope, solve for 𝑦, getting 𝑦 = 3 1 1 5 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = (−1) − ! "! " = −1 − = − 4 3 4 4 3 5 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 − 4 4

3 3 𝑥 − 2. Thus, 𝑚 = . 4 4

73. Through (0.5, 5) and parallel to the line 4𝑥 − 2𝑦 = 11 Point: (0.5, 5)

Slope: Same as slope of 4𝑥 − 2𝑦 = 11. To find the slope, solve for 𝑦, getting 𝑦 = 2𝑥 − Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 5 − (2)(0.5) = 5 − 1 = 4 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 2𝑥 + 4

74. Through (1/3, 0) and parallel to the line 6𝑥 − 2𝑦 = 11 Point: (1∕3, 0)

Slope: Same as slope of 6𝑥 − 2𝑦 = 11. To find the slope, solve for 𝑦, getting 𝑦 = 3𝑥 − 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − (3)! " = −1 3 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 3𝑥 − 1 75. Through (0, 0) and (𝑝, 𝑞)

𝑦2 − 𝑦1 𝑞−0 𝑞 = = 𝑥2 − 𝑥1 𝑝 − 0 𝑝 𝑞 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − (0) = 0 𝑝 𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 𝑝 Point: (0, 0)

Slope: 𝑚 =

76. Through (𝑝, 𝑞) parallel to 𝑦 = 𝑟𝑥 + 𝑠 Point: (𝑝, 𝑞) Slope: Since the line has the same slope as 𝑦 = 𝑟𝑥 + 𝑠, 𝑚 = 𝑟 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − 𝑟𝑝 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑟𝑥 + 𝑞 − 𝑟𝑝 or 𝑦 = 𝑟(𝑥 − 𝑝) + 𝑞 77. Through (𝑝, 𝑞) and (𝑟, 𝑞) (𝑝 ≠ 𝑟) 𝑦 − 𝑦1 𝑞−𝑞 Point: (𝑝, 𝑞) Slope: 𝑚 = 2 = =0 𝑥2 − 𝑥1 𝑟 − 𝑝

11 . Thus, 𝑚 = 2. 2

Solutions Section 1.3 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − (0)𝑝 = 𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏; that is, 𝑦 = 𝑞 78. Through (𝑝, 𝑞) and (−𝑝, −𝑞) (𝑝 ≠ 0) 𝑦 − 𝑦1 −𝑞 − 𝑞 −2𝑞 𝑞 Point: (𝑝, 𝑞) Slope: 𝑚 = 2 = = = 𝑥2 − 𝑥1 −𝑝 − 𝑝 −2𝑝 𝑝 𝑞 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − 𝑝 = 𝑞 − 𝑞 = 0 𝑝 𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 𝑝

79. Through (−𝑝, 𝑞) and (𝑝, −𝑞) (𝑝 ≠ 0) 𝑦 − 𝑦1 𝑞 −𝑞 − 𝑞 −2𝑞 Point: (−𝑝, 𝑞) Slope: 𝑚 = 2 = = =− 𝑥2 − 𝑥1 𝑝 − (−𝑝) 2𝑝 𝑝 𝑞 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − !− "(−𝑝) = 𝑞 − 𝑞 = 0 𝑝 𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥 𝑝

80. Through (𝑝, 𝑞) and (𝑟, 𝑠) (𝑝 ≠ 𝑟) 𝑦 − 𝑦1 𝑠−𝑞 Point: (𝑝, 𝑞) Slope: 𝑚 = 2 = 𝑥2 − 𝑥1 𝑟 − 𝑝 𝑠−𝑞 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 𝑞 − ! "𝑝 𝑟−𝑝 𝑠−𝑞 𝑠−𝑞 𝑠−𝑞 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = ! "𝑥 + 𝑞 − ! "𝑝, or 𝑦 = ! "(𝑥 − 𝑝) + 𝑞 𝑟−𝑝 𝑟−𝑝 𝑟−𝑝 81. We are given two points on the graph of the linear cost function: (100, 10,500) and (120, 11,000) (𝑥 is the number of bicycles, and the second coordinate is the cost 𝐶). Marginal cost: 𝐶 − 𝐶1 11,000 − 10,500 500 𝑚= 2 = $25 per bicycle = = 𝑥2 − 𝑥1 120 − 100 20 Fixed cost: 𝑏 = 𝐶1 − 𝑚𝑥1 = 10,500 − (25)(100) = 10,500 − 2,500 = $8,000 82. We are given two points on the graph of the linear cost function: (1,000, 6,000) and (1,500, 8,500) (𝑥 is the number of cases, and the second coordinate is the cost 𝐶). Marginal cost: 𝐶 − 𝐶1 8,500 − 6,000 2,500 𝑚= 2 = $5 per case. = = 𝑥2 − 𝑥1 1,500 − 1,000 500 Fixed cost: 𝑏 = 𝐶1 − 𝑚𝑥1 = 6,000 − (5)(1,000) = $1,000 83. We are given two points on the graph of the linear cost function: (10, 5,920) and (20, 11,820) (𝑥 is the number of iPhones made in an hour, and the second coordinate is the cost 𝐶). Marginal cost: 𝐶 − 𝐶1 11,820 − 5,920 5,900 𝑚= 2 = $590 per iPhone = = 𝑥2 − 𝑥1 20 − 10 10 Fixed cost: 𝑏 = 𝐶1 − 𝑚𝑥1 = 5,920 − (590)(10) = $20 Thus, the cost equation is 𝐶 = 𝑚𝑥 + 𝑏 = 590𝑥 + 20. The cost to manufacture each additional iPhone is the marginal cost: $590.

Solutions Section 1.3 The cost to manufacture 40 iPhones in an hour is obtained by setting 𝑥 = 40 in the cost equation: 𝐶(40) = 590(40) + 20 = $23,620 84. We are given two points on the graph of the linear cost function: (8, 4,110) and (16, 8,190) (𝑥 is the number of consoles made in an hour, and the second coordinate is the cost 𝐶). Marginal cost: 𝐶 − 𝐶1 8,190 − 4,110 4,080 𝑚= 2 = $510 per unit = = 𝑥2 − 𝑥1 16 − 8 8 Fixed cost: 𝑏 = 𝐶1 − 𝑚𝑥1 = 4,110 − (510)(8) = $30 Thus, the cost equation is 𝐶 = 𝑚𝑥 + 𝑏 = 510𝑥 + 30. The cost to manufacture each additional Kinect is the marginal cost: $510. The cost to manufacture 30 Kinects in an hour is obtained by setting 𝑥 = 30 in the cost equation: 85. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏 (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: (1, 1,960) and (5, 1,800). Slope: 𝑞 − 𝑞1 1,800 − 1,960 −160 𝑚= 2 = = = −40 𝑝2 − 𝑝1 5−1 4 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 1,960 − (−40)(1) = 1,960 + 40 = 2,000 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −40𝑝 + 2,000 86. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: (5, 3,950) and (10, 3,700). Slope: 𝑞 − 𝑞1 3,700 − 3,950 −250 𝑚= 2 = = = −50 𝑝2 − 𝑝1 10 − 5 5 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 3,950 − (−50)5 = 4,200 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −50𝑝 + 4,200

87. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: 2020: (𝑝, 𝑞) = (650, 220) 2024: (𝑝, 𝑞) = (450, 700) 𝑞 − 𝑞1 700 − 220 480 Slope: 𝑚 = 2 = = = −2.4 𝑝2 − 𝑝1 450 − 650 −200 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 220 − (−2.4)650 = 1,780 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −2.4𝑝 + 1,780. If 𝑝 = $400, then 𝑞 = −2.4(400) + 1,780 = 820 million phones. b. Since the slope is −2.4 million phones per unit increase in price, we interpret the slope as follows: For every $1 increase in price, sales of smartphones decrease by 2.4 million units. 88. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: 2013: (335, 1,010) 2017: (265, 1,710)

Solutions Section 1.3 𝑞 − 𝑞1 1,710 − 1,010 700 Slope: 𝑚 = 2 = = = −10 𝑝2 − 𝑝1 265 − 335 −70 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 1,010 − (−10)335 = 4,360 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −10𝑝 + 4,360. If 𝑝 = $385, then 𝑞 = −10(385) + 4,360 = 510 million phones. b. Since the slope is −10 million phones per unit increase in price, we interpret the slope as follows: For every $1 increase in price, sales of smartphones decrease by 10 million units.

89. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: (3, 28,000) and (5, 19,000). 𝑞 − 𝑞1 19,000 − 28,000 −9,000 Slope: 𝑚 = 2 = = = −4,500 𝑝2 − 𝑝1 5−3 2 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 28,000 − (−4,500)3 = 28,000 + 13,500 = 41,500 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −4,500𝑝 + 41,500. b. The units of measurement of the slope are generally units of 𝑦 per unit of 𝑥. In this case: Units of 𝑞 per unit of 𝑝. That is, Rides per day per $1 increase in the fare. Since the slope is −4,500 rides/day per $1 increase in the price, we interpret it as saying that ridership decreases by 4,500 rides per day for every $1 increase in the fare. c. From part (a), the demand equation is 𝑞 = −4,500𝑝 + 41,500 If the fare is $6, we have 𝑝 = 6, so 𝑞 = −4500(6) + 41,500 = −27,000 + 41,500 = 14,500 rides/day.

90. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). We are given two points on its graph: (5, 14) and (3, 18). 𝑞 − 𝑞1 18 − 14 4 Slope: 𝑚 = 2 = = = −2 𝑝2 − 𝑝1 3−5 −2 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 14 − (−2)(5) = 24 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −2𝑝 + 24. b. The units of measurement of the slope are generally units of 𝑦 per unit of 𝑥. In this case: Units of 𝑞 per unit of 𝑝. That is, Millions of rides/day per Z1 increase in the fare. Since the slope is −2 million rides/day per Z1 increase in the price, we interpret it as saying that ridership decreases by 2 million rides per day for every Z1 increase in the fare. c. From part (a), the demand equation is 𝑞 = −2𝑝 + 24 If the fare is Z10, we have 𝑝 = 10, so 𝑞 = −2(10) + 24 = 4 million rides/day.

91. a. In a linear model of 𝑦 versus time 𝑡, the slope is the number of units of 𝑦 per unit time, and we are given this quantity: 40 million pounds/year. Thus, working in millions of pounds, we can take 𝑚 = 40. We are also given the 𝑦-intercept (the value of 𝑦 at 𝑡 = 0) as 𝑏 = 290. Thus, the model is 𝑦 = 40𝑡 + 290 million pounds of pasta b. In 2005, 𝑡 = 15, and so 𝑦(15) = 40(15) + 290 = 890 million pounds. 92. a. In a linear model of 𝑦 versus time 𝑡, the slope is the number of units of 𝑦 per unit time, and we are given this quantity: 60 million kg/year. Thus, working in millions of kilograms, we can take 𝑚 = 60. We are also given the 𝑦-intercept (the value of 𝑦 at 𝑡 = 0) as 𝑏 = 550. Thus, the model is 𝑦 = 60𝑡 + 550 million kg of mercury b. The year 2230 corresponds to 𝑡 = 20, and so 𝑦 = 60(20) + 550 = 1,750 million kg.

Solutions Section 1.3 93. a. The desired linear model has the form 𝑛 = 𝑚𝑡 + 𝑏, where 𝑡 is time in years since 2013. We are given two points on its graph: 2015 data: (2, 90); 2019 data: (6, 270) 𝑛 − 𝑛1 270 − 90 180 Slope: 𝑚 = 2 = = = 45 𝑡2 − 𝑡1 6−2 4 Intercept: 𝑏 = 𝑛1 − 𝑚𝑡1 = 90 − (45)(2) = 0 Thus, the linear model is 𝑛 = 𝑚𝑝 + 𝑏 = 45𝑡. b. The units of measurement of the slope are units of 𝑛 per unit of 𝑡; that is, thousands of (daily) transactions per year. The number of daily Bitcoin transactions increased at a rate of 45,000 per year. c. The year 2021 corresponds to 𝑡 = 8, so 𝑛 = 45(8) = 360 thousand, somewhat higher than the actual number of transactions. 94. a. The desired linear model has the form 𝐸 = 𝑚𝑡 + 𝑏, where 𝑡 is time in years since 2010. We are given two points on its graph: 2016 data: (6, 213); 2021 data: (11, 210) 𝐸 − 𝐸1 210 − 213 −3 Slope: 𝑚 = 2 = = = −0.6 𝑡2 − 𝑡1 11 − 6 5 Intercept: 𝑏 = 𝐸1 − 𝑚𝑡1 = 213 − (−0.6)(6) = 216.6 Thus, the linear model is 𝐸 = 𝑚𝑝 + 𝑏 = −0.6𝑡 + 216.6 b. The units of measurement of the slope are units of 𝐸 per unit of 𝑡; that is, billions of dollars per year. Exxon Mobil's operating expenses decreased at a rate of $0.6 billion per year. c. The year 2018 corresponds to 𝑡 = 8, and so 𝐸 = −0.6(8) + 216.6 = $211.8 billion, which differs quite significantly from the actual operating expenses. 95. 𝑠(𝑡) = 2.5𝑡 + 10 a. Velocity = slope = 2.5 feet/sec. b. After 4 seconds, 𝑡 = 4, so 𝑠(4) = 2.5(4) + 10 = 10 + 10 = 20 Thus the model train has moved 20 feet along the track. c. The train will be 25 feet along the track when 𝑠 = 25. Substituting gives 25 = 2.5𝑡 + 10 Solving for time t gives 2.5𝑡 = 25 − 10 = 15 15 𝑡= = 6 seconds 2.5 96. 𝑠(𝑡) = −1.8𝑡 + 9 a. Velocity = slope = −1.8 feet/sec. b. 𝑠(4) = −1.8(4) + 9 = 1.8 feet from the ground. c. 0 = −1.8𝑡 + 9, giving 𝑡 = 5 seconds

97. a. Take 𝑠 to be displacement from Jones Beach, and 𝑡 to be time in hours. We are given two points: (𝑡, 𝑠) = (10, 0) (𝑡, 𝑠) = (10.1, 13)

𝑠 = 0 for Jones Beach. 6 minutes = 0.1 hours

We are asked for the speed, which equals the magnitude of the slope. 𝑠 − 𝑠1 13 − 0 13 𝑚= 2 = = = 130 𝑡2 − 𝑡1 10.1 − 10 0.1 Units of slope = units of 𝑠 per unit of 𝑡 = miles per hour Thus, the police car was traveling at 130 mph. b. For the displacement from Jones Beach at time 𝑡, we want to express 𝑠 as a linear function of 𝑡; namely, 𝑠 = 𝑚𝑡 + 𝑏. We already know 𝑚 = 130 from part (a). For the intercept, use

Solutions Section 1.3 𝑏 = 𝑠1 − 𝑚𝑡1 = 0 − 130(10) = −1,300 Therefore, the displacement at time 𝑡 is 𝑠 = 𝑚𝑡 + 𝑏 = 130𝑡 − 1,300

98. a. Take 𝑠 to be displacement from Jones Beach, and 𝑡 to be time in hours. We are given two points: (𝑡, 𝑠) = (9.9, 0) and (10.1, 13) We are asked for the speed, which equals the magnitude of the slope. 𝑠 − 𝑠1 13 − 0 𝑚= 2 = = 65 miles per hour 𝑡2 − 𝑡1 10.1 − 9.9 Thus, the perp was traveling at 65 mph. b. 𝑠 = 𝑚𝑡 + 𝑏, where 𝑚 = 65 from part (a), and 𝑏 = 𝑠1 − 𝑚𝑡1 = 0 − 65(9.9) = −643.5 Therefore, 𝑠 = 𝑚𝑡 + 𝑏 = 65𝑡 − 643.5

99. a. The desired linear model has the form 𝐿 = 𝑚𝑛 + 𝑏. We are given two points on its graph: Second edition data: (2, 585); Seventh edition data: (7, 782) 𝐿 − 𝐿1 782 − 585 197 Slope: 𝑚 = 2 = = = 39.4 𝑛2 − 𝑛1 7−2 5 Intercept: 𝑏 = 𝐿1 − 𝑚𝑛1 = 585 − (39.4)(2) = 506.2 Thus, the linear model is 𝐿 = 𝑚𝑛 + 𝑏 = 39.4𝑛 + 506.2 b. The units of measurement of the slope are units of 𝐿 per unit of 𝑛; that is, pages per edition; Applied Calculus is growing at a rate of 39.4 pages per edition. c. The length 𝐿 will equal 1,500 when 39.4𝑛 + 506.2 = 1,500. Solving for 𝑛 gives 39.4𝑛 = 1,500 − 506.2 = 993.8 993.8 𝑛= ≈ 25.2 39.4 Thus, by the 26th edition, the book will be over 1,500 pages long.

100. a. The desired linear model has the form 𝐿 = 𝑚𝑛 + 𝑏. We are given two points on its graph: Second edition data: (2, 603); Fifth edition data: (5, 690) 𝐿 − 𝐿1 690 − 603 87 Slope: 𝑚 = 2 = = = 29 𝑛2 − 𝑛1 5−2 3 Intercept: 𝑏 = 𝐿1 − 𝑚𝑛1 = 603 − (29)(2) = 545 Thus, the linear model is 𝐿 = 𝑚𝑛 + 𝑏 = 29𝑛 + 545. b. The units of measurement of the slope are units of 𝐿 per unit of 𝑛; that is, pages per edition; Finite Mathematics is growing at a rate of 29 pages per edition. 1,000 − 545 c. 𝐿 = 29𝑛 + 545 = 1,000 when 𝑛 = ≈ 15.7. Thus, by the 16th edition, the book will be over 29 1,000 pages long. 101. 𝐹 = Fahrenheit temperature, 𝐶 = Celsius temperature, and we want 𝐹 as a linear function of 𝐶. That is, 𝐹 = 𝑚𝐶 + 𝑏. (𝐹 plays the role of 𝑦 and 𝐶 plays the role of 𝑥.) We are given two points: (𝐶, 𝐹 ) = (0, 32) (𝐶, 𝐹 ) = (100, 212)

Slope: 𝑚 =

Freezing point Boiling point

𝐹2 − 𝐹1 212 − 32 180 = = = 1.8 𝐶2 − 𝐶1 100 − 0 100

Solutions Section 1.3 Intercept: 𝑏 = 𝐹1 − 𝑚𝐶1 = 32 − 1.8(0) = 32 Thus, the linear relation is 𝐹 = 𝑚𝐶 + 𝑏 = 1.8𝐶 + 32 When 𝐶 = 30 ∘ 𝐹 = 1.8(30) + 32 = 54 + 32 = 86 ∘ When 𝐶 = 22 ∘ 𝐹 = 1.8(22) + 32 = 39.6 + 32 = 71.6 ∘. Rounding to the nearest degree gives 72 ∘F. When 𝐶 = −10 ∘, 𝐹 = 1.8(−10) + 32 = −18 + 32 = 14 ∘. When 𝐶 = −14 ∘, 𝐹 = 1.8(−14) + 32 = −25.2 + 32 = 6.8 ∘. Rounding to the nearest degree gives 7°F.

102. 𝐹 = Fahrenheit temperature, 𝐶 = Celsius temperature, and we want 𝐹 as a linear function of 𝐶. That is, 𝐹 = 𝑚𝐶 + 𝑏. (𝐹 plays the role of 𝑦 and 𝐶 plays the role of 𝑥.) We are given two points: (𝐹 , 𝐶) = (32, 0) and (212, 100) 𝐶 − 𝐶1 100 − 0 100 5 Slope: 𝑚 = 2 = = = 𝐹2 − 𝐹1 212 − 32 180 9 5 160 Intercept: 𝑏 = 𝐶1 − 𝑚𝐹1 = 0 − (32) = − 9 9 Thus, the linear relation is 5 160 𝐶 = 𝑚𝐹 + 𝑏 = 𝐹 − 9 9 5 160 360 𝐶(104) = (104) − = = 40 ∘ 9 9 9 5 160 225 𝐶(77) = (77) − = = 25 ∘ 9 9 9 5 160 −90 𝐶(14) = (14) − = = −10 ∘ 9 9 9 5 160 −360 𝐶(−40) = (−40) − = = −40 ∘ 9 9 9 103. a. 𝑆 = Southwest Airlines net income (in $ millions), 𝐽 = JetBlue Airways net income (in $ millions), and we want 𝐽 as a linear function of 𝑆. That is, 𝐽 = 𝑚𝑆 + 𝑏 𝐽 plays the role of 𝑦 and 𝑆 plays the role of 𝑥. We are given two points: (𝑆, 𝐽) = (2,000, 1,100) (𝑆, 𝐽) = (−3,000, −1,400)

2017 data 2020 data

𝐽2 − 𝐽1 −1,400 − 1,100 −2,500 = = = 0.5 𝑆2 − 𝑆1 −3,000 − 2,000 −5,000 Intercept: 𝑏 = 𝐽1 − 𝑚𝑆1 = 1,100 − (0.5)(2,000) = 100 Thus, the linear relation is 𝐽 = 𝑚𝑆 + 𝑏 = 0.5𝑆 + 100. b. In 2019, Southwest Airlines' net income was 𝑆 = 2,300, so 𝐽 = 0.5(2,300) + 100 = 1,250, predicting a $1,250 million net income for JetBlue, $650 million higher than the actual $600 million net income JetBlue earned in 2019. c. The units of measurement of the slope are units of 𝐽 per unit of 𝑆; that is, millions of dollars of JetBlue Airways net income per million dollars of Southwest Airlines net income; JetBlue Airways earned an additional net income of $0.50 per $1 additional net income earned by Southwest Airlines. Slope: 𝑚 =

Solutions Section 1.3 104. a. 𝐴 = Alaska Air Group net income (in $ millions), 𝐽 = JetBlue Airways net income (in $ millions), and we want 𝐽 as a linear function of 𝐴. That is, 𝐽 = 𝑚𝐴 + 𝑏 𝐽 plays the role of 𝑦 and 𝐴 plays the role of 𝑥. We are given two points: (𝐴, 𝐽) = (800, 600) (𝐴, 𝐽) = (−1,200, −1,400)

2016 data 2020 data

𝐽2 − 𝐽1 −1,400 − 600 −2,000 = = =1 𝐴2 − 𝐴1 −1,200 − 800 −2,000 Intercept: 𝑏 = 𝐽1 − 𝑚𝐴1 = 600 − (1)(800) = −200 Thus, the linear relation is 𝐽 = 𝑚𝐴 + 𝑏 = 𝐴 − 200. b. 2017: 𝐴 = 1,000; 𝐽 = 1,000 − 200 = $800 million 2018: 𝐴 = 400; 𝐽 = 400 − 200 = $200 million 2019: 𝐴 = 800; 𝐽 = 800 − 200 = $600 million comparing these values with those in the table shows that the model gives a perfect prediction for every year except 2017 ($300 million lower than the actual JetBlue income). c. The units of measurement of the slope are units of 𝐽 per unit of 𝐴; that is, millions of dollars of JetBlue Airways net income per million dollars of Alaska Air Group net income; JetBlue Airways earned an additional net income of $1 per $1 additional net income earned by Alaska Air Group. Slope: 𝑚 =

105. Income = royalties + screen rights

𝐼 = 5\% of net profits + 50,000 𝐼 = 0.05𝑁 + 50,000 Equation notation 𝐼(𝑁) = 0.05𝑁 + 50,000 Function notation

For an income of $100,000,

100,00 = 0.05𝑁 + 50,000 0.05𝑁 = 50,000 50,000 𝑁= = 1,000,000 0.05

Her marginal income is her increase in income per $1 increase in net profit. This is the slope, 𝑚 = 0.05 dollars of income per dollar of net profit, or 5\hbox{\rm\rlap/c} per dollar of net profit. 106. 𝐼 = 𝑚𝑁 + 𝑏 𝑏 = 100,000, 𝑚 = 0.08, and so 𝐼 = 0.08𝑁 + 100,000 For an income of $1,000,000, 1,000,000 = 0.08𝑁 + 100,000 0.08𝑁 = 900,000 𝑁 = 11,250,000 Marginal income is 𝑚 = 8 hbox rlap ∕c per dollar of net profit.

107. The year 2000 corresponds to 𝑡 = 10, which is in the range 6 ≤ 𝑡 < 15, so we use the first equation: 𝑣(𝑡) = 400𝑡 − 2,200. The slope is 400 MHz/year, telling us that the speed of a processor was increasing by 400 MHz/year. 108. The year 2000 corresponds to 𝑡 = 25, which is in the range 20 ≤ 𝑡 ≤ 30, so we use the second equation: 𝑣(𝑡) = 174𝑡 − 3,420. The slope is 174 MHz/year, telling us that the speed of a processor was increasing by 174 MHz/year.

Solutions Section 1.3 109. a. The data are 𝑡

𝑦

2,100 2,950

1970–1990 (first two data points): 𝑦 − 𝑦1 700 − 78 Slope: 𝑚 = 2 = = 31.1 𝑡2 − 𝑡1 20 − 0 Intercept: 𝑏 = 78, specified in first data point Thus, the linear model is 𝑦 = 𝑚𝑡 + 𝑏 = 31.1𝑡 + 78. b. 1990–2010 (second and third data points): 𝑦 − 𝑦1 2,950 − 700 Slope: 𝑚 = 2 = = 112.5 𝑡2 − 𝑡1 40 − 20 Intercept: 𝑏 = 𝑦1 − 𝑚𝑡1 = 700 − (112.5)20 = −1,550 Thus, the linear model is 𝑦 = 𝑚𝑡 + 𝑏 = 112.5𝑡 − 1,550. c. Since the first model is valid for 0 ≤ 𝑡 ≤ 20 and the second one for 20 ≤ 𝑡 ≤ 40, we put them together as 31.1𝑡 + 78 if 0 ≤ 𝑡 < 20 𝑦= {112.5𝑡 − 1,550 if 20 ≤ 𝑡 ≤ 40. Notice that, since both formulas agree at 𝑡 = 20, we can also say 31.1𝑡 + 78 if 0 ≤ 𝑡 ≤ 20 𝑦= {112.5𝑡 − 1,550 if 20 < 𝑡 ≤ 40. d. Since 2004 is represented by 𝑡 = 34, we use the second formula to obtain 𝑦 = 112.5(34) − 1,550 = 2,275, or $2,275,000, in good agreement with the actual value shown in the graph. 110. a. The data are 𝑡

𝑦 222 2,100 5,600

1980–2000 (first two data points): 𝑦 − 𝑦1 2,100 − 222 Slope: 𝑚 = 2 = = 93.9 𝑡2 − 𝑡1 20 − 0 Intercept: 𝑏 = 222, specified in first data point Thus, the linear model is 𝑦 = 𝑚𝑡 + 𝑏 = 93.9𝑡 + 222. b. 2000–2020 (second and third data points): 𝑦 − 𝑦1 5,600 − 2,100 Slope: 𝑚 = 2 = = 175 𝑡2 − 𝑡1 40 − 20 Intercept: 𝑏 = 𝑦1 − 𝑚𝑡1 = 2,100 − (175)20 = −1,400 Thus, the linear model is 𝑦 = 𝑚𝑡 + 𝑏 = 175𝑡 − 1,400. c. Since the first model is valid for 0 ≤ 𝑡 ≤ 20 and the second one for 20 ≤ 𝑡 ≤ 40, we put them together as 93.9𝑡 + 222 if 0 ≤ 𝑡 < 20 𝑦= {175𝑡 − 1,400 if 20 ≤ 𝑡 ≤ 40. Notice that, since both formulas agree at 𝑡 = 20, we can also say

93.9𝑡 + 222 if 0 ≤ 𝑡 ≤ 20 {175𝑡 − 1,400 if 20 < 𝑡 ≤ 40. d. Since 1992 is represented by 𝑡 = 12, we use the first formula to obtain 𝑦 = 93.9(12) + 222 = 1,348.8, or $1,348,800, considerably higher than the actual value shown on the graph. The actual cost is not linear in the range 1980–2000. Solutions Section 1.3

𝑦=

111. 2006–2012: Points: 2006 data: (𝑡, 𝑆) = (0, 531) 2012 data: (𝑡, 𝑆) = (6, 547) 𝑆 − 𝑆1 547 − 531 8 Slope: 𝑚 = 2 = = 𝑡2 − 𝑡1 6−0 3 Intercept: We are given the 𝑆-intercept as 531. 8 Thus, the linear model is 𝑆 = 𝑚𝑡 + 𝑏 = 𝑡 + 531. 3 2012–2018: Points: 2000 data: (𝑡, 𝑆) = (6, 547) 2004 data: (𝑡, 𝑆) = (12, 529) 𝑆 − 𝑆1 529 − 547 Slope: 𝑚 = 2 = = −3 𝑡2 − 𝑡1 12 − 6 Intercept: 𝑏 = 𝑆1 − 𝑚𝑡1 = 547 − (−3)6 = 565 Thus, the linear model is 𝑆 = 𝑚𝑡 + 𝑏 = −3𝑡 + 565 Putting them together gives 𝑆=

8 𝑡 + 531 3

{−3𝑡 + 565

if 0 ≤ 𝑡 < 6

if 6 ≤ 𝑡 ≤ 12. The score in 2009 is 𝑆(3), so we use the first formula to obtain 8 𝑆(3) = (3) + 531 = 539, in exact agreement with the actual score. 3 112. 2001–2004: Points: 2001 data: (𝑡, 𝑃 ) = (0, 9.7) 2004 data: (𝑡, 𝑃 ) = (3, 4.3) 𝑃 − 𝑃1 4.3 − 9.7 Slope: 𝑚 = 2 = = −1.8 𝑡2 − 𝑡1 3−0 Intercept: 𝑏 = 9.7, specified in first data point Thus, the linear model is 𝑁 = 𝑚𝑡 + 𝑏 = −1.8𝑡 + 9.7. 2004–2007: Points: 2004 data: (𝑡, 𝑃 ) = (3, 4.3) 2007 data: (𝑡, 𝑃 ) = (6, 10.3) 𝑃 − 𝑃1 10.3 − 4.3 Slope: 𝑚 = 2 = =2 𝑡2 − 𝑡1 6−3 Intercept: 𝑏 = 𝑃1 − 𝑚𝑡1 = 4.3 − (2)(3) = −1.7 Thus, the linear model is 𝑁 = 𝑚𝑡 + 𝑏 = 2𝑡 − 1.7. Putting them together gives −1.8𝑡 + 9.7 if 0 ≤ 𝑡 ≤ 3 𝑃= {2𝑡 − 1.7 if 3 < 𝑡 ≤ 6. The percentage of delinquent borrowers in 2006 is 𝑃 (5), so we use the second formula to obtain 𝑃 (5) = 2(5) − 1.7 = 8.3%.

Solutions Section 1.3 113. Compute the corresponding successive changes Δ𝑥 in 𝑥 and Δ𝑦 in 𝑦, and compute the ratios Δ𝑦∕Δ𝑥. If the answer is always the same number, then the values in the table come from a linear function. 114. The desired equation has the form 𝑦 = 𝑚𝑥 + 𝑏. The slope 𝑚 is given by 𝑚 = Δ𝑦∕Δ𝑥, where Δ𝑥 and Δ𝑦 are corresponding changes in 𝑥 and 𝑦. The intercept 𝑏 is given by the 𝑦-value corresponding to 𝑥 = 0, if supplied. If it is not supplied, choose any point (𝑥1 , 𝑦1 ) and use the formula 𝑏 = 𝑦1 − 𝑚𝑥1 . 115. To find the linear function, solve the equation 𝑎𝑥 + 𝑏𝑦 = 𝑐 for 𝑦 : 𝑏𝑦 = −𝑎𝑥 + 𝑐 𝑎 𝑐 𝑦 =− 𝑥+ 𝑏 𝑏

𝑎 𝑐 Thus, the desired function is 𝑓(𝑥) = − 𝑥 + . 𝑏 𝑏 𝑎 𝑐 If 𝑏 = 0, then and are undefined, and 𝑦 cannot be specified as a function of 𝑥. (The graph of the 𝑏 𝑏 resulting equation would be a vertical line.)

116. The slope of the line with equation 𝑦 = 𝑚𝑥 + 𝑏 is the number of units that 𝑦 increases per unit increase in 𝑥 . 117. The slope of the line is 𝑚 = as fast as 𝑥, then its slope is 3 .

Δ𝑦 3 = = 3. Therefore, if, in a straight line, 𝑦 is increasing three times Δ𝑥 1

118. The slope is 𝑚 = −4∕3 units of 𝑦 per unit of 𝑥. We do not have enough information to compute the intercept. 119. If 𝑚 is positive, then 𝑦 will increase as 𝑥 increases; if 𝑚 is negative, then 𝑦 will decrease as 𝑥 increases; if 𝑚 is zero, then 𝑦 will not change as 𝑥 changes. 120. Since Δ𝑦 = −Δ𝑥, the function is linear with slope Δ𝑦 −Δ𝑥 𝑚= = = −1. Δ𝑥 Δ𝑥 121. The slope computed in cell C2 is given by 𝑦 − 𝑦1 −1 − 2 𝑚= 2 = = −1.5 𝑥2 − 𝑥1 3−1 If we increase the 𝑦-coordinate in cell B3, this increases 𝑦2 , and thus increases the numerator Δ𝑦 = 𝑦2 − 𝑦1 without affecting the denominator Δ𝑥. Thus the slope will increase. 122. The slope increases: Δ𝑦 is negative, and stays the same, while Δ𝑥 becomes a larger positive number, so the (negative) slope decreases in absolute value, meaning that it actually increases. 123. The units of the slope 𝑚 are units of 𝑦 (bootlags) per unit of 𝑥 (zonars). The intercept 𝑏 is on the

Solutions Section 1.3 𝑦-axis, and is thus measured in units of 𝑦 (bootlags). Thus, 𝑚 is measured in bootlags per zonar and 𝑏 is measured in bootlags. 124. Units of slope = units of 𝑦 per unit of 𝑥 = miles per dollar. Thus, the independent variable is measured in dollars and the dependent variable is measured in miles.

125. If a quantity changes linearly with time, it must change by the same amount for every unit change in time. Thus, since it increases by 10 units in the first day, it must increase by 10 units each day, including the third. 126. Since 𝑄(0) is positive, 𝑏 is positive. Since 𝑄 decreases with increasing 𝑇 , 𝑚 is negative.

127. Since the slope is 0.1, the velocity is increasing at a rate of 0.1 m/sec every second. Since the velocity is increasing, the object is accelerating (choice B). 128. Velocity = slope = 0.2 units of position per unit time. Thus, the object is moving with constant speed (choice A). 129. Write 𝑓(𝑥) = 𝑚𝑥 + 𝑏 and 𝑔(𝑥) = 𝑛𝑥 + 𝑐. Then 𝑓(𝑥) + 𝑔(𝑥) = 𝑚𝑥 + 𝑏 + (𝑛𝑥 + 𝑐) = (𝑚 + 𝑛)𝑥 + (𝑏 + 𝑐), also a linear function with slope 𝑚 + 𝑛.

130. Not necessarily; for instance, if 𝑓(𝑥) = 2𝑥 + 1 and 𝑔(𝑥) = 𝑥, then their ratio has values 2𝑥 + 1 1 = 2 + , not a linear function of 𝑥. (Only if 𝑔 is a nonzero constant will the ratio will be linear.) 𝑥 𝑥 131. Answers may vary. For example, 𝑓(𝑥) = 𝑥 1∕3, 𝑔(𝑥) = 𝑥 2∕3 gives 𝑓(𝑥)𝑔(𝑥) = 𝑥 1∕3𝑥 2∕3 = 𝑥.

132. Answers may vary. For example, 𝑓(𝑥) = 𝑥 3 + 2𝑥 2, 𝑔(𝑥) = 𝑥 2 gives 𝑓(𝑥) 𝑥 3 + 2𝑥 2 = = 𝑥 + 2 (with domain all real numbers other than 0). 𝑔(𝑥) 𝑥2 133. Increasing the number of items from the break-even number results in a profit: Because the slope of the revenue graph is larger than the slope of the cost graph, it is higher than the cost graph to the right of the point of intersection, and hence corresponds to a profit.

134. You should solve your equation for 𝑞 to obtain 𝑞 as a function of 𝑝. Simply switching 𝑝 and 𝑞 will result in the wrong equation, and starting from scratch would be a less efficient way of obtaining the result you want.

Solutions Section 1.4 Section 1.4

1. (1, 1), (2, 2), (3, 4); 𝑦 = 𝑥 − 1

𝒚̂ =𝒙 − 𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

SSE = Sum of squares of residuals (last column) = 1 + 1 + 4 = 6 2. (0, 1), (1, 1), (2, 2); 𝑦 = 𝑥 + 1

𝒚̂ =𝒙 + 𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

−1

SSE = Sum of squares of residuals (last column) = 0 + 1 + 1 = 2 3. (0, −1), (1, 3), (4, 6), (5, 0); 𝑦 = −𝑥 + 2 𝒙 0

𝒚

−1

𝒚̂ = − 𝒙 + 𝟐 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐 2

−3

−2

SSE = Sum of squares of residuals (last column) = 9 + 4 + 64 + 9 = 86 4. (2, 4), (6, 8), (8, 12), (10, 0); 𝑦 = 2𝑥 − 8 𝒙

𝒚

𝒚̂ =𝟐𝒙 − 𝟖 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

−4

−12

144

SSE = Sum of squares of residuals (last column) = 240 5. (1, 1), (2, 2), (3, 4) a. 𝑦 = 1.5𝑥 − 1

Solutions Section 1.4

𝒚̂ =𝟏.𝟓𝒙 − 𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

0.5

0.25

3.5

0.5

0.25

SSE = Sum of squares of residuals = 0.5 b. 𝑦 = 2𝑥 − 1.5

𝒚̂ =𝟐𝒙 − 𝟏.𝟓 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

0.5

2.5

4.5

0.5

0.25

−0.5

0.25

−0.5

0.25

SSE = Sum of squares of residuals = 0.75 The model that gives the better fit is (a) because it gives the smaller value of SSE. 6. (0, 1), (1, 1), (2, 2) a. 𝑦 = 0.4𝑥 + 1.1

𝒚̂ =𝟎.𝟒𝒙 + 𝟏.𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

1.1

1.5

1.9

−0.1 −0.5

0.01

0.1

0.01

0.25

SSE = Sum of squares of residuals = 0.27 b. 𝑦 = 0.5𝑥 + 0.9

𝒚̂ =𝟎.𝟓𝒙 + 𝟎.𝟗 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

0.9

1.4

1.9

0.1

−0.4

0.01

0.1

0.01

0.16

SSE = Sum of squares of residuals = 0.18 The model that gives the better fit is (b) because it gives the smaller value of SSE. 7. (0, −1), (1, 3), (4, 6), (5, 0) a. 𝑦 = 0.3𝑥 + 1.1 𝒙 0

𝒚

−1

𝒚̂ =𝟎.𝟑𝒙 + 𝟏.𝟏 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐 1.1

−2.1

4.41

1.4

1.6

2.56

2.3

3.7

13.69

2.6

−2.6

6.76

SSE = Sum of squares of residuals = 27.42

b. 𝑦 = 0.4𝑥 + 0.9 𝒙

𝒚

−1

Solutions Section 1.4

𝒚̂ =𝟎.𝟒𝒙 + 𝟎.𝟗 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐 −1.9

0.9

3.61

1.3

1.7

2.89

2.5

3.5

12.25

2.9

−2.9

8.41

SSE = Sum of squares of residuals = 27.16 The model that gives the better fit is (b) because it gives the smaller value of SSE. 8. (2, 4), (6, 8), (8, 12), (10, 0) a. 𝑦 = −0.1𝑥 + 7

𝒚̂ = − 𝟎.𝟏𝒙 + 𝟕 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

6.8

6.4

1.6

2.56

6.2

5.8

33.64

−2.8 −6

7.84

SSE = Sum of squares of residuals = 80.04 b. 𝑦 = −0.2𝑥 + 6

𝒚̂ = − 𝟎.𝟐𝒙 + 𝟔 𝒚 − 𝒚̂ (𝒚 − 𝒚̂ ) 𝟐

𝒙

𝒚

5.6

4.8

3.2

10.24

4.4

7.6

57.76

−1.6 −4

2.56

SSE = Sum of squares of residuals = 86.56 The model that gives the better fit is (a) because it gives the smaller value of SSE. 9. Data points (𝑥, 𝑦) : (1, 1), (2, 2), (3, 4) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

1 3 6

1 4 7

12 17

1 9

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(17) − (6)(7) 9 𝑚= = = = 1.5 6 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(14) − 6 2

Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 7 − 1.5(6) −2 ≈ −0.6667 = = 𝑛 3 3

Solutions Section 1.4

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 ≈ 1.5𝑥 − 0.6667. Graph:

6 5 4 3 2 1

x -1

𝒙

𝒚

𝒙𝒚

𝒙𝟐

10. Data points (𝑥, 𝑦) : (0, 1), (1, 1), (2, 2) 0 2 3

1 2 4

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(5) − (3)(4) 3 𝑚= = = = 0.5 6 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(5) − 3 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 4 − 0.5(3) 2.5 Intercept: 𝑏 = ≈ 0.8333 = = 𝑛 3 3

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 ≈ 0.5𝑥 + 0.8333. Graph:

2.5

2 1.5 1 0.5 -1

Solutions Section 1.4 11. Data points (𝑥, 𝑦) : (0, −1), (1, 3), (3, 6), (4, 1) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

−1

1 4

0 4

0 9

16 26

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(25) − (8)(9) 28 𝑚= = = = 0.7 40 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(26) − 8 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 9 − 0.7(8) 3.4 Intercept: 𝑏 = = = = 0.85 𝑛 4 4

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = 0.7𝑥 + 0.85. Graph:

6 5 4 3 2 1

x -1

𝒙

𝒚

𝒙𝒚

𝒙𝟐

100

12. Data points (𝑥, 𝑦) : (2, 4), (4, 8), (8, 12), (10, 0) 2

96 136

184

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(136) − (24)(24) −32 𝑚= = = = −0.2 160 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(184) − 24 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 24 − (−0.2)(24) 28.8 Intercept: 𝑏 = = = = 7.2 𝑛 4 4

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = −0.2𝑥 + 7.2.

Solutions Section 1.4 Graph:

12 10 8 6 4 2

𝒙

𝒚

𝒙𝒚

𝒙𝟐

𝒚𝟐

13. a. (1, 3), (2, 4), (5, 6)

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(41) − (8)(13) 𝑟= = 2 2 2 2 √3(30) − 8 2√3(61) − 13 2 𝑛(∑ 𝑥 ) − (∑ 𝑥) ⋅ 𝑛(∑ 𝑦 ) − (∑ 𝑦) √ √ 19 ≈ ≈ 0.9959 19.078784

b. (0, −1), (2, 1), (3, 4) 𝒙 0

𝒚

−1

𝒙𝒚

𝒙𝟐

𝒚𝟐

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(14) − (5)(4) 𝑟= = 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅ 𝑛(∑ 𝑦 2) − (∑ 𝑦) 2 √3(13) − 5 2√3(18) − 4 2 √ √ 22 ≈ ≈ 0.9538 23.0651252

c. (4, −3), (5, 5), (0, 0) 𝒙

𝒚

𝒙𝒚

Solutions Section 1.4 𝒙𝟐

𝒚𝟐

−3

−12

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(13) − (9)(2) 𝑟= = 2 2 2 2 √3(41) − 9 2√3(34) − 2 2 𝑛(∑ 𝑥 ) − (∑ 𝑥) ⋅ 𝑛(∑ 𝑦 ) − (∑ 𝑦) √ √ 21 ≈ ≈ 0.3273 64.1560597

The value of 𝑟 in part (a) has the largest absolute value. Therefore, the regression line for the data in part (a) is the best fit. The value of 𝑟 in part (c) has the smallest absolute value. Therefore, the regression line for the data in part (c) is the worst fit. Since 𝑟 is not ±1 for any of these lines, none of them is a perfect fit. 14. a. (1, 3), (−2, 9), (2, 1) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

𝒚𝟐

-2

−18

1 4

−13

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(−13) − (1)(13) 𝑟= = 2 2 2 2 √3(9) − 1 2√3(91) − 13 2 𝑛(∑ 𝑥 ) − (∑ 𝑥) ⋅ 𝑛(∑ 𝑦 ) − (∑ 𝑦) √ √ −52 = = −1 (Best, perfect fit) 52

b. (0, 1), (1, 0), (2, 1) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

𝒚𝟐

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(2) − (3)(2) 𝑟= = = 0 (Worst) 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅ 𝑛(∑ 𝑦 2) − (∑ 𝑦) 2 √3(5) − 3 2√3(2) − 2 2 √ √ Solutions Section 1.4

c. (0, 0), (5, −5), (2, −2.1) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

𝒚𝟐

4.41

29.41

−5

−25

−2.1 −4.2

−7.1 −29.2

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(−29.2) − (7)(−7.1) 𝑟= = 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅ 𝑛(∑ 𝑦 2) − (∑ 𝑦) 2 √3(29) − 7 2√3(29.41) − (−7.1) 2 √ √ −37.9 ≈ ≈ −0.9997 37.9098932

15. Data points (𝑥, 𝑦) : (0, 3.9), (4, 4.8), (8, 5.7) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

4.8

19.2

14.4

64.8

3.9

5.7

4 12

45.6

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(64.8) − (12)(14.4) 21.6 𝑚= ≈ 0.23 = = 96 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(80) − 12 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 14.4 − 0.225(12) 11.7 Intercept: 𝑏 = = = = 3.9 𝑛 3 3

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 ≈ 0.23𝑥 + 3.9. Since 2018 corresponds to 𝑥 = 6, the prediction for 2018 is 𝑦 = 0.23(6) + 3.9 ≈ 5.3 million subscribers.

Solutions Section 1.4 16. Data points (𝑥, 𝑦) : (0, 35), (10, 33), (20, 32) 𝒙

𝒚

𝒙𝒚

𝒙𝟐

330

100

970

500

10 30

640

400

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(970) − (30)(100) −90 𝑚= = = = −0.15 600 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(500) − 30 2 ∑ 𝑦 − 𝑚(∑ 𝑥) 100 − (−0.15)(30) 104.5 Intercept: 𝑏 = ≈ 34.83 = = 𝑛 3 3

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 ≈ −0.15𝑥 + 34.83. Since 2022 corresponds to 𝑥 = 22, the prediction for 2022 is 𝑦 = −0.15(22) + 34.83. ≈ 31.5 million subscribers.

17. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). Data points (𝑝, 𝑞) : (6, 1.2), (5, 1.4), (4.5, 1.5) 𝒑

𝒒

𝒑𝒒

𝒑𝟐

1.2

7.2

4.5

1.5

6.75 20.25

15.5

1.4 4.1

36 25

20.95 81.25

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑝𝑞) − (∑ 𝑝)(∑ 𝑞) 3(20.95) − (15.5)(4.1) −0.7 𝑚= = = = −0.2 3.5 𝑛(∑ 𝑝 2) − (∑ 𝑝) 2 3(81.25) − 15.5 2 ∑ 𝑞 − 𝑚(∑ 𝑝) 4.1 − (−0.2)(15.5) 7.2 Intercept: 𝑏 = = = = 2.4 𝑛 3 3

Thus, the regression line is 𝑞 = 𝑚𝑝 + 𝑏 = −0.2𝑝 + 2.4. When the selling price is $550, 𝑝 = 5.5, and so 𝑞 ≈ −0.2(5.5) + 2.4 = 1.3 billion smartphones.

Solutions Section 1.4 18. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏. (𝑝 is the price, and 𝑞 is the demand). Data points (𝑝, 𝑞) : (4, 0.7), (3, 1), (2.5, 1.5) 𝒑

𝒒

𝒑𝒒

𝒑𝟐

0.7

2.8

2.5

1.5

3.75

6.25

9.5

3.2

9.55 31.25

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑝𝑞) − (∑ 𝑝)(∑ 𝑞) 3(9.55) − (9.5)(3.2) −1.75 𝑚= = = = −0.5 3.5 𝑛(∑ 𝑝 2) − (∑ 𝑝) 2 3(31.25) − 9.5 2 ∑ 𝑞 − 𝑚(∑ 𝑝) 3.2 − (−0.5)(9.5) 7.95 Intercept: 𝑏 = ≈ 2.7 = = 𝑛 3 3

Thus, the regression line is 𝑞 = 𝑚𝑝 + 𝑏 ≈ −0.5𝑝 + 2.7. When the selling price is $350, 𝑝 = 3.5, and so 𝑞 ≈ −0.5(3.5) + 2.7 = 0.95 billion, or 950 million smartphones. 19. Following is the table we use to compute the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

400

240

1,600

720

6,400

100

1,500 10,000

240

2,520 18,400

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(2,520) − (240)(33) 2,160 Slope: 𝑚 = = = = 0.135 16,000 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(18,400) − 240 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 33 − (0.135)(240) 0.6 = = = 0.15 𝑛 4 4

The regression model is therefore 𝑦 = 𝑚𝑥 + 𝑏 = 0.135𝑥 + 0.15. 𝑦(50) = 0.135(50) + 0.15 = 6.9 million jobs

Solutions Section 1.4 20. Following is the table we use to compute the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

200

2,000

100

900

36,000

1,600

1,000 50,000

2,500

2,000 160,000 6,400

180 4,100 248,000 10,600 (The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(248,000) − (180)(4,100) 254,000 Slope: 𝑚 = = = = 25.4 10,000 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(10,600) − 180 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 4100 − (25.4)(180) −472 = = = −118 𝑛 4 4

The regression model is therefore 𝑦 = 𝑚𝑥 + 𝑏 = 25.4𝑥 − 118. 𝑦(70) = 25.4(70) − 118 = $1,660 billion

21. a. Data points (𝑅, 𝑃 ) : (190, 4), (240, 12), (300, 12), (420, 27) 𝑹

𝑷

𝑹𝑷

240

2,880

420

11,340 176,400

190 300 1,150

12 55

𝑹𝟐

760

36,100

3,600

90,000

57,600

18,580 360,100

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑅𝑃 ) − (∑ 𝑅)(∑ 𝑃 ) 4(18,580) − (1,150)(55) 11,070 𝑚= ≈ 0.094 = = 2 2 2 117,900 𝑛(∑ 𝑅 ) − (∑ 𝑅) 4(360,100) − 1,150 ∑ 𝑃 − 𝑚(∑ 𝑅) 55 − 0.09389(1,150) −52.9771 Intercept: 𝑏 = ≈ ≈ −13.244 = 𝑛 4 4

Thus, the regression line is 𝑃 = 𝑚𝑅 + 𝑏 ≈ 0.094𝑅 − 13.244.

Solutions Section 1.4 Graph:

30 25 20 15 10 5 0 R 100 200 300 400 500 (Independent variable is 𝑅 and dependent variable is 𝑃 ) b. The units of measurement of the slope are units of profit per unit of revenue: billions of dollars of profit per billion dollars of revenue, or just dollars of profit per dollar of revenue. Thus, Amazon earned $0.094 in profit per additional $1 in revenue. c. 𝑃 = 0.094𝑅 − 13.244, and we are given 𝑃 = 15. Substituting gives 15 = 0.094𝑅 − 13.244 Solving for 𝑅 gives 28.244 𝑅= ≈ 300 0.094 Thus, the company would need to earn about $300 billion in revenue. d. The graph shows a good fit, so the linear model seems reasonable. 22. a. Data points (𝑅, 𝐼) : (190, 5), (240, 15), (300, 14), (420, 28) 𝑹

𝑰

240

3,600

420

11,760 176,400

190 300 1,150

14 62

𝑹𝑰

𝑹𝟐

950

36,100

4,200

90,000

57,600

20,510 360,100

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑅𝐼) − (∑ 𝑅)(∑ 𝐼) 4(20,510) − (1,150)(62) 10,740 𝑚= ≈ 0.091 = = 117,900 𝑛(∑ 𝑅 2) − (∑ 𝑅) 2 4(360,100) − 1,150 2 ∑ 𝐼 − 𝑚(∑ 𝑅) 62 − 0.09109(1,150) −42.75827 Intercept: 𝑏 = ≈ ≈ −10.69 = 𝑛 4 4

Thus, the regression line is 𝐼 = 𝑚𝑅 + 𝑏 ≈ 0.091𝑅 − 10.69.

Solutions Section 1.4 Graph:

30 25 20 15 10 5 0 R 180 230 280 330 380 430 (Independent variable is 𝑅 and dependent variable is 𝐼) b. The units of measurement of the slope are units of operating income per unit of revenue: billions of dollars of operating operating income per billion dollars of revenue, or just dollars of operating income per dollar of revenue. Thus, Amazon had operating income of $0.091 per $1 in revenue. c. 𝐼 = 0.091𝑅 − 10.69, and we are given 𝐼 = 10. Substituting gives 10 = 0.091𝑅 − 10.69 Solving for 𝑅 gives 20.69 𝑅= ≈ 227 0.091 Thus, the company would need to earn approximately $227 billion in revenue. d. The graph shows a good fit, so the linear model seems reasonable. 23. a. The following result and plot were obtained using the Function Evaluator and Grapher on the Web site with the setup shown. Regression equation: 𝐿 = 39.29𝑛 + 528.71

b. The units of measurement of the slope are units of 𝐿 per unit of 𝑛; that is, pages per edition; Applied Calculus is growing at a rate of 39.29 pages per edition. 24. a. The following result and plot were obtained using the Function Evaluator and Grapher on the Web site with the setup shown. Regression equation: 𝐿 = 25.43𝑛 + 552.24

Solutions Section 1.4

b. The units of measurement of the slope are units of 𝐿 per unit of 𝑛; that is, pages per edition; Finite Mathematics is growing at a rate of 25.43 pages per edition.

25. a. Since production is a function of cultivated area, we take 𝑥 as cultivated area, and 𝑦 as production: 𝑥 𝑦

See the technology note accompanying Example 2 for the use of technology to obtain regression lines. We obtained the following regression line and plot in Excel. (coefficients rounded to two decimal places):𝑦 = 1.62𝑥 − 23.87

b. To interpret the slope 𝑚 = 1.62, recall that units of 𝑚 are units of 𝑦 per unit of 𝑥; that is, millions of tons of production of soybeans per million acres of cultivated land. Thus, production increases by 1.62 million tons of soybeans per million acres of cultivated land. More simply, each acre of cultivated land produces about 1.62 tons of soybeans.

26. a. Since production is a function of cultivated area, we take 𝑥 as cultivated area, and 𝑦 as production: 𝑥 𝑦

b. To interpret the slope 𝑚 = 1.38, recall that units of 𝑚 are units of 𝑦 per unit of 𝑥; That is, millions of tons of production of soybeans per million acres of cultivated land. Thus, production increases by 1.38 million tons of soybeans per million acres of cultivated land. More simply, each acre of cultivated land produces about 1.38 tons of soybeans. 27. a. 𝑦 = Continental net income as a function of 𝑥 = Price of oil. See the technology notes

Solutions Section 1.4 accompanying Example 2 and 3 for the use of technology to obtain regression lines and correlation coefficients. The following result and plot were obtained using the Function Evaluator and Grapher on the Web site with the setup shown:

Regression equation: 𝑦 = −11.85𝑥 + 797.71 Correlation coefficient: 𝑟 ≈ −0.414b. As |𝑟| ≈ 0.414 is significantly less than 0.8, the values of 𝑥 and 𝑦 are not strongly correlated, so that Continental's net income does not appear correlated to the price of oil.c. The points in the graph are nowhere near the regression line, confirming the conclusion in (b).

28. a. 𝑦 = Continental net income as a function of 𝑥 = Price of oil. See the technology notes accompanying Examples 2 and 3 for the use of technology to obtain regression lines and correlation coefficients. The following result and plot were obtained using the Function Evaluator and Grapher on the Web site with the setup shown:

Regression equation: 𝑦 = −28.90𝑥 + 1208.01 Correlation coefficient: 𝑟 ≈ −0.408b. As |𝑟| ≈ 0.408 is significantly less than 0.8, the values of 𝑥 and 𝑦 are not strongly correlated, so that American's net income does not appear correlated to the price of oil.c. The points in the graph are nowhere near the regression line, confirming the conclusion in (b).

29. a. Using 𝑥 = Number of natural science doctorates and 𝑦 = Number of engineering doctorates gives us the following table of values: 𝑥 5,000 5,400 5,900 6,200 6,300

6,700

𝑦 7,600 8,400 9,600 9,500 10,200 10,800

Using one of the technology methods of Example 2 (see the marginal note on using technology), we obtain the following regression line and plot (coefficients rounded to three significant digits): 𝑦 = 1.85𝑥 − 1570

Solutions Section 1.4 Graph:

12000 10000 8000 6000 4000 2000 0 x 4500 5500 6500 b. To interpret the slope, recall that units of the slope are units of 𝑦 (engineering doctorates) per unit of 𝑥 (natural science doctorates). Thus, 𝑚 = 1.85 engineering doctorates per natural science doctorate, indicating that there are around 1.85 additional doctorates in engineering per additional doctorate in the natural sciences. c. Using the technology method of Example 3, we can use technology to show the value of 𝑟 2 : 𝑟 2 ≈ 0.9649 𝑟 = √𝑟 2 ≈ √0.9649 ≈ 0.982 Since 𝑟 is close to 1, the correlation between 𝑥 and 𝑦 is a strong one. d. Yes; the data points are close to and randomly scattered about the regression line.

30. a. Using 𝑥 = Number of social science doctorates and 𝑦 = Number of education doctorates gives us the following table of values: 𝑥 7,900 8,500 8,700 9,000 8,900 8,900 𝑦 5,300 4,800 4,800 5,100 4,800 4,700

5500

5000

4500 x 7500 8000 8500 9000 9500 b. To interpret the slope, recall that units of the slope are units of 𝑦 (education doctorates) per unit of 𝑥 (social science doctorates). Thus, 𝑚 ≈ −0.35 education doctorates per social science doctorate, indicating that there are about 0.35 fewer doctorates in education per additional doctorate in the social sciences. c. Using the technology method of Example 3, we can use technology to show the value of 𝑟 2 : 𝑟 2 ≈ 0.3914 𝑟 = √𝑟 2 ≈ √0.3914 ≈ 0.626 Since 𝑟 is not close to 1, the correlation between 𝑥 and 𝑦 is a poor one. d. No; the data points suggest no relationship in particular; they are far from the regression line and haphazardly scattered.

Solutions Section 1.4 31. a. As 𝑡 is time in years since 2010, we use the following set of data for the regression: 𝑡

𝑦 5,000 5,400 5,900 6,200 6,300 6,700 Using one of the technology methods of Example 2 (see the marginal note on using technology), we obtain the following regression line and plot (coefficients rounded to three significant digits): 𝑦 = 164𝑡 + 5100 Graph:

7000

6000 5000 4000

𝑟 ≈ 0.985 b. Units of the slope are units of 𝑦 (natural science doctorates) per unit of 𝑡 (years); thus, doctorates per year. So, the number of natural science doctorates has been increasing at a rate of about 164 per year. c. The slopes of successive pairs of points do not show an increasing nor decreasing trend as we go from left to right, so the number of natural science doctorates is increasing at a more-or-less constant rate. d. Yes: If 𝑟 had been equal to 1, then the points would lie exactly on the regression line, which would indicate that the number of doctorates is growing at a constant rate. 32. a. As 𝑡 is time in years since 2010, we use the following set of data for the regression: 𝑡

𝑦 7,900 8,500 8,700 9,000 8,900 8,900 Using one of the technology methods of Example 2 (see the marginal note on using technology), we obtain the following regression line and plot (coefficients rounded to three significant digits): 𝑦 = 92.9𝑡 + 8190 Graph:

10000

9000 8000 7000

0 2 4 6 8 10 𝑟 ≈ 0.916, indicating a reasonably good fit. b. Units of the slope are units of 𝑦 (social science doctorates) per unit of 𝑡 (years); thus, doctorates per year. Thus, the number of social science doctorates has been increasing at a rate of about 93 per year. c. The data points suggest a concave-down curve rather than a straight line, indicating that the number of

Solutions Section 1.4 doctorates has been growing at a slower and slower rate (the slopes of successive pairs of points increase as we go from left to right). d. No: If 𝑟 had been equal to 1, then the points would lie exactly on the regression line, which would indicate that the number of doctorates is growing at a constant rate. 33. a. More-or-less constant rate; Exercise 29 suggests a roughly linear relationship between the number of natural science doctorates and the number of engineering doctorates, and Exercise 31 suggests that the number of natural science doctorates has been increasing at a more-or less constant rate. Therefore, the number of engineering doctorates is also increasing at a more-or-less constant rate. b. No; 𝑟 = 1 in Exercise 29 would indicate an exactly linear relationship between the number of natural science doctorates and the number of engineering doctorates, and so the conclusion would be the same. c. No; 𝑟 = 1 in Exercise 31 would indicate that the number of natural science doctorates has been increasing at a constant rate, and so the conclusion would be the same. 34. a. Haphazard; The graph in Exercise 30 suggests that the number of education doctorates has no particular relationship to the number of social science doctorates, and Exercise 32 suggests that the number of social science doctorates has been increasing with time. Therefore, the number of education doctorates would be expected to have no particular relationship to time either; that is, to behave haphazardly with respect to time as well. b. Yes; 𝑟 = 1 in Exercise 30 would indicate an exactly linear relationship between the number of social science doctorates and the number of education doctorates, and so the conclusion would be that the number of education doctorates increased at a faster and faster rate. c. No; 𝑟 = 1 in Exercise 32 would indicate that the number of social science doctorates has been increasing at a constant rate, and so the conclusion would be the same, as the number of education doctorates would still be haphazard. 35. a. Using the method of Example 3, we obtain the following regression line and plot (coefficients rounded to two decimal places): 𝑝 = 0.13𝑡 + 0.22; 𝑟 ≈ 0.97 Graph:

b. The first and last points lie above the regression line, while the central points lie below it, suggesting a curve. c. Here is a worksheet showing the computation of the residuals (based on Example 1 in the text):

Notice that the residuals are positive at first, become negative, and then become positive, confirming the impression from the graph.

Solutions Section 1.4 36. a. Using the method of Example 3, we obtain the following regression line and plot (coefficients rounded to two decimal places): 𝑐 = 35.5𝑡 − 21; 𝑟 ≈ 0.92 Graph:

Notice that the residuals are positive at first, become negative, and then become positive, confirming the impression from the graph. 37. The regression line is defined to be the line that gives the lowest sum-of-squares error, SSE. If we are given two points, (𝑎, 𝑏) and (𝑐, 𝑑) with 𝑎 ≠ 𝑐, then there is a line that passes through these two points, giving SSE = 0. Since 0 is the smallest value possible, this line must be the regression line. 38. SSE = 0; the straight line passing through the given points has predicted values equal to the observed values. Hence, the residuals are zero, giving SSE = 0.

39. If the points (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ), ..., (𝑥𝑛 , 𝑦𝑛 ) lie on a straight line, then the sum-of-squares error, SSE, for this line is zero. Since 0 is the smallest value possible, this line must be the regression line. 40. No. The regression line may pass through none of the given points. 41. Calculation of the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

−𝑎 𝑎 0

𝑎 𝑎

2𝑎

−𝑎 2

𝑎2

2𝑎 2

𝑎2

(The bottom row contains the column sums.) 𝑛 = 3 (number of data points)}

Slope: 𝑚 =

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

3(0) − (0)(2𝑎) =0 3(2𝑎 2) − 0 2

Solutions Section 1.4

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2

Correlation coefficient 𝑟 =

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅

𝑛(∑ 𝑦 2) − (∑ 𝑦) 2

√ √ we have just seen that this numerator is zero. Hence, 𝑟 = 0.

has the same numerator as 𝑚, and

42. Calculation of the regression line: 𝑥 0 0

𝑎

𝑦

𝑥𝑦

𝑥2

−𝑎

𝑎

𝑎2 𝑎2

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points)} 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(0) − (𝑎)(0) Slope: 𝑚 = = =0 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3𝑎 2 − 0 2

Correlation coefficient 𝑟 =

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅

𝑛(∑ 𝑦 2) − (∑ 𝑦) 2

√ √ we have just seen that this numerator is zero. Hence, 𝑟 = 0.

has the same numerator as 𝑚, and

43. No. The regression line through (−1, 1), (0, 0), and (1, 1) passes through none of these points. 44. A mathematical model may only be valid for a limited range of values of the variables concerned, and extrapolation can lead to absurd results. 45. (Answers may vary.) The data in Exercise 35 give 𝑟 ≈ 0.97, yet the plotted points suggest a curve, not a straight line.

46. (Answers may vary.) If 𝑟 is not close to 1, then the points are not close to the regression line; they may be scattered randomly above and below the line in a manner not suggesting a parabola.

Solutions Chapter 1 Review Chapter 1 Review

1. a. 1 b. −2 c. 0 d. 𝑓(2) − 𝑓(−2) = 0 − 1 = −1

2. a. −1 b. −3 c. 0 d. 𝑓(2) − 𝑓(−2) = 0 − (−1) = 1

3. a. 1 b. 0 c. 0 d. 𝑓(1) − 𝑓(−1) = 0 − 1 = −1

4. a. 2 b. −1 c. 0 d. 𝑓(1) − 𝑓(−1) = 0 − 2 = −2

5. 𝑦 = −2𝑥 + 5 𝑦-intercept = 5, slope = −2

6. 2𝑥 − 3𝑦 = 12 Solving for 𝑦 gives

−3𝑦 = −2𝑥 + 12 𝑦 = 23 𝑥 − 4 : 𝑦-intercept = −4, slope = 23

7. 𝑦 =

1 𝑥 2

{𝑥 − 1

if − 1 ≤ 𝑥 ≤ 1

Solutions Chapter 1 Review

if 1 < 𝑥 ≤ 3

8. 𝑓(𝑥) = 4𝑥 − 𝑥 2 with domain [0, 4] Technology formula: 4*x-x^2

9. The graph of the function has a V-shape, indicating an absolute value function.

10. The graph of the function is a straight line, indicating a linear function.

11. The graph of the function is a straight line, indicating a linear function.

12. In the graph, 𝑦 doubles for each 1-unit increase in 𝑥, indicating an exponential model.

14. In the graph, 𝑦 is halved for each 1-unit increase in 𝑥, indicating an exponential model.

Solutions Chapter 1 Review 13. The parabolic shape of the graph indicates a quadratic model.

15. Through (3, 2) with slope −3 Point: (3, 2) Slope: 𝑚 = −3 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 2 − (−3)(3) = 2 + 9 = 11 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −3𝑥 + 11. 16. Through (−2, 4) with slope −1 Point: (−2, 4) Slope: 𝑚 = −2 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 4 − (−1)(−2) = 2 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −𝑥 + 2. 17. Through (1, −3) and (5, 2)

𝑦2 − 𝑦1 2 − (−3) 5 = = = 1.25 𝑥2 − 𝑥1 5−1 4 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = −3 − (1.25)(1) = −4.25 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 1.25𝑥 − 4.25. Point: (1, −3)

Slope: 𝑚 =

18. Through (−1, 2) and (1, 0)

𝑦2 − 𝑦1 0−2 −2 = = = −1 𝑥2 − 𝑥1 1 − (−1) 2 𝑏 = 𝑦1 − 𝑚𝑥1 = 2 − (−1)(−1) = 1 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = −𝑥 + 1. Point: (−1, 2)

Slope: 𝑚 =

Intercept:

19. Through (1, 2) parallel to 𝑥 − 2𝑦 = 2 Point: (1, 2) Slope: Same as slope of 𝑥 − 2𝑦 = 2. To find the slope, solve for 𝑦 : −2𝑦 = −𝑥 − 2 1 1 𝑦 = 𝑥 + 1, so that 𝑚 = . 2 2 1 3 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 2 − (1) = 2 2 1 3 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 + . 2 2 20. Through (−3, 1) parallel to −2𝑥 − 4𝑦 = 5 Point: (−3, 1)

1 5 Slope: Same as slope of −2𝑥 − 4𝑦 = 5. To find the slope, solve for 𝑦, getting 𝑦 = − 𝑥 − . 2 4 1 Thus, 𝑚 = − . 2 1 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 1 + (−3) = − 2 2 1 1 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = − 𝑥 − . 2 2

Solutions Chapter 1 Review 21. With slope 4 crossing 2𝑥 − 3𝑦 = 6 at its 𝑥-intercept We need the 𝑥-intercept of 2𝑥 − 3𝑦 = 6. This is given by setting 𝑦 = 0 and solving for 𝑥 : 2𝑥 − 0 = 6 𝑥=3 Thus, the point is (3, 0) because 𝑦 = 0 on the 𝑥-axis. Slope: 𝑚 = 4 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − 4(3) = −12 Thus, the equation is 𝑦 = 𝑚𝑥 + 𝑏 = 4𝑥 − 12 22. With slope 1/2 crossing 3𝑥 + 𝑦 = 6 at its 𝑥-intercept We need the x-intercept of 3𝑥 + 𝑦 = 6. This is given by setting 𝑦 = 0 and solving for 𝑥 : 3𝑥 + 0 = 6 𝑥=2 Thus, the point is (2, 0) because 𝑦 = 0 on the 𝑥-axis. 1 Slope: 𝑚 = 2 1 Intercept: 𝑏 = 𝑦1 − 𝑚𝑥1 = 0 − (2) = −1 2 Thus, the equation is 1 𝑦 = 𝑚𝑥 + 𝑏 = 𝑥 − 1 2 23. 𝑦 = −𝑥∕2 + 1 : 𝑥

−1

Observed 𝑦 Predicted 𝑦 Residual2 1

1.5

0.25

0.5

2.25

SSE:

2.5

𝑦 = −𝑥∕4 + 1 : 𝑥

−1

Observed 𝑦 Predicted 𝑦 Residual2 1

1.25

0.0625

0.75

1.5625

0.5

0.25

SSE:

1.875

The second line, 𝑦 = −𝑥∕4 + 1, is a better fit.

24. 𝑦 = 𝑥 + 1 : 𝑥

−2

Solutions Chapter 1 Review

Observed 𝑦 Predicted 𝑦 Residual2 −1

−1

SSE:

−1

𝑦 = 𝑥∕2 + 1 :

𝑥

Observed 𝑦 Predicted 𝑦 Residual2

−1

0.5

0.25

1.5

0.25

2.5

0.25

SSE:

5.75

−2

−1

The first line, 𝑦 = 𝑥 + 1, is the better fit. 25.

𝑥

𝑦

−1

𝑥𝑦

−1

𝑥2

𝑦2

(The bottom row contains the column sums.)

𝑛 = 3 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 3(1) − (2)(3) −3 Slope: 𝑚 = ≈ −0.214 = = 14 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 3(6) − 2 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 3 − (−0.214)(2) ≈ 1.14 = 𝑛 3

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = −0.214𝑥 + 1.14. 𝑟=

√

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅

√

𝑛(∑ 𝑦 2) − (∑ 𝑦) 2

3(1) − (2)(3)

√3(6) − 2 2√3(5) − 3 2

≈ −0.33

Solutions Chapter 1 Review 26.

𝑥

𝑦

−2

−1

𝑥𝑦

𝑥2

𝑦2

−1

(The bottom row contains the column sums.)

𝑛 = 6 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 6(20) − (3)(10) 90 Slope: 𝑚 = ≈ 0.857 = = 2 2 2 105 𝑛(∑ 𝑥 ) − (∑ 𝑥) 6(19) − 3 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 10 − (0.857)(3) ≈ 1.24 = 𝑛 6

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = 0.857𝑥 + 1.24. 𝑟=

√

𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 ⋅

√

𝑛(∑ 𝑦 2) − (∑ 𝑦) 2

6(20) − (3)(10)

√6(19) − 3 2√6(32) − 10 2

≈ 0.92

27. a. Graph:

Since the data definitely suggests a curve, we rule out a linear function, leaving us with a choice of quadratic or exponential. Of the two, an exponential function would fit best, given the leveling off we see on the left; the graph of a quadratic function would not flatten out, but instead form a low point and begin rising again toward the left.b. The ratios (rounded to 1 decimal place) are: 𝑉 (1)∕𝑉 (0) 𝑉 (2)∕𝑉 (1) 300 =3 100

𝑉 (3)∕𝑉 (2)

𝑉 (4)∕𝑉 (3)

𝑉 (5)∕𝑉 (4)

𝑉 (6)∕𝑉 (5)

1,000 3,300 10,500 33,600 107,400 ≈ 3.3 ≈ 3.2 ≈ 3.2 ≈ 3.2 = 3.3 300 1,000 3,300 10,500 33,600

They are close to 3.2. c. The data suggest that website traffic is increasing by a factor of around 3.2 per year, so the prediction for next year (year 6) would be around 3.2×107,400 ≈ 343,700 visits per day.

Solutions Chapter 1 Review 28. a. Graph:

Since the data definitely suggest a curve, we rule out a linear function, leaving us with a choice of quadratic or exponential. Of the two, a quadratic function would fit best, given the parabolic shape of the graph. b. The differences (rounded to 1 decimal place) are: 𝐶(1) − 𝐶(0)

−0.32 ≈ −0.3

𝐶(2) − 𝐶(1) 𝐶(3) − 𝐶(2) 𝐶(4) − 𝐶(3) 𝐶(5) − 𝐶(4) −0.1

0.12 ≈ 0.1

0.28 ≈ 0.3

0.48 ≈ 0.5

The rounded differences increase linearly with slope 0.2. c. Assuming the linear trend of differences continue, the next difference 𝐶(6) − 𝐶(5) will be around 0.7, so that the cost of a paperback will be about $5.88 + 0.70 = $6.58.

0.03𝑥 + 2 if 0 ≤ 𝑥 ≤ 50 {0.05𝑥 + 1 if 𝑥 > 50 Notice that 𝑥 is thousands of visit per day, so 10,000 visits corresponds to 𝑥 = 10, and the servers will crash an average of 𝑐(10) = 0.03(10) + 2 = 2.3 times per day. (We used the first formula because 10 is in the interval [0, 50].) For 50,000 visitors, 𝑐(50) = 0.03(50) + 2 = 3.5 crashes per day. (We again used the first formula because 50 is still in the interval [0, 50].) For 100,000 visitors, 𝑐(100) = 0.05(100) + 1 = 6 crashes per day. (We used the second formula because 100 is in the interval (50, +∞).) b. The coefficient 0.03 is the slope of the first formula, indicating that, for website traffic of up to 50,000 visits per day (0 ≤ 𝑥 ≤ 50), the number of crashes is increasing by 0.03 per additional thousand visits. c. To experience 8 crashes in a day, we desire 𝑐(𝑥) = 8. If we try the first formula, we get 0.03𝑥 + 2 = 8 giving 𝑥 = (8 − 2)∕0.03 = 200 , which is not in the domain of the first formula. So, we try the second formula: 0.05𝑥 + 1 = 8 7 0.05𝑥 = 7, so 𝑥 = = 140, 0.05 which is in the domain of the second formula. Thus, we estimate that there were 140,000 visitors that day. 29. a. 𝑐(𝑥) =

1.55𝑥 if 0 ≤ 𝑥 ≤ 100 {1.75𝑥 − 20 if 100 < 𝑥 ≤ 250 𝑠(60) = 1.55(60) = 93 books per day (We used the first formula, since 60 is in [0, 100].) 𝑠(100) = 1.55(100) = 155 books per day (We used the first formula, since 100 is in [0, 100].) 𝑠(160) = 1.75(160) − 20 = 260 books per day (We used the second formula, since 160 is in (100, 250].) b. The coefficient 1.75 is the slope of the second formula, measured in books sold per thousand visitors. Thus, book sales are increasing at a rate of 1.75 books per thousand new visitors when the number of visitors is between 100,000 and 250,000 per day. c. To sell an average of 300 books per day, we desire 𝑛(𝑥) = 300. If we try the first formula, we get 1.55𝑥 = 300, giving 𝑥 ≈ 194, which is not in the domain of the first formula. So, we try the second formula: 30. a.

𝑠(𝑥) =

1.75𝑥 − 20 = 300

Solutions Chapter 1 Review

320 ≈ 182.9 thousand visitors, 1.75 which is in the domain of the second formula. Thus, about 182,900 visitors per day will result in average sales of 300 books per day. 1.75𝑥 = 320, so 𝑥 =

31.

𝑡

𝑛(𝑡) 12.5

37.5

62.5

72.0

74.5

75.0

(a) Technology formulas: (A): 300/(4+100*5^(-t)) (C): -2.3*t^2+30.0*t-3.3

(B): 13.3*t+8.0 (D): 7*3^(0.5*t)

Here are the values for the four given models (rounded to 1 decimal place): 𝑡

(A) 12.5

37.5

62.5

72.1

74.4

74.9

(B) 21.3

34.6

47.9

61.2

74.5

87.8

47.5

66.0

79.9

89.2

93.9

(D) 12.1

21.0

36.4

63.0 109.1 189.0

Model (A) gives an almost perfect fit, whereas the other models are not even close. b. Looking at the table, we see the following behavior as 𝑡 increases: (A) Leveling off (B) Rising (C) Rising (begins to fall after 7 months, however) (D) Rising 32.

𝑡

𝑛(𝑡) 1,330 520

520 1,340 2,980

(a) Technology formulas: (A): 3000/(1+12*2^(-t)) (C): 300*1.6^t

(B): 2000/(4.2-0.7*t) (D): 100*(4.1*t^2-20.4*t+29.5)

Here are the values for the three given models (rounded to the nearest integer): 𝑡

(A)

429

750 1,200 1,714 2,182

(B)

571

714

(C)

480

768 1,229 1,966 3,146

(D) 1,320 510

952

520

1,429 2,857

1,350 3,000

Model (D) gives a close fit, whereas the other models are not even close. b. If you extrapolate the models, you find the following behavior: (A) Leveling off (B) Becomes undefined and then negative (C) Rising (D) Rising

Solutions Chapter 1 Review 33. a. Using 𝑣(𝑐) = −0.000005𝑐 + 0.085𝑐 + 1,750, we get 2

𝑣(5,000) = −0.000005(5,000) 2 + 0.085(5,000) + 1,750 = −125 + 425 + 1,750 = 2,050 𝑣(6,000) = −0.000005(6,000) 2 + 0.085(6,000) + 1,750 = −180 + 510 + 1,750 = 2,080

Thus, increasing monthly advertising from $5,000 to $6,000 per month would result in 2,080 − 2,050 = 30 more visits per day. b. The following table shows the result of increasing expenditure by steps of $1,000: Tech formula: -0.000005*x^2+0.085*x+1750 𝑐 5,000 6,000 7,000 8,000 9,000 10,000

𝑣(𝑐) 2,050 2,080 2,100 2,110 2,110 2,100

The successive changes in the numbers of visits are: 2,080 − 2,050 = 30; 2,100 − 2,080 = 20; 2,110 − 2,100 = 10; 2,110 − 2,110 = 0; 2,100 − 2,110 = −10, showing that the numbers of visits would increase at a slower and slower rate and then begin to decrease. c. Here is a portion of the graph of 𝑣 :

For 𝑐 = 8,500 or larger, we see that website traffic is projected to decrease as advertising increases, and then drop toward zero. Thus, the model does not appear to give a reasonable prediction of traffic at expenditures larger than $8,500 per month. 34. a. Using 𝑐(𝑛) = 0.0008𝑛 2 − 72𝑛 + 2,000,000, we get

𝑐(20,000) = 0.0008(20,000) 2 − 72(20,000) + 2,000,000 = 880,000 𝑐(30,000) = 0.0008(30,000) 2 − 72(30,000) + 2,000,000 = 560,000

Thus, increasing the run size from 20,000 to 30,000 per month would result in a savings of 880, 000 − 560, 000 = 320,000 dollars. b. The following table shows the result of increasing run size in steps of 10,000: Tech formula: 0.0008*x^2-72*x+2000000 𝑛 20,000

30,000

40,000

50,000

60,000

70,000

𝑐(𝑛) 880,000 560,000 400,000 400,000 560,000 880,000 Change

-320,000 -160,000

160,000 320,000

The table shows that the cost decreases at a slower and slower rate and then begins to increase. Going from 30,000 to 40,000 decreases the cost by $160,000—considerably less than going from 20,000 to 30,000.

Solutions Chapter 1 Review

c. Here is a portion of the graph of 𝑣 :

The graph shows that the cost is a minimum for a print run size of around 45,000.

𝑣2 − 𝑣1 2,100 − 2,050 50 = = = 0.05 𝑐2 − 𝑐1 6,000 − 5,000 1,000 Intercept: 𝑏 = 𝑣1 − 𝑚𝑐1 = 2,050 − (0.05)(5,000) = 1,800 Thus, the equation is 𝑣 = 𝑚𝑐 + 𝑏 = 0.05𝑐 + 1,800. b. A budget of $7,000 per month for banner ads corresponds to 𝑣 = 7,000. 𝑣(7,000) = 0.05(7,000) + 1,800 = 2,150 new visitors per day c. We are given 𝑣 = 2,500 and want 𝑐. 2,500 = 0.05𝑐 + 1,800 0.05𝑐 = 2,500 − 1,800 = 700 700 Thus, 𝑐 = = $14,000 per month. 0.05 35. a. Point: (5,000, 2,050)

Slope: 𝑚 =

𝑐2 − 𝑐1 550,000 − 880,000 330,000 = =− = −16.5 𝑛2 − 𝑛1 40,000 − 20,000 20,000 Intercept: 𝑏 = 𝑐1 − 𝑚𝑛1 = 880,000 − (−16.5)(20,000) = 1,210,000 Thus, the equation is 𝑐 = 𝑚𝑛 + 𝑏 = −16.5𝑛 + 1,210,000. b. 𝑐(25,000) = −16.5(25,000) + 1,210,000 = $797,500 c. We are given 𝑐 = 418,000 and want 𝑛. 418,000 = −16.5𝑛 + 1,210,000 −16.5𝑛 = 418,000 − 1,210,000 = −792,000 −792,000 Thus, 𝑛 = = 48,000. −16.5 36. a. Point: (20,000, 880,000)

Slope: 𝑚 =

𝑑2 − 𝑑1 93.5 − 74.5 19 = = = 0.95 𝑤2 − 𝑤1 90 − 70 20 Intercept: 𝑏 = 𝑑1 − 𝑚𝑤1 = 74.5 − (0.95)(70) = 8 Thus, the equation is 𝑑 = 𝑚𝑤 + 𝑏 = 0.95𝑤 + 8. OHagan dropped 90 m, so 𝑑 = 90, and we want 𝑤. 90 = 0.95𝑤 + 8 0.95𝑤 = 90 − 8 = 82 82 Thus, 𝑤 = ≈ 86 kg. .95 37. Point: (𝑤, 𝑑) = (70, 74.5)

Slope: 𝑚 =

𝑇2 − 𝑇1 75 − 80 −5 = = = 0.25 𝑟2 − 𝑟1 120 − 140 −20 Intercept: 𝑏 = 𝑇1 − 𝑚𝑟1 = 80 − (0.25)(140) = 45 Thus, the equation is 𝑇 = 𝑚𝑟 + 𝑏 = 0.25𝑟 + 45. 𝑇 = 65, and we want 𝑟. 65 = 0.25𝑟 + 45 0.25𝑟 = 65 − 45 = 20 38. Point: (𝑟, 𝑇 ) = (140, 80)

Slope: 𝑚 =

Thus, 𝑟 =

20 = 80 chirps/min. 0.25

Solutions Chapter 1 Review

39. a. Cost function: 𝐶 = 𝑚𝑥 + 𝑏, where 𝑏 = fixed cost = $500 per week, and 𝑚 = marginal cost = $5.50 per album Thus, the linear cost function is 𝐶 = 5.5𝑥 + 500. Revenue function: 𝑅 = 𝑚𝑥 + 𝑏, where 𝑏 = fixed revenue = 0, and 𝑚 = marginal revenue = $9.50 per album Thus, the linear revenue function is 𝑅 = 9.5𝑥. Profit function: 𝑃 = 𝑅 − 𝐶 𝑃 = 9.5𝑥 − (5.5𝑥 + 500) = 4𝑥 − 500 b. For breakeven, 𝑃 = 0 4𝑥 − 500 = 0 4𝑥 = 500 500 𝑥= = 125 albums per weekTo make a profit, the company should sell more than this number. 4 c. 𝑅 = 8.00𝑥 𝑃 = 8𝑥 − (5.5𝑥 + 500) = 2.5𝑥 − 500 For breakeven, 500 2.5𝑥 − 500 = 0, so 𝑥 = = 200 2.5 To make a profit, the company should sell more than this number.

40. a. Cost function: 𝐶 = 𝑚𝑥 + 𝑏, where 𝑏 = fixed cost = $900 per month, and 𝑚 = marginal cost = $4 per novel Thus, the linear cost function is 𝐶 = 4𝑥 + 900. Revenue function: 𝑅 = 𝑚𝑥 + 𝑏, where 𝑏 = fixed revenue = 0, and 𝑚 = marginal revenue = $5.50 per novel Thus, the linear revenue function is 𝑅 = 5.50𝑥. Profit function: 𝑃 = 𝑅 − 𝐶 𝑃 = 5.50𝑥 − (4𝑥 + 900) = 5.50𝑥 − 4𝑥 − 900 = 1.50𝑥 − 900b. For breakeven, 𝑃 = 0 1.50𝑥 − 900 = 0 1.50𝑥 = 900 900 𝑥= = 600 novels per monthc. 𝑅 = 5.00𝑥 1.50 𝑃 = 5.00𝑥 − (4𝑥 + 900) = 5𝑥 − 4𝑥 − 900 = 𝑥 − 900 For breakeven, 𝑥 − 900 = 0, so 𝑥 = 900 novels per month 41. a. Demand: We are given two points: (𝑝, 𝑞) = (7, 500) and (9.5, 300) 𝑞 − 𝑞1 300 − 500 −200 Slope: 𝑚 = 2 = = = −80 𝑝2 − 𝑝1 9.5 − 7 2.5 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 500 − (−80)(7) = 500 + 560 = 1,060 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −80𝑝 + 1,060. b. When 𝑝 = $12, the demand is 𝑞 = −80(12) + 1,060 = 100 albums per week c. From Exercise 39, the cost function is 𝐶 = 5.5𝑞 + 500 = 5.5(−80𝑝 + 1,060) + 500 = −440𝑝 + 5,830 + 500 𝐶 = −440𝑝 + 6,330

We use 𝑞 for the monthly sales rather than 𝑥.

We want everything expressed in terms of 𝑝, so we used the demand equation.

To compute the profit in terms of price, we need the revenue as well:

Solutions Chapter 1 Review 𝑅 = 𝑝𝑞 = 𝑝(−80𝑝 + 1,060) = −80𝑝 2 + 1,060𝑝Profit: 𝑃 = 𝑅 − 𝐶𝑃 = −80𝑝 2 + 1,060𝑝 − (−440𝑝 + 6,330) = −80𝑝 2 + 1,500𝑝 − 6,330 Now we compare profits for the three prices: 𝑃 (7.00) = −80(7) 2 + 1,500(7) − 6,330 = $250 𝑃 (9.50) = −80(9.5) 2 + 1,500(9.5) − 6,330 = $700 𝑃 (12) = −80(12) 2 + 1,500(12) − 6,330 = $150 Thus, charging $9.50 will result in the largest weekly profit of $700. 42. a. Demand: We are given two points: (𝑝, 𝑞) = (10, 350) and (5.5, 620) 𝑞 − 𝑞1 620 − 350 270 Slope: 𝑚 = 2 = = = −60 𝑝2 − 𝑝1 5.5 − 10 −4.5 Intercept: 𝑏 = 𝑞1 − 𝑚𝑝1 = 350 − (−60)(10) = 350 + 600 = 950 Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −60𝑝 + 950. b. When 𝑝 = $15, the demand is 𝑞 = −60(15) + 950 = −900 + 950 = 50 novels per month c. From Exercise 40, the cost function is 𝐶 = 4𝑞 + 900 = 4(−60𝑝 + 950) + 900

= −240𝑝 + 3,800 + 900 𝐶 = −240𝑝 + 4,700

We use 𝑞 for the monthly sales rather than 𝑥.

We want everything expressed in terms of 𝑝, so we used the demand equation.

To compute the profit in terms of price, we need the revenue as well: 𝑅 = 𝑝𝑞 = 𝑝(−60𝑝 + 950) = −60𝑝 2 + 950𝑝 Profit: 𝑃 = 𝑅 − 𝐶 𝑃 = −60𝑝 2 + 950𝑝 − (−240𝑝 + 4,700) = −60𝑝 2 + 1,190𝑝 − 4,700 Now we compare profits for the three prices: 𝑃 (5.50) = −60(5.5) 2 + 1,190(5.5) − 4,700 = $30 𝑃 (10) = −60(10) 2 + 1,190(10) − 4,700 = $1,200 𝑃 (15) = −60(15) 2 + 1,190(15) − 4,700 = −$350 (loss) Thus, charging $10 will result in the largest monthly profit of $1,200. 43. a. Calculation of the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

440

3,520

8.5

380

3,230

72.25

250

2,500

100

11.5

180

2,070 132.25

1,250 11,320 368.5

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(11,320) − (38)(1,250) −2,220 Slope: 𝑚 = = = = −74 30 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(368.5) − 38 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 1,250 − (−74)(38) = = 1,015.5 𝑛 4

Solutions Chapter 1 Review Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = −74𝑥 + 1,015.5. Using the variable names 𝑝 and 𝑞 makes this equation 𝑞 = −74𝑝 + 1,015.5. b. 𝑞(10.50) = −74(10.50) + 1,015.5 = 238.5 ≈ 239 albums per week 44. a. Calculation of the regression line: 𝑥

𝑦

𝑥𝑦

𝑥2

5.5

620

3,410

30.25

350

3,500

100

11.5

350

4,025 132.25

300

3,600

1,620 14,535 406.5

144

(The bottom row contains the column sums.)

𝑛 = 4 (number of data points) 𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦) 4(14,535) − (39)(1,620) −5,040 Slope: 𝑚 = = = = −48 105 𝑛(∑ 𝑥 2) − (∑ 𝑥) 2 4(406.5) − 39 2 Intercept: 𝑏 =

∑ 𝑦 − 𝑚(∑ 𝑥) 1,620 − (−48)(39) = = 873 𝑛 4

Thus, the regression line is 𝑦 = 𝑚𝑥 + 𝑏 = −48𝑥 + 873. Using the variable names 𝑝 and 𝑞 makes this equation 𝑞 = −48𝑝 + 873. b. 𝑞(8) = −48(8) + 873 = 489 novels per month

Solutions Chapter 1 Case Study Chapter 1 Case Study

1. Here is the given data, with 𝑡 = 0 representing 2020: 𝑡

𝑦

0.3

1.5

2.6

3.4

4.3

5.0

Using technology, we obtain the following linear regression model: 𝑦 = 0.8893𝑡 − 0.225; 𝑟 ≈ 0.9948 Since 𝑚 ≈ 0.889, MeTube viewership is increasing at a rate of about 0.889 billion views per year. 2. Graph:

The graph does not suggest a quadratic model because the plotted points do not suggest a parabola. 3. Graph with regression parabola:

Regression equation: 𝑦 = −0.001190𝑡 2 + 0.8964𝑡 − 0.2310 The parabola appears to fit no better than the regression line, suggesting that the quadratic model is not appropriate. 4. Here is the tabulated data together with the result of =LINEST(A2:A8,B2:C8,,TRUE):

𝑝 is computed using =TDIST(ABS(E1/E2),F4,2), and returns the value 𝑝 ≈ 0.9662. There is a very low degree of confidence, 1 − 𝑝 ≈ 0.0338, or 3.38%, we can have in asserting that the coefficient of 𝑥 2 is not zero (or that a quadratic model is needed) and so a quadratic model is not appropriate.

Solutions Section 2.1 Section 2.1

1. 𝑓(𝑥) = 2𝑥 2 − 𝑥 − 2 a. 𝑎 = 2, 𝑏 = −1, 𝑐 = −2 b. Table of values: 𝑥

𝑓(𝑥)

−3 19

−2 8

−1 1

−2

−1

c. 𝑓(𝑎 + ℎ) = 2(𝑎 + ℎ) 2 − (𝑎 + ℎ) − 2 = 2(𝑎 2 + 2𝑎ℎ + ℎ 2) − (𝑎 + ℎ) − 2 = 2𝑎 2 + 4𝑎ℎ + 2ℎ 2 − 𝑎 − ℎ − 2 d. 2x^2-x-2 2. 𝑓(𝑥) = −2𝑥 2 + 𝑥 + 2 a. 𝑎 = −2, 𝑏 = −1, 𝑐 = 2 b. Table of values: 𝑥

−3

𝑓(𝑥) −19

−2 −8

−1 −1

−4

−13

c. 𝑓(𝑎 + ℎ) = −2(𝑎 + ℎ) 2 + (𝑎 + ℎ) + 2 = −2(𝑎 2 + 2𝑎ℎ + ℎ 2) + (𝑎 + ℎ) + 2 = −2𝑎 2 − 4𝑎ℎ − 2ℎ 2 + 𝑎 + ℎ + 2 d. -2x^2+x+2 3. 𝑓(𝑥) = 10𝑥 2 − 5𝑥 a. 𝑎 = 10, 𝑏 = −5, 𝑐 = 0 b. Table of values: 𝑥

−3

𝑓(𝑥) 105

−2 50

−1 15

0 0

c. 𝑓(𝑎 + ℎ) = 10(𝑎 + ℎ) 2 − 5(𝑎 + ℎ) = 10(𝑎 2 + 2𝑎ℎ + ℎ 2) − 5(𝑎 + ℎ) = 10𝑎 2 + 20𝑎ℎ + 10ℎ 2 − 5𝑎 − 5ℎ d. 10x^2-5x 4. 𝑓(𝑥) = −𝑥 2 − 50 a. 𝑎 = −1, 𝑏 = 0, 𝑐 = −50 b. Table of values: 𝑥

−3

𝑓(𝑥) −59

−2

−54

−1

−51

−50

−51

−54

−59

c. 𝑓(𝑎 + ℎ) = −(𝑎 + ℎ) 2 − 50 = −(𝑎 2 + 2𝑎ℎ + ℎ 2) − 50 = −𝑎 2 − 2𝑎ℎ − ℎ 2 − 50 d. -(x^2)-50 (See the margin note next to Quick Example 2 in the text as to the reason for the parentheses.) 5. 𝑓(𝑥) = −𝑥 2 − 𝑥 − 1 a. 𝑎 = −1, 𝑏 = −1, 𝑐 = −1

Solutions Section 2.1 b. Table of values: 𝑥

𝑓(𝑥)

−3 −7

−2 −3

−1 −1

−1

−3

−7

−13

c. 𝑓(𝑎 + ℎ) = −(𝑎 + ℎ) 2 − (𝑎 + ℎ) − 1 = −(𝑎 2 + 2𝑎ℎ + ℎ 2) − (𝑎 + ℎ) − 1 = −𝑎 2 − 2𝑎ℎ − ℎ 2 − 𝑎 − ℎ − 1 d. -(x^2)-x-1 (See margin note next to Quick Example 2 in the textbook as to the reason for the parentheses.) 6. 𝑓(𝑥) = −3𝑥 2 + 3𝑥 − 1 a. 𝑎 = −3, 𝑏 = 3, 𝑐 = −1 b. Table of values: 𝑥

−3

𝑓(𝑥) −37

−2

−19

−1 −7

−1

−7

−19

c. 𝑓(𝑎 + ℎ) = −3(𝑎 + ℎ) 2 + 3(𝑎 + ℎ) − 1 = −3(𝑎 2 + 2𝑎ℎ + ℎ 2) + 3(𝑎 + ℎ) − 1 = −3𝑎 2 − 6𝑎ℎ − 3ℎ 2 + 3𝑎 + 3ℎ − 1 d. -3x^2+3x-1

7. 𝑓(𝑥) = 𝑥 2 + 3𝑥 + 2. 𝑎 = 1, 𝑏 = 3, 𝑐 = 2; −𝑏∕(2𝑎) = −3∕2, 𝑓(−3∕2) = −1∕4, so: vertex: (−3∕2, −1∕4), 𝑦-intercept = 𝑐 = 2. 𝑥 2 + 3𝑥 + 2 = (𝑥 + 1)(𝑥 + 2), so: 𝑥-intercepts: −2, −1. 𝑎 > 0 so the parabola opens upward.

8. 𝑓(𝑥) = −𝑥 2 − 𝑥. 𝑎 = −1, 𝑏 = −1, 𝑐 = 0; −𝑏∕(2𝑎) = −1∕2, 𝑓(−1∕2) = 1∕4, so: vertex: (−1∕2, 1∕4), 𝑦-intercept = 𝑐 = 0. −𝑥 2 − 𝑥 = −𝑥(𝑥 + 1), so: 𝑥-intercepts: −1, 0. 𝑎 < 0 so the parabola opens downward.

9. 𝑓(𝑥) = −𝑥 2 + 4𝑥 − 4. 𝑎 = −1, 𝑏 = 4, 𝑐 = −4; −𝑏∕(2𝑎) = 2, 𝑓(2) = 0, so: vertex: (2, 0), 𝑦-intercept = 𝑐 = −4. −𝑥 2 + 4𝑥 − 4 = −(𝑥 − 2) 2, so: 𝑥-intercept: 2. 𝑎 < 0 so the parabola opens downward.

Solutions Section 2.1

10. 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 1. 𝑎 = 1, 𝑏 = 2, 𝑐 = 1; −𝑏∕(2𝑎) = −1, 𝑓(−1) = 0, so: vertex: (−1, 0), 𝑦-intercept = 𝑐 = 1. 𝑥 2 + 2𝑥 + 1 = (𝑥 + 1) 2, so: 𝑥-intercept: −1. 𝑎 > 0 so the parabola opens upward.

11. 𝑓(𝑥) = −𝑥 2 − 40𝑥 + 500. 𝑎 = −1, 𝑏 = −40, 𝑐 = 500; −𝑏∕(2𝑎) = −20, 𝑓(−20) = 900, so: vertex: (−20, 900), 𝑦-intercept = 𝑐 = 500. −𝑥 2 − 40𝑥 + 500 = −(𝑥 + 50)(𝑥 − 10), so: 𝑥-intercepts: −50, 10. 𝑎 < 0 so the parabola opens downward.

12. 𝑓(𝑥) = 𝑥 2 − 10𝑥 − 600. 𝑎 = 1, 𝑏 = −10, 𝑐 = −600; −𝑏∕(2𝑎) = 5; 𝑓(5) = −625, so: vertex: (5, −625), 𝑦-intercept = 𝑐 = −600. 𝑥 2 − 10𝑥 − 600 = (𝑥 + 20)(𝑥 − 30), so: 𝑥-intercepts: −20, 30. 𝑎 > 0 so the parabola opens upward.

13. 𝑓(𝑥) = 𝑥 2 + 𝑥 − 1. 𝑎 = 1, 𝑏 = 1, 𝑐 = −1; −𝑏∕(2𝑎) = −1∕2; 𝑓(−1∕2) = −5∕4, so: vertex: (−1∕2, −5∕4), 𝑦-intercept = 𝑐 = −1. From the quadratic formula: 𝑥-intercepts: −1∕2 ± √5∕2. 𝑎 > 0 so the parabola opens upward.

Solutions Section 2.1

14. 𝑓(𝑥) = 𝑥 2 + √2𝑥 + 1. 𝑎 = 1, 𝑏 = √2, 𝑐 = 1; −𝑏∕(2𝑎) = −√2∕2; 𝑓(−√2∕2) = 1∕2, so: vertex: (−√2∕2, 1∕2), 𝑦-intercept = 𝑐 = 1. 𝑏 2 − 4𝑎𝑐 = −2 < 0, so no 𝑥-intercept. 𝑎 > 0 so the parabola opens upward.

15. 𝑓(𝑥) = 𝑥 2 + 1. 𝑎 = 1, 𝑏 = 0, 𝑐 = 1; −𝑏∕(2𝑎) = 0, 𝑓(0) = 1, so: vertex: (0, 1), 𝑦-intercept = 𝑐 = 1. 𝑏 2 − 4𝑎𝑐 = −4 < 0, so no 𝑥-intercept. 𝑎 > 0 so the parabola opens upward.

16. 𝑓(𝑥) = −𝑥 2 + 5. 𝑎 = −1, 𝑏 = 0, 𝑐 = 5; −𝑏∕(2𝑎) = 0, 𝑓(0) = 5, so vertex: (0, 5), 𝑦-intercept = 𝑐 = 1. From the quadratic formula: 𝑥-intercepts: −√5, √5. 𝑎 < 0 so the parabola opens downward.

17. 𝑞 = −4𝑝 + 100, 𝑅 = 𝑝𝑞 = 𝑝(−4𝑝 + 100) = −4𝑝 2 + 100𝑝; maximum revenue when 𝑝 = −𝑏∕(2𝑎) = $12.50

Solutions Section 2.1

18. 𝑞 = −3𝑝 + 300, 𝑅 = 𝑝𝑞 = 𝑝(−3𝑝 + 300) = −3𝑝 2 + 300𝑝, maximum revenue when 𝑝 = −𝑏∕(2𝑎) = $50

19. 𝑞 = −2𝑝 + 400, 𝑅 = 𝑝𝑞 = 𝑝(−2𝑝 + 400) = −2𝑝 2 + 400𝑝, maximum revenue when 𝑝 = −𝑏∕(2𝑎) = $100

20. 𝑞 = −5𝑝 + 1200, 𝑅 = 𝑝𝑞 = 𝑝(−5𝑝 + 1200) = −5𝑝 2 + 1200𝑝, maximum revenue when 𝑝 = −𝑏∕(2𝑎) = $120

21. 𝑦 = −0.7955𝑥 2 + 4.4591𝑥 − 1.6000 22. 𝑦 = −0.7955𝑥 2 − 4.4591𝑥 − 1.6000

23. 𝑦 = −1.1667𝑥 − 6.1667𝑥 − 3.0000 2

Solutions Section 2.1

24. 𝑦 = −0.3333𝑥 2 + 1.6667𝑥 + 3.0000 25. a. Positive because the data suggest a curve that is concave up. b. The data suggest a parabola that is concave up (𝑎 positive) and with 𝑦-intercept around 1,840. Only choice (C) has both properties. c. −𝑏∕(2𝑎) = 44∕12 ≈ 4, which is 2014. Extrapolating in the positive direction leads one to predict more and more steeply rising military expenditure, which may or may not occur; extrapolating in the negative direction predicts more and more steeply increasing military expenditure as we go back in time, contradicting history — military expenditures have risen and fallen many times. 26. a. Negative because the data suggest a curve that is concave down. b. The data suggest a parabola that is concave down (𝑎 negative) and with 𝑦-intercept around 72. Only choice (B) has both properties. c. −𝑏∕(2𝑎) = 0.7∕0.16 = 4.375, which is closest to 2014. Extrapolating in either direction leads one to predict eventually negative values for the funding, which doesn't make sense. 27. The given function 𝐼(𝑡) = 11𝑡 2 − 170𝑡 + 1,300 is quadratic with graph a concave-up parabola (as 𝑎 = 11 is positive). Thus, its vertex is the lowest point on the graph and occurs when 𝑡 = −𝑏∕(2𝑎) = 170∕22 ≈ 7.73 ≈ 8, corresponding to 2018. At that time, imports were about

𝐼(10.26) = 11(7.73) 2 − 170(7.73) + 1,300 ≈ 640 thousand barrels/day (to two significant digits).

28. The given function 𝑃 (𝑡) = −0.01𝑡 2 + 0.01𝑡 + 2.5 is quadratic with graph a concave-down parabola (as 𝑎 = −0.01 is negative). Thus, its vertex is the highest point on the graph and occurs when 𝑡 = −𝑏∕(2𝑎) = 0.01∕0.02 = 0.5, midway through 2010. At that time, production was approximately 𝑃 (0.5) = −0.01(0.5) 2 + 0.01(0.5) + 2.5 ≈ 2.5 million barrels/day (to two significant digits).

29. a. The vertex of 𝑦 = 0.4𝑥 2 + 𝑥 + 26.5 occurs when 𝑥 = −𝑏∕(2𝑎) = −1∕0.8 = −1.25 ≈ −1, which would correspond to 2019. b. 𝑟(𝑥) = 0.4𝑥 2 + 𝑥 + 26.5 has domain [0, 5], and its value increases as 𝑥 increases from 0 (all the coefficients are positive). So, its lowest value occurs at 𝑥 = 0, corresponding to 2020. (One can also graph the given function to verify that its lowest value occurs at 𝑥 = 0.) c. The 𝑥-coordinate of the vertex in part (a) is outside the domain [0, 5] of 𝑟(𝑥) and so the model does not apply to that value of 𝑥. d. From the answer to part (b), the lowest revenue is 𝑟(0) = 0.4(0) 2 + 0 + 26.5 = 26.5 billion dollars. 30. a. The vertex of 𝑦 = −0.2𝑥 2 + 3𝑥 + 18 occurs when 𝑥 = −𝑏∕(2𝑎) = 3∕0.4 = 7.5 ≈ 8, which would correspond to 2028. b. 𝑟(𝑥) = −0.2𝑥 2 + 3𝑥 + 18 has domain [0, 5], and its graph is seen to increase as 𝑥 increases from 0 to 5. So, its highest value occurs at 𝑥 = 5, corresponding to 2025. c. The 𝑥-coordinate of the vertex in part (a) is outside the domain [0, 5] of 𝑟(𝑥) and so the model does not apply to that value of 𝑥. d. From the answer to part (b), the highest revenue is 𝑟(5) = −0.2(5) 2 + 3(5) + 18 = 28 billion dollars. 31. 𝑞 = −0.5𝑝 + 140. Revenue is 𝑅 = 𝑝𝑞 = −0.5𝑝 2 + 140𝑝. Maximum revenue occurs when 𝑝 = −𝑏∕(2𝑎) = $140; the corresponding revenue is 𝑅 = $9, 800. 32. 𝑞 = −2𝑝 + 320. Revenue is 𝑅 = 𝑝𝑞 = −2𝑝 2 + 320𝑝. Maximum revenue occurs when

Solutions Section 2.1 𝑝 = −𝑏∕(2𝑎) = $80; the corresponding revenue is 𝑅 = $12, 800.

33. The given data points are (𝑝, 𝑞) = (40, 200, 000) and (60, 160, 000). The line passing through these points is 𝑞 = −2000𝑝 + 280, 000. Revenue is 𝑅 = 𝑝𝑞 = −2000𝑝 2 + 280, 000𝑝. Maximum revenue occurs when 𝑝 = −𝑏∕(2𝑎) = 70 houses; the corresponding revenue is 𝑅 = $9, 800, 000. 34. The given data points are (𝑝, 𝑞) = (50, 190, 000) and (70, 170, 000). The line passing through these points is 𝑞 = −1000𝑝 + 240, 000. Revenue is 𝑅 = 𝑝𝑞 = −1000𝑝 2 + 240, 000𝑝. Maximum revenue occurs when 𝑝 = −𝑏∕(2𝑎) = 120 houses; the corresponding revenue is 𝑅 = $14, 400, 000. 35. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏 (𝑥 is the price 𝑝 and 𝑦 is the demand 𝑞). We are given two points on its graph: (3, 28, 000) and (5, 19, 000). Slope: 𝑚 = 𝑝2 −𝑝1 = 19,000−28,000 = −9,000 = −4, 500 5−3 2 𝑞 −𝑞 2

Intercept:

𝑏 = 𝑞1 − 𝑚𝑝1 = 28, 000 − (−4, 500)(3) = 28, 000 + 13, 500 = 41, 500

Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −4, 500𝑝 + 41, 500. b. Revenue: 𝑅= 𝑝𝑞 = 𝑝(−4, 500𝑝 + 41, 500)= −4, 500𝑝 2 + 41, 500𝑝

For maximum revenue,

𝑝 = −𝑏 ≈$4.61. = −41,500 2𝑎 −900

The corresponding daily revenue is

𝑅 = −4500(4.61) 2 + 41, 500(4.61) = $95, 680.55.

c. The maximum annual revenue the company could have earned was 95, 680.55 × 365≈$34, 923, 400, which is about $10 million short of what it needed to break even. Therefore, it would not have been possible to break even.

36. a. A linear demand function has the form 𝑞 = 𝑚𝑝 + 𝑏 (𝑥 is the price 𝑝 and 𝑦 is the demand 𝑞). We are given two points on its graph: (5, 14) and (3, 18). Slope: 4 𝑚 = 𝑝2 −𝑝1 = 18−14 = −2 = −2 3−5 𝑞 −𝑞 2

Intercept:

𝑏 = 𝑞1 − 𝑚𝑝1 = 14 − (−2)(5) = 24

Thus, the demand equation is 𝑞 = 𝑚𝑝 + 𝑏 = −2𝑝 + 24. b. Revenue: 𝑅= 𝑝𝑞 = 𝑝(−2𝑝 + 24)= −2𝑝 2 + 24𝑝

For maximum revenue,

𝑝 = −𝑏 = −24 =Ż6. 2𝑎 −4

The corresponding daily revenue is

𝑅 = −2(6) 2 + 24(6) =Ż72 million.

Solutions Section 2.1

c. The maximum annual revenue the company could have earned was 72, 000, 000 × 670 =Ż48,240,000,000, which is about Ż240 million more than it needed to finance the lab. Therefore, it would have been possible to finance the lab.

37. a. The data points are (𝑥, 𝑞) = (2, 280) and (1.5, 560). Thus, 𝑞 = −560𝑥 + 1, 400 and 𝑅 = 𝑥𝑞 = −560𝑥 2 + 1, 400𝑥. b. 𝑃 = 𝑅 − 𝐶 = −560𝑥 2 + 1400𝑥 − 30. The largest monthly profit occurs when 𝑥 = −𝑏∕(2𝑎) = $1.25 and then 𝑃 = $845 per month.

38. a. The data points are (𝑥, 𝑞) = (8, 400) and (4, 600). Thus, 𝑞 = −50𝑥 + 800 and 𝑅 = −50𝑥 2 + 800𝑥. b. 𝑃 = 𝑅 − 𝐶 = −50𝑥 2 + 800𝑥 − 500. The largest weekly profit occurs when 𝑥 = −𝑏∕(2𝑎) = $8 per T-shirt and then 𝑃 = $2700 per week. 39. As a function of 𝑞, 𝐶 = 0.5𝑞 + 20. Substituting 𝑞 = −400𝑥 + 1, 200, we get 𝐶 = 0.5(−400𝑥 + 1, 200) + 20 = −200𝑥 + 620.

The profit is 𝑃 = 𝑅 − 𝐶 = 𝑥𝑞 − 𝐶 = −400𝑥 2 + 1400𝑥 − 620. The profit is largest when 𝑥 = −𝑏∕(2𝑎) =$1.75 per log-on; the corresponding profit is 𝑃 =$605 per month. 40. As a function of 𝑞, 𝐶 = 5𝑞 + 400. Substituting for 𝑞, we get 𝐶 = 5(−40𝑥 + 600) + 400 = −200𝑥 + 3400.

The profit is 𝑃 = 𝑅 − 𝐶 = 𝑥𝑞 − 𝐶 = −40𝑥 2 + 800𝑥 − 3400. The profit is largest when 𝑥 = −𝑏∕(2𝑎) =$10 per T-shirt; the corresponding profit is 𝑃 =$600 per week.

41. a. The data points are (𝑝, 𝑞) = (10, 300) and (15, 250), so 𝑞 = −10𝑝 + 400. b. 𝑅 = 𝑝𝑞 = −10𝑝 2 + 400𝑝 c. 𝐶 = 3𝑞 + 3000 = 3(−10𝑝 + 400) + 3, 000 = −30𝑝 + 4, 200 d. 𝑃 = 𝑅 − 𝐶 = −10𝑝 2 + 430𝑝 − 4, 200. The maximum profit occurs when 𝑝 = −𝑏∕(2𝑎) =$21.50.

42. a. The data points are (𝑝, 𝑞) = (2500, 15) and (2000, 20), so 𝑞 = −0.01𝑝 + 40. b. 𝑅 = 50𝑝𝑞 = 50𝑝(−0.01𝑝 + 40) = −0.5𝑝2 + 2000𝑝 c. (i) 𝐶 = 120, 000 + 80, 000𝑞, so (ii) 𝐶 = 120, 000 + 80, 000(−0.01𝑝 + 40) = −800𝑝 + 3, 320, 000. d. 𝑃 = 𝑅 − 𝐶 = −0.5𝑝 2 + 2800𝑝 − 3, 320, 000. The maximum profit occurs when 𝑝 = −𝑏∕(2𝑎) =$2800 per hour.

Solutions Section 2.1 43. Here is the Excel tabulation of the data, together with the scatter plot and the quadratic (polynomial order 2) trendline (with the option "Display equation on chart" checked):

From the trendline, the quadratic model is 𝑓(𝑡) = 6𝑡 2 − 46𝑡 + 1,840. To estimate world military expenditure in 2017, substitute the corresponding value 𝑡 = 7, to obtain 𝑓(7) = 6(7) 2 − 46(7) + 1,840 = 1,812 representing $1,812 billion. The actual figure (from Exercise 25) is $1,850 billion, so the predicted value is $38 billion lower than the actual value. 44. Here is the Excel tabulation of the data, together with the scatter plot and the quadratic (polynomial order 2) trendline (with the option "Display equation on chart" checked):

From the trendline, the quadratic model is 𝑓(𝑡) = −0.0731𝑡 2 + 0.616𝑡 + 72.08. To estimate the amount of funding in 2015, substitute the corresponding value 𝑡 = 5, to obtain 𝑓(5) = −0.0731(5) 2 + 0.616(5) + 72.08 ≈ 73.33, or $73.33 billion. The actual amount of funding (from Exercise 26) was $73.33 billion which agrees with the predicted figure to two decimal places.

Solutions Section 2.1 45. a. Here is the Excel tabulation of the data, together with a scatter plot and quadratic (polynomial order 2) trendline. The option "Display equation on chart" has been checked.

We round the coefficients to two significant digits to get 𝑆(𝑡) = −0.70𝑡 2 − 6.0𝑡 + 50. b. To predict the sales in 2015 we substitute 𝑡 = 5 and compute 𝑆(5) = −0.70(5) 2 − 6.0(5) + 50 = 2.5 ≈ 3 million units. \\Likewise, the 2016 prediction is 𝑆(6) = −0.70(6) 2 − 6.0(6) + 50 = −11.2 ≈ −11 million units. Even though we expected 2015 sales to be much lower than the 2014 sales, the 2016 prediction, being negative, shows the danger of extrapolating curve-fitting models. 46. a. Here is the Excel tabulation of the data, together with a scatter plot and quadratic (polynomial order 2) trendline. The option "Display equation on chart" has been checked.

We round the coefficients to two significant digits to get 𝑆(𝑡) = −3.2𝑡 2 + 20𝑡 + 22. b. To predict the sales in 2010 we substitute 𝑡 = 5 and compute 𝑆(5) = −3.2(5) 2 + 20(5) + 22 = 42 million units. \\Likewise, the 2011 prediction is 𝑆(6) = −3.2(6) 2 + 20(6) + 22 = 26.8 ≈ 27 million units. These values are far from the figures of 50.4 and 42.6 million units given in the table in the preceding exercise. This shows the danger of shows the danger of extrapolating curve-fitting models.

Solutions Section 2.1 47. If 𝑎 = 0, then 𝑓(𝑥) = 𝑏𝑥 + 𝑐, a linear function, so its graph is a straight line.

48. Since 𝑐 is the 𝑦-intercept of the graph, 𝑐 = 0 implies that the graph has a 𝑦-intercept of zero, meaning that it passes through the origin.

49. Since the curve is concave up, 𝑎 is positive. Since the 𝑦-intercept is negative, 𝑐 is negative. Hence the correct choice is (C). 50. Since the curve is concave down, 𝑎 is negative. Since the 𝑦-intercept is positive, 𝑐 is positive. Hence the correct choice is (B).

51. Positive; The 𝑥-coordinate of the vertex is negative, so −𝑏∕(2𝑎) must be negative. Since 𝑎 is positive (the parabola is concave up), this means that 𝑏 must also be positive to make −𝑏∕(2𝑎) negative. 52. Positive; The 𝑥-coordinate of the vertex is positive, so −𝑏∕2𝑎 must be positive. Since 𝑎 is negative (the parabola is concave down), this means that 𝑏 must be positive to make −𝑏∕2𝑎 positive.

53. The 𝑥-coordinate of the vertex represents the unit price that leads to the maximum revenue, the 𝑦-coordinate of the vertex gives the maximum possible revenue, the 𝑥-intercepts give the unit prices that result in zero revenue, and the 𝑦-intercept gives the revenue resulting from zero unit price (which is obviously zero). 54. The 𝑥-coordinate of the vertex gives the time at which the stone reaches its highest point, the 𝑦-coordinate of the vertex gives the maximum height, the 𝑥-intercepts give the times that the stone is at zero height, while the 𝑦-intercept gives the height of the stone at time zero. 55. Graph the data to see whether the points suggest a curve rather than a straight line. If the curve suggested by the graph is concave up or concave down, then a quadratic model would be a likely candidate. 56. Graphing the data shows that neither a linear model nor a quadratic model is appropriate. The curve suggested by the graph is concave down on [0, 4] and concave up on [4, 8]. This is unlike a parabola, which is either always concave up or concave down. 57. No; The graph of a quadratic function is a parabola. In the case of a concave-up parabola, the curve would unrealistically predict sales increasing extremely fast and becoming unrealistically large in the future. In the case of a concave-down parabola, the curve would predict "negative" sales from some point on. 58. Answers may vary. Some uses are: using the model to interpolate; that is, predict missing values inside the domain; using the model for short-term prediction; using a quadratic model to determine maximum possible revenue or profit.

59. If 𝑞 = 𝑚𝑝 + 𝑏 (with 𝑚 < 0), then the revenue is given by 𝑅 = 𝑝𝑞 = 𝑚𝑝 2 + 𝑏𝑝. This is the equation of a parabola with 𝑎 = 𝑚 < 0 and so is concave down. Thus, the vertex is the highest point on the parabola, showing that there is a single highest value for 𝑅, namely, the 𝑦-coordinate of the vertex. 60. The given equation is the equation of a concave-up parabola, and thus has the vertex as its lowest point. The 𝑦-coordinate of the vertex therefore gives the lowest possible average cost, which occurs when 𝑥 is assigned the value of the 𝑥-coordinate of the vertex.

Solutions Section 2.1

61. Since 𝑅 = 𝑝𝑞, the demand must be given by 𝑞 = 𝑅𝑝 = −50𝑝 𝑝+60𝑝 = −50𝑝 + 60. 2

62. Since 𝑅 = 𝑝𝑞, the demand is given by 𝑞 = 𝑅𝑝 = −50𝑝 +60𝑝+50 . This is not linear. = −50𝑝 + 60 + 50 𝑝 𝑝

Solutions Section 2.2 Section 2.2 1. 4^x 𝑥

𝑓(𝑥)

2. 3^x 𝑥

𝑓(𝑥)

−3

−2

−1

1 4

−3

−2

−1

2 9

−2

−1

2 1 9

1 27

−2

−1

1 1 4

1 16

1 64

2 8

1 64

1 16

1 27

1 9

3. 3^(-x) 𝑥

𝑓(𝑥)

−3 27

4. 4^(-x) 𝑥

𝑓(𝑥)

−3 64

1 4

1 3

5. 2*2^x or 2*(2^x) 𝑥

𝑔(𝑥)

−3

−2

1 4

1 2

−1 1

6. 2*3^x or 2*(3^x) 𝑥

𝑔(𝑥)

−3 2 27

𝑥

−3

ℎ(𝑥) −24

1 3

−2

−1

1 6

−2

−1

2 9

7. -3*2^(-x)

−12

2 3

−6

−3

−

3 2

−

3 4

−

3 8

Solutions Section 2.2 8. -2*3^(-x) 𝑥

−3

ℎ(𝑥) −54 9. 2^x-1 𝑥

𝑟(𝑥)

−3

−

7 8

10. 2^(-x)+1 𝑥

𝑟(𝑥)

−3 9

11. 2^(x-1) 𝑥

𝑠(𝑥)

−3 1 16

12. 2^(1-x) 𝑥

𝑠(𝑥) 13.

−3 16

−2

−1 −6

−2

−

2 3

−

2 9

−

−2 3 4

−1

−

−2

−1

−2

−1

−2

−1

−18

−

1 8

1 2

1 4

1 2

3 2

5 4

1 2

2 27

9 8

1 4 14.

Solutions Section 2.2 15.

16.

17.

18.

19.

𝑥

𝑓(𝑥) 𝑔(𝑥)

−2

0.5 8

−1

1.5 4

4.5 2

13.5 1

40.5 1 2

For every increase in 𝑥 by one unit, the value of 𝑓 is multiplied by 3, so 𝑓 is exponential. Since 𝑓(0) = 4.5, the exponential model is 𝑓(𝑥) = 4.5(3 𝑥). For every increase in 𝑥 by one unit, the value of 𝑔 is multiplied by 1/2, so 𝑔 is exponential. Since 𝑔(0) = 2, the exponential model is 𝑔(𝑥) = 2(1∕2) 𝑥, or 2(2 −𝑥). 20.

𝑥

−2

−1

𝑔(𝑥)

−1

𝑓(𝑥)

1 2

For every increase in 𝑥 by one unit, the value of 𝑓 is multiplied by 2, so 𝑓 is exponential. The values of 𝑔 decrease and then increase, so 𝑔 is not exponential. Since 𝑓(0) = 2, the exponential model is 𝑓(𝑥) = 2(2 𝑥).

𝑥

−2

𝑔(𝑥)

0.3

𝑥

−2

21.

−1

𝑓(𝑥) 22.5

7.5 0.9

Solutions Section 2.2

2.5

7.5

2.7

8.1

22.5 16.2

When 𝑥 increases from −1 to 0, the value of 𝑓 is multiplied by 1/3, but when 𝑥 is increased from 0 to 1, the value of 𝑓 is multiplied by 3. So 𝑓 is not exponential. When 𝑥 increases from −1 to 0, the value of 𝑔 is multiplied by 3, but when 𝑥 is increased from 1 to 2, the value of 𝑔 is multiplied by 2. So 𝑔 is not exponential. 22.

𝑓(𝑥) 𝑔(𝑥)

−1

0.3

0.9

2.7

8.1

24.3

1.5

0.75 0.375 0.1875

−1

For every increase in 𝑥 by one unit, the value of 𝑓 is multiplied by 3, so 𝑓 is exponential. Since 𝑓(0) = 2.7, the exponential model is 𝑓(𝑥) = 2.7(3 𝑥). For every increase in 𝑥 by one unit, the value of 𝑔 is multiplied by 0.5, so 𝑔 is exponential. Since 𝑔(0) = 0.75, the exponential model is 𝑔(𝑥) = 0.75(0.5) 𝑥, or 0.75(2 −𝑥). 𝑥

23.

−2

𝑓(𝑥) 100 𝑔(𝑥) 100

200 20

400

600

800

0.8

0.16

The values of 𝑓(𝑥) double for every one-unit increase in 𝑥 except when 𝑥 increases from 1 to 2, when the value of 𝑓 is multiplied by 4/3. Hence 𝑓 is not exponential. On the other hand, 𝑔 is multiplied by 0.2 for every increase by one unit in 𝑥, so 𝑔 is exponential. Since 𝑔(0) = 4, the exponential model is 𝑔(𝑥) = 4(0.2) 𝑥. 𝑥

24.

𝑓(𝑥) 𝑔(𝑥)

−2

−1

0.8

0.2

0.1 20

0.05 0.025

The value of 𝑓(𝑥) is multiplied by 1/4 when 𝑥 increases from −2 to −1 but halved 𝑥 when increases from −1 to 0. Hence 𝑓 is not exponential. The value of 𝑔(𝑥) is multiplied by 1/2 when 𝑥 increases from −2 to −1 but multiplied by 0.2 when 𝑥 increases from 1 to 2. Hence 𝑔 is not exponential. 25. 2.718^(-2*x) 𝑥

−3

−2

−1

𝑓(𝑥) 403.2 54.58 7.388

0 1

0.1354 0.01832 0.002480

Solutions Section 2.2 26. 2.718^(x/5) 𝑥

−3

−2

−1

𝑔(𝑥) 0.5488 0.6703 0.8187 27. 1.01*2.02^(-4*x) 𝑥

−3

−2

1.221 1.492 1.822

−1

ℎ(𝑥) 4 662 280.0 16.82 1.01 0.06066 0.003643 0.0002188 28. 3.42*3^(-x/5) 𝑥

−3

−2

ℎ(𝑥) 6.612 5.308 4.261 3.42 2.745 2.204 1.769 29. 50*(1+1/3.2)^(2*x) 𝑥

−3

−2

−1

𝑟(𝑥) 9.781 16.85 29.02

30. 0.043*(4.5-5/1.2)^(-x) 𝑥

−3

−2

−1

86.13 148.4 255.6 0

𝑟(𝑥) 0.001592 0.004778 0.01433 0.043 0.129 0.387 1.161 31. The following solutions also show some common errors you should avoid. 2^(x-1) not 2^x-1 32. 2^(-4*x) not 2^-4*x 33. 2/(1-2^(-4*x)) not 2/1-2^-4*x and not 2/1-2^(-4*x) 34. 2^(3-x)/(1-2^x) or (2^(3-x))/(1-2^x) not 2^3-x/1-2^x and not 2^3-x/(1-2^x) 35. (3+x)^(3*x)/(x+1) or ((3+x)^(3*x))/(x+1)not (3+x)^(3*x)/x+1 and not (3+x^(3*x))/(x+1) 36. 20.3^(3*x)/(1+20.3^(2*x)) or (20.3^(3*x))/(1+20.3^(2*x))not 3^(3*x)/1+20.3^(2*x) and not (20.3^3*x)/(1+20.3^2*x) 37. 2*e^((1+x)/x) or 2*EXP((1+x)/x)not 2*e^1+x/x and not 2*e^(1+x)/x and not 2*EXP(1+x)/x 38. 2*e^(2/x)/x or (2*e^(2/x))/x or 2*EXP(2/x)/x or (2*EXP(2/x))/xnot 2*e^((2/x)/x) and not 2*e^2/x/x and not 2*EXP((2/x)/x)

39. In these solutions, 𝑓1 is black and 𝑓2 is gray.

40.

y1 = 1.6^x y2 = 1.8^x

y1 = 2.2^x y2 = 2.5^x

41. (Note that the 𝑥-axis shown here crosses at 200, not 0.)

42. (Note that the 𝑥-axis shown here crosses at 95.)

y1 = 300*1.1^x y2 = 300*1.1^(2*x)

y1 = 100*1.01^(2*x) y2 = 100*1.01^(3*x)

43.

44.

y1 = 2.5^(1.02*x) y2 = e^(1.02*x) or exp(1.02*x)

y1 = 2.5^(-1.02*x) y2 = e^(-1.02*x) or exp(-1.02*x)

45. (Note that the 𝑥-axis shown here crosses at 900.)

46. (Note that the 𝑥-axis here crosses at 1,100.)

Solutions Section 2.2

y1 = 1000*1.045^(-3*x) y2 = 1000*1.045^(3*x)

y1 = 1202*1.034^(-3*x) y2 = 1202*1.034^(3*x)

47. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 0.5. Also, 𝑓(0) = 500. So, 𝑓(𝑥) = 500(0.5) 𝑥.

Solutions Section 2.2 48. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 2. Also, 𝑓(0) = 500. So, 𝑓(𝑥) = 500(2) 𝑥. 49. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 3. Also, 𝑓(0) = 10. So, 𝑓(𝑥) = 10(3) 𝑥.

50. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 1/3. Also, 𝑓(0) = 90. So, 𝑓(𝑥) = 90(1∕3) 𝑥. 51. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 225∕500 = 0.45. Also, 𝑓(0) = 500. So, 𝑓(𝑥) = 500(0.45) 𝑥.

52. Each time 𝑥 increases by 1, 𝑓(𝑥) is multiplied by 3∕5 = 0.6. Also, 𝑓(0) = 5. So, 𝑓(𝑥) = 5(0.6) 𝑥. 53. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 1 = −110 and 𝐴𝑏 2 = −121. Dividing, 𝑏 = −121∕(−110) = 1.1. Substituting, 𝐴(1.1) = −110, so 𝐴 = −100. Thus, 𝑓(𝑥) = −100(1.1) 𝑥.

54. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 1 = −41 and 𝐴𝑏 2 = −42.025. Dividing, 𝑏 = −42.025∕(−41) = 1.025. Substituting, 𝐴(1.025) = −41, so 𝐴 = −41∕1.025 = −40. Thus, 𝑓(𝑥) = −40(1.025) 𝑥. 55. We want an equation of the form 𝑦 = 𝐴𝑏 𝑥. Substituting the coordinates of the given points gives 36 = 𝐴𝑏 2 324 = 𝐴𝑏 4

Dividing the second equation by the first gives 324∕36 = 9 = 𝑏 2, so 𝑏 = 3. Substituting into the first equation now gives 36 = 𝐴(3) 2 = 9𝐴, so 𝐴 = 36∕9 = 4. Hence the model is 𝑦 = 𝐴𝑏 𝑥 = 4(3 𝑥). 56. We want an equation of the form 𝑦 = 𝐴𝑏 𝑥. Substituting the coordinates of the given points gives −4 = 𝐴𝑏 2 −16 = 𝐴𝑏 4.

Dividing the second equation by the first gives −16∕(−4) = 4 = 𝑏 2, so 𝑏 = 2. Substituting into the first equation now gives −4 = 𝐴(2) 2 = 4𝐴, so 4𝐴 = −14. Hence the model is 𝑦 = 𝐴𝑏 𝑥 = −1(2 𝑥). 57. We want an equation of the form 𝑦 = 𝐴𝑏 𝑥. Substituting the coordinates of the given points gives −25 = 𝐴𝑏 −2 −0.2 = 𝐴𝑏.

Dividing the second equation by the first gives −0.2∕(−25) = 0.008 = 𝑏 3, so 𝑏 = 0.2. Substituting into the second equation now gives −0.2 = 0.2𝐴, so 𝐴 = −1. Hence the model is 𝑦 = 𝐴𝑏 𝑥 = −1(0.2 𝑥). 58. We want an equation of the form 𝑦 = 𝐴𝑏 𝑥. Substituting the coordinates of the given points gives 1.2 = 𝐴𝑏 0.108 = 𝐴𝑏 3.

Dividing the second equation by the first gives 0.108∕1.2 = 0.09 = 𝑏 2, so 𝑏 = 0.3. Substituting into the first equation now gives 1.2 = 0.3𝐴, so 𝐴 = 4. Hence the model is 𝑦 = 𝐴𝑏 𝑥 = 4(0.3 𝑥). 59. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 1 = 3 and 𝐴𝑏 3 = 6. Dividing, 𝑏 2 = 6∕3 = 2, so 𝑏 = √2 ≈ 1.4142. Substituting, 𝐴√2 = 3, so 𝐴 = 3∕√2 ≈ 2.1213. Thus, 𝑦 = 2.1213(1.4142 𝑥).

60. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 1 = 2 and 𝐴𝑏 4 = 6. Dividing, 𝑏 3 = 6∕2 = 3, so 𝑏 = √3 ≈ 1.4422. Solutions Section 2.2

Substituting, 𝐴 √3 = 2, so 𝐴 = 2∕ √3 ≈ 1.3867. Thus, 𝑦 = 1.3867(1.4422 𝑥).

61. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 2 = 3 and 𝐴𝑏 6 = 2. Dividing, 𝑏 4 = 2∕3, so 𝑏 = √2∕3 ≈ 0.9036. 4

Substituting, 𝐴( √2∕3) 2 = 3, so 𝐴 = 3∕( √2∕3) 2 ≈ 3.6742. Thus, 𝑦 = 3.6742(0.9036 𝑥). 4

62. Write 𝑓(𝑥) = 𝐴𝑏 𝑥. We have 𝐴𝑏 −1 = 2 and 𝐴𝑏 3 = 1. Dividing, 𝑏 4 = 1∕2, so 𝑏 = √1∕2 ≈ 0.8409. Substituting, 𝐴∕ √1∕2 = 2, so 𝐴 = 2 √1∕2 ≈ 1.6818. Thus, 𝑦 = 1.6818(0.8409 𝑥). 4

63. 𝑦 = 1.0442(1.7564) 𝑥 64. 𝑦 = 1.0442(0.5694) 𝑥

65. 𝑦 = 15.1735(1.4822) 𝑥 66. 𝑦 = 0.4782(1.8257) 𝑥

67. We want a model of the form 𝑓(𝑡) = 𝐴𝑏 𝑡. We are given two points on the graph: (0, 300) and (2, 75). Substituting the coordinates, we get 300 = 𝐴𝑏 0 = 𝐴 75 = 𝐴𝑏 . 2

Substitute (0, 300). Substitute (2, 75).

This gives 𝐴 = 300, 75 = 300𝑏 2, so 𝑏 2 = 75∕300 = 0.25, giving 𝑏 = 0.25 1∕2 = 0.5, and the model is 𝑓(𝑡) = 300(0.5) 𝑡. After 5 hours, 𝑓(5) = 300(0.5) 5 = 9.375 mg. 68. We want a model of the form 𝑓(𝑡) = 𝐴𝑏 𝑡. We are given two points on the graph: (0, 200) and (2, 112.5). Substituting the coordinates, we get 200 = 𝐴𝑏 0 = 𝐴 112.5 = 𝐴𝑏 2.

Substitute (0, 200).

Substitute (2, 112.5).

This gives 𝐴 = 200, 112.5 = 200𝑏 2, so 𝑏 2 = 112.5∕200 = 0.5625, giving 𝑏 = 0.5625 1∕2 = 0.75, and the model is 𝑓(𝑡) = 200(0.75) 𝑡. After 4 hours, 𝑓(4) = 200(0.75) 4 ≈ 63 mg/dL. 69. a. Linear model: 𝑆 = 𝑚𝑡 + 𝑏 Points: (0, 320) and (5, 950) 𝑆 − 𝑆1 950 − 320 Slope: 𝑚 = 2 = = 126 𝑡2 − 𝑡1 5−0 Intercept: 𝑏 = 𝑆1 − 𝑚𝑡1 = 320 − (126)(0) = 320 Model: 𝑆 = 𝑚𝑡 + 𝑏 = 126𝑡 + 320 Exponential model: 𝑆 = 𝐴𝑏 𝑡 Substitute (0, 320): 320 = 𝐴𝑏 0 = 𝐴; substitute (5, 950): 950 = 𝐴𝑏 5 = 320𝑏 5. Thus 950 = 𝑏5 320

𝑏 =(

950 1∕5 ≈ 1.24. 320 )

Solutions Section 2.2

Model: 𝑆 = 𝐴𝑏 𝑡 = 320(1.24) 𝑡 The exponential model is applicable: The successive ratios of the values of 𝑆 are not too far from 𝑏 = 1.24. The successive differences climb steadily, making a linear model not appropriate. b. 2020 corresponds to 𝑡 = 4, and so 𝑆 = 𝐴𝑏 𝑡 = 320(1.24) 4 ≈ 757 GW, in reasonable agreement with the actual figure of 770 GW. 70. a. Linear model: 𝑊 = 𝑚𝑡 + 𝑏 Points: (0, 200) and (10, 710) 𝑊 − 𝑊1 710 − 200 Slope: 𝑚 = 2 = = 51 𝑡2 − 𝑡1 10 − 0 Intercept: 𝑏 = 𝑊1 − 𝑚𝑡1 = 200 − (51)(0) = 200 Model: 𝑊 = 𝑚𝑡 + 𝑏 = 51𝑡 + 200 Exponential model: 𝑊 = 𝐴𝑏 𝑡 Substitute (0, 200): 200 = 𝐴𝑏 0 = 𝐴; substitute (10, 710): 710 = 𝐴𝑏 10 = 200𝑏 10. Thus 710 = 3.55 = 𝑏 10 200 𝑏 = 3.55 1∕10 ≈ 1.14.

Model: 𝑊 = 𝐴𝑏 𝑡 = 200(1.14) 𝑡 The exponential model is applicable: The successive ratios of the values of 𝑊 are not too far from 𝑏 2 ≈ 1.30. (Note that 𝑏 2 is the ratio we expect to see every two years, as in the values given.) The successive differences climb fairly steadily, making a linear model not appropriate. b. 2018 corresponds to 𝑡 = 8, and so 𝑊 = 𝐴𝑏 𝑡 = 200(1.14) 8 ≈ 571 GW, in reasonable agreement with the actual figure of 590 GW. 71. a. The successive ratios are 𝑡

Ratios

210 229 252 282 309 331 ≈ 1.12 ≈ 1.09 ≈ 1.10 ≈ 1.12 ≈ 1.10 ≈ 1.07 187 210 229 252 282 309

As the ratios are all close to 1.1 with no systematic change, we conclude that the growth is approximately exponential. b. The 1970 and 1980 data together with the answer to part (a) give us: (10, 210): 𝐴𝑏 10 = 210 (20, 229): 𝐴𝑏 20 = 229 229 𝑏 10 = ≈ 1.09047619 210 𝑏 ≈ 1.09047619 1∕10 ≈ 1.00870 to 6 significant digits 210 210 2 𝐴𝑏 10 = 210 gives 𝐴 = 10 = ≈ 192.576, 229 𝑏

So the model is 𝑈(𝑡) = 192.576(1.00870) 𝑡, and predicts 𝑈(60) = 192.576(1.00870) 60 ≈ 324 million for 2020, reasonably close to the actual figure of 331 million.

Solutions Section 2.2 72. a. The successive ratios are 𝑡

Ratios

3.70 4.46 5.33 6.14 6.96 7.79 ≈ 1.22 ≈ 1.21 ≈ 1.20 ≈ 1.15 ≈ 1.13 ≈ 1.12 3.03 3.70 4.46 5.33 6.14 6.96

As the ratios show a definite downward trend from 1.22 to 1.12, we conclude that the growth is not exponential, as exponential growth would require the ratios to be approximately constant. b. The 1970 and 1980 data together with the answer to part (a) give us: (10, 3.70): 𝐴𝑏 10 = 3.70 (20, 4.46): 𝐴𝑏 20 = 4.46 4.46 𝑏 10 = ≈ 1.205405405 3.70 𝑏 ≈ 1.205405405 1∕10 ≈ 1.01886 to 6 significant digits 3.70 3.70 2 𝐴𝑏 10 = 3.70 gives 𝐴 = 10 = ≈ 3.06951, 4.46 𝑏

So the model is 𝑊 (𝑡) = 3.06951(1.01886) 𝑡, and predicts 𝑊 (60) = 3.06951(1.01886) 60 ≈ 9.42 billion in 2020, more than 20% higher than the actual figure of 7.79 billion. 73. If 𝑦 represents the size of the culture at time 𝑡, then 𝑦 = 𝐴𝑏 𝑡. We are told that the initial size is 1,000, so 𝐴 = 1, 000. We are told that the size doubles every 3 hours, so 𝑏 3 = 2, or 𝑏 = 2 1∕3. Thus, 𝑦 = 1, 000(2 1∕3) 𝑡 = 1, 000(2 𝑡∕3). There will be 1, 000(2 48∕3) = 65, 536, 000 bacteria after 2 days. 74. If 𝑦 represents the size of the culture at time 𝑡, then 𝑦 = 𝐴𝑏 𝑡. We are told that the initial size is 1,000, so 𝐴 = 1, 000. We are told that 𝑦 = 1, 500 when 𝑡 = 2, so 1, 500 = 1, 000𝑏 2, giving 𝑏 = 1.5 1∕2. Thus, 𝑦 = 1, 000(1.5 1∕2) 𝑡 = 1, 000(1.5 𝑡∕2). There will be 1, 000(1.5 48∕2) ≈ 16, 800, 000 bacteria after 2 days.

75. The desired model is 𝐶(𝑡) = 𝐴𝑏 𝑡. At time 𝑡 = 0 (April 1, 2021) the number of active cases was 615,000, so 𝐴 = 615,000. Since the number was increasing by 6.9% each day, that number is multiplied by 1.069 each day, so 𝑏 = 1.069. Hence, the model is 𝐶(𝑡) = 615,000(1.069) 𝑡. As April 21, 2021 corresponds to 𝑡 = 20, the estimated number of cases was about 𝐶(20) = 615,000(1.069) 20 ≈ 2,335,765. 76. The desired model is 𝑆(𝑡) = 𝐴𝑏 𝑡. At time 𝑡 = 0 (April 1, 2003) the number of cases was 1,804, so 𝐴 = 1,804. Since the number was increasing by 4% each day, the number of cases is multiplied by 1.04 each day, so 𝑏 = 1.04. Hence, the model is 𝑆(𝑡) = 1,804(1.04) 𝑡. Since April 30, 2003 corresponds to 𝑡 = 29, the number of cases was about 𝑆(14) = 1,804(1.04) 29 ≈ 5,626. 77. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = 5,000, 𝑟 = 0.05∕100 = 0.0005, and 𝑛 = 12. We get the model 𝐴(𝑡) = 5,000(1 + 0.0005∕12) 12𝑡 In July 2028 (𝑡 = 7), the deposit would be worth 5,000(1 + 0.0005∕12) 12(7) ≈ $5,018. 78. Apply the formula 𝑟 𝑛𝑡 𝐴(𝑡) = 𝑃 !1 + " 𝑛 with 𝑃 = 4,000, 𝑟 = 0.0061, and 𝑛 = 365. We get the model 𝐴(𝑡) = 4,000(1 + 0.0061∕365) 365𝑡

In July 2029 (𝑡 = 8), the deposit would be worth 4,000(1 + 0.0061∕365) 365(8) ≈ $4,200. Solutions Section 2.2

79. Substitute 𝐴 = 15,000, 𝑃 = 10,000, 𝑟 = 0.025, and 𝑛 = 1 into the compound interest formula: 15,000 = 10,000(1 + 0.025) 𝑡, (1.025) 𝑡 = 1.5, 𝑡 = log(1.5)∕ log(1.025) ≈ 16 years.

80. Substitute 𝐴 = 11,000, 𝑃 = 10,000, 𝑟 = 0.00025, and 𝑛 = 1 into the compound interest formula: 11,000 = 10,000(1 + 0.00025) 𝑡, (1.00025) 𝑡 = 1.1, 𝑡 = log(1.1)∕ log(1.00025) ≈ 381 years. 81. Substitute 𝐴 = 20,000, 𝑃 = 10,400, 𝑟 = 0.010, and 𝑛 = 12 into the compound interest formula: 20,000 = 10,400(1 + 0.010∕12) 12𝑡, (1 + 0.010∕12) 12𝑡 = 200∕104, 12𝑡 = log(200∕104)∕ log(1 + 0.010∕12) ≈ 785 months. 82. Substitute 𝐴 = 20,000, 𝑃 = 10,400, 𝑟 = 0.025, and 𝑛 = 12 into the compound interest formula: 20,000 = 10,400(1 + 0.025∕12) 12𝑡, (1 + 0.025∕12) 12𝑡 = 200∕104, 12𝑡 = log(200∕104)∕ log(1 + 0.025∕12) ≈ 314 months.

83. Substitute 𝐴 = 2𝑃 , 𝑟 = 0.035, and 𝑛 = 2 into the compound interest formula: 2𝑃 = 𝑃 (1 + 0.035∕2) 2𝑡, (1.0175) 2𝑡 = 2, 2𝑡 = log 2∕ log 1.0175, 𝑡 = (log 2∕ log 1.0175)∕2 ≈ 20 years.

84. Substitute 𝐴 = 2𝑃 , 𝑟 = 0.0425, and 𝑛 = 2 into the compound interest formula: 2𝑃 = 𝑃 (1 + 0.0425∕2) 2𝑡, (1.02125) 2𝑡 = 2, 2𝑡 = log 2∕ log 1.02125, 𝑡 = (log 2∕ log 1.02125)∕2 ≈ 16 years. 85. If 99.95% has decayed, then 0.05% remains, so 𝐶(𝑡) = 0.0005𝐴. Therefore, 0.0005𝐴 = 𝐴(0.999879) 𝑡, (0.999879) 𝑡 = 0.0005, so 𝑡 = log(0.0005)∕ log(0.999879) ≈ 63,000 years old. 86. If 30% has decayed, then 70% remains, so 𝐶(𝑡) = 0.70𝐴. Therefore, 0.70𝐴 = 𝐴(0.999879) 𝑡, (0.999879) 𝑡 = 0.70, 𝑡 = log(0.70)∕ log(0.999879) ≈ 3,000 years old.

87. a. Let 𝑦 be the number of frogs in year 𝑡, with 𝑡 = 0 representing 2 years ago; we seek a model of the form 𝑦 = 𝐴𝑏 𝑡. We are given the initial value of 𝐴 = 50,000 and are told that 50,000𝑏 2 = 32,000. This gives 𝑏 = (32,000∕50,000) 1∕2 = 0.64 1∕2 = 0.8. Thus, 𝑦 = 50,000(0.8) 𝑡. b. When 𝑡 = 3, 𝑦 = 50,000(0.8) 3 = 25,600 tags. 1,000 c. We want 𝑡 so that 50,000(0.8) 𝑡 = 1,000 so 0.8 𝑡 = = 0.02. Taking the logarithm of both sides 50,000 gives log(0.8 𝑡) = log 0.02 ⟹ 𝑡 log 0.8 = log 0.02, so log 0.02 𝑡= ≈ 17.53 years, log 0.8 so by 𝑡 = 18 (16 years from now) the number will have dropped to less than 1,000.

88. a. Let 𝑦 be the number of flies in year 𝑡, with 𝑡 = 0 representing 3 years ago; we seek a model of the form 𝑦 = 𝐴𝑏 𝑡. We are given the initial value of 𝐴 = 4,000 and we are told that 4,000𝑏 3 = 1,372. This gives 𝑏 = (1,372∕4,000) 1∕3 = 0.343 1∕3 = 0.7. Thus, 𝑦 = 4,000(0.7) 𝑡. b. When 𝑡 = 4, 𝑦 = 4,000(0.7) 4 = 960.4 ≈ 960 flies. 500 c. We want 𝑡 so that 4,000(0.7) 𝑡 = 500 so 0.7 𝑡 = = 0.125. Taking the logarithm of both sides gives 4,000 log(0.7 𝑡) = log 0.125 ⟹ 𝑡 log 0.7 = log 0.125, so log 0.125 𝑡= ≈ 5.8 years, log 0.7 so by 𝑡 = 6 (three years from now) the number will have dropped to less than 500.

Solutions Section 2.2 89. a. Here is the Excel tabulation of the data, together with the scatter plot and the exponential trendline (with the option "Display equation on chart" checked):

Note that the displayed model has the form 𝑃 (𝑡) = 0.3394𝑒 0.1563𝑡. 𝑒 is the number 2.71828183...... discussed in the next section, and built in to all calculators. To express this in the form 𝐴(𝑏) 𝑡 we write 0.153𝑒 0.0863𝑡 = 0.3394(𝑒 0.1563) 𝑡 ≈ 0.3394(1.169) 𝑡

𝑒 0.1563 is EXP(0.1563) in Excel

so the model is

𝑃 (𝑡) = 0.339(1.169) 𝑡.

b. In 2005, 𝑡 = 11, so the predicted cost is 𝑃 (15 = 0.339(1.169) 11 ≈ $1.9 million.

90. a. Here is the Excel tabulation of the data, together with the scatter plot and the exponential trendline (with the option "Display equation on chart" checked):

Note that the displayed model has the form 𝑃 (𝑡) = 0.1533𝑒 0.0863𝑡. 𝑒 is the number 2.71828183...... discussed in the next section, and built in to all calculators. To express this in the form 𝐴(𝑏) 𝑡 we write 0.153𝑒 0.0863𝑡 = 0.153(𝑒 0.0863𝑡) 𝑡 ≈ 0.153(1.090) 𝑡

so the model is

𝑒 0.0863 is EXP(0.0863) in Excel

𝑃 (𝑡) = 0.153(1.090) 𝑡.

Solutions Section 2.2

b. In 2005, 𝑡 = 15, so the predicted cost is 𝑃 (15) = 0.153(1.090) 15 ≈ $0.56 million.

91. a. Here is the Excel tabulation of the data, together with the scatter plot and the exponential trendline (with the option "Display equation on chart" checked; we have enlarged the graph in order to see more clearly how the points are scattered around the regression curve):

Note that the displayed model has the form 𝑈(𝑡) = 189.17𝑒 0.0096𝑡. 𝑒 is the number 2.71828183 . . . discussed in the next section, and built in to all calculators. To express this in the form 𝐴(𝑏) 𝑡 we write 189.17𝑒 0.0096𝑡 = 189.17(𝑒 0.0096) 𝑡 ≈ 189.2(1.010) 𝑡

so the model is

𝑈(𝑡) = 189.2(1.010) 𝑡.

𝑒 0.0096 is EXP(0.0096) in Excel

Solutions Section 2.2 b. Yes; the points are very close to, and scattered randomly around, the regression curve. 92. a. Here is the Excel tabulation of the data, together with the scatter plot and the exponential trendline (with the option "Display equation on chart" checked; we have enlarged the graph in order to see more clearly how the points are scattered around the regression curve):

Note that the displayed model has the form 𝑊 (𝑡) = 3.1722𝑒 0.0158𝑡 𝑒 is the number 2.71828183 . . . discussed in the next section, and built in to all calculators. To express this in the form 𝐴(𝑏) 𝑡 we write 3.1722𝑒 0.0158𝑡 = 3.1722(𝑒 0.0158) 𝑡 ≈ 3.172(1.016) 𝑡

𝑒 0.0158 is EXP(0.0158) in Excel

so the model is

𝑊 (𝑡) = 3.172(1.016) 𝑡.

b. No; the points suggest a graph that is less curved than the regression curve. 93. (B) An exponential function eventually becomes larger than any polynomial. 94. (B) An exponential decay function eventually becomes smaller than the reciprocal of any polynomial (see Exercise 99).

Solutions Section 2.2 95. Exponential functions of the form 𝑓(𝑥) = 𝐴𝑏 𝑥 (𝑏 > 1) increase rapidly for large values of 𝑥. In reallife situations, such as population growth, this model is reliable only for relatively short periods of growth. Eventually, population growth tapers off because of pressures such as limited resources and overcrowding. 96. If an investment earns 5% compounded continuously, it is as if interest is being added every moment. Thus, the interest for 1 month will have been calculated on a growing amount, rather than on the original fixed (smaller) amount used for interest compounded monthly. 97. Linear functions are better for cost models where there is a fixed cost and a variable cost and for simple interest, where interest is paid only on the original amount invested. Exponential models are better for compound interest and population growth. In both of these latter examples, the rate of growth depends on the current number of items, rather than on a fixed initial quantity. 98. Quadratic models are better for revenue and profit functions where demand depends linearly on the price. Exponential models are better for compound interest and population growth. 99. Take the ratios 𝑦2 ∕𝑦1 and 𝑦3 ∕𝑦2 . If they are the same, the points fit on an exponential curve.

100. For the points to fit on an exponential curve we must have 𝑦2 ∕𝑦1 = 𝑦3 ∕𝑦2 , hence 𝑦3 = 𝑦22∕𝑦1 .

Solutions Section 2.3 Section 2.3

1. Use the formula 𝑓(𝑡) = 𝑃 𝑒 𝑟𝑡 with 𝑃 = 5,000 and 𝑟 = 0.10, giving 𝑓(𝑡) = 5,000𝑒 0.10𝑡.

2. Use the formula 𝑓(𝑡) = 𝑃 𝑒 𝑟𝑡 with 𝑃 = 2,000 and 𝑟 = 0.053, giving 𝑓(𝑡) = 2,000𝑒 0.053𝑡.

3. Use the formula 𝑓(𝑡) = 𝑃 𝑒 𝑟𝑡 with 𝑃 = 1,000 and 𝑟 = −0.063, giving 𝑓(𝑡) = 1,000𝑒 −0.063𝑡.

4. Use the formula 𝑓(𝑡) = 𝑃 𝑒 𝑟𝑡 with 𝑃 = 10,000 and 𝑟 = −0.60, giving 𝑓(𝑡) = 10,000𝑒 −0.60𝑡. 5. 𝑓(𝑥) = 4𝑒 2𝑥 = 4(𝑒 2) 𝑥 ≈ 4(7.389) 𝑥

6. 𝑓(𝑥) = 2.1𝑒 −0.1𝑥 = 2.1(𝑒 −0.1) 𝑥 ≈ 2.1(0.9048) 𝑥

7. 𝑓(𝑡) = 2.1(1.001) 𝑡 = 2.1𝑒 (ln 1.001)𝑡 ≈ 2.1𝑒 0.0009995𝑡

8. 𝑓(𝑡) = 2.1(0.991) 𝑡 = 23.4𝑒 (ln 0.991)𝑡 ≈ 23.4𝑒 −0.009041𝑡 9. 𝑓(𝑡) = 10(0.987) 𝑡 = 10𝑒 (ln 0.987)𝑡 ≈ 10𝑒 −0.01309𝑡 10. 𝑓(𝑡) = 2.3(2.2) 𝑡 = 2.3𝑒 (ln 2.2)𝑡 ≈ 2.3𝑒 0.7885𝑡

11. 𝑄 = 1,000 when 𝑡 = 0; half-life = 1. We want a model of the form 𝑄 = 𝑄0 𝑒 −𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 1,000. For 𝑘, we use the formula 𝑡ℎ 𝑘 = ln 2 with 𝑡ℎ = half-life = 1, so 𝑘 = ln 2, and the model is 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 1,000𝑒 −𝑡 ln 2.

12. 𝑄 = 2,000 when 𝑡 = 0; half-life = 5. We want a model of the form 𝑄 = 𝑄0 𝑒 −𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 2,000. For 𝑘, we use the formula 𝑡ℎ 𝑘 = ln 2 with 𝑡ℎ = half-life = 5, so 2𝑘 = ln 2, giving 𝑘 = (ln 2)∕5 and the model is 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 2,000𝑒 −𝑡(ln 2)∕5.

13. 𝑄 = 500 when 𝑡 = 0; half-life = 50. We want a model of the form 𝑄 = 𝑄0 𝑒 −𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 500. For 𝑘, we use the formula 𝑡ℎ 𝑘 = ln 2 with 𝑡ℎ = half-life = 50, so 𝑘 = ln 2∕50, and the model is 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 500𝑒 −𝑡 ln 2∕50. 14. 𝑄 = 50 when 𝑡 = 0; half-life = 500. We want a model of the form 𝑄 = 𝑄0 𝑒 −𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 50. For 𝑘, we use the formula 𝑡ℎ 𝑘 = ln 2 with 𝑡ℎ = half-life = 500, so 𝑘 = ln 2∕500, and the model is 𝑄 = 𝑄0 𝑒 −𝑘𝑡 = 50𝑒 −𝑡 ln 2∕500. 15. 𝑄 = 1,000 when 𝑡 = 0; doubling time = 2. We want a model of the form 𝑄 = 𝑄0 𝑒 𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 1,000. For 𝑘, we use the formula 𝑡𝑑 𝑘 = ln 2 with 𝑡𝑑 = doubling time = 2, so 2𝑘 = ln 2, giving 𝑘 = (ln 2)∕2 and the model is 𝑄 = 𝑄0 𝑒 𝑘𝑡 = 1,000𝑒 𝑡(ln 2)∕2.

Solutions Section 2.3 16. 𝑄 = 2,000 when 𝑡 = 0; doubling time = 5. We want a model of the form 𝑄 = 𝑄0 𝑒 𝑘𝑡 for suitable 𝑄0 and 𝑘. We are given 𝑄0 = 2,000. For 𝑘, we use the formula 𝑡𝑑 𝑘 = ln 2 with 𝑡𝑑 = doubling time = 5, so 5𝑘 = ln 2, giving 𝑘 = (ln 2)∕5 and the model is 𝑄 = 𝑄0 𝑒 𝑘𝑡 = 2,000𝑒 𝑡(ln 2)∕5.

17. 𝑄 = 1,000𝑒 0.5𝑡. Since the exponent is positive, the model represents exponential growth. We use the formula 𝑡𝑑 𝑘 = ln 2, giving 𝑡𝑑 (0.5) = ln 2, giving doubling time = 𝑡𝑑 = (ln 2)∕0.5 = 2 ln 2.

18. 𝑄 = 1,000𝑒 −0.025𝑡. Since the exponent is negative, the model represents exponential decay. We use the formula 𝑡ℎ 𝑘 = ln 2, giving 𝑡ℎ (0.025) = ln 2, giving half-life = 𝑡ℎ = (ln 2)∕0.025 = 40 ln 2. 19. 𝑄 = 100𝑒 −𝑡. Since the exponent is negative, the model represents exponential decay. We use the formula 𝑡ℎ 𝑘 = ln 2, giving 𝑡ℎ (1) = ln 2, giving half-life = 𝑡ℎ = ln 2.

20. 𝑄 = 5 000𝑒 𝑡∕3. Since the exponent is positive, the model represents exponential growth. We use the formula 𝑡𝑑 𝑘 = ln 2, giving 𝑡𝑑 (1∕3) = ln 2, giving doubling time = 𝑡𝑑 = 3 ln 2. 21. 𝑄 = 𝑄0 𝑒 −4𝑡. Since the exponent is negative, the model represents exponential decay. We use the formula 𝑡ℎ 𝑘 = ln 2, giving 𝑡ℎ (4) = ln 2, giving half-life = 𝑡ℎ = (ln 2)∕4. 22. 𝑄 = 𝑄0 𝑒 𝑡. Since the exponent is positive, the model represents exponential growth. We use the formula 𝑡𝑑 𝑘 = ln 2, giving 𝑡𝑑 (1) = ln 2, giving doubling time = 𝑡𝑑 = ln 2.

23. a. From the continuous compounding formula, 1,000𝑒 0.04×10 = $1,491.82, of which $491.82 is interest. b. 𝑃 = 1,000𝑒 0.04𝑡 = 1,000(𝑒 0.04) 𝑡 ≈ 1,000(1.040811) 𝑡, which corresponds to an annually compounded interest rate of 4.08% 24. a. From the continuous compounding formula, 2,000𝑒 0.31×10 = $44,395.90, of which $42,395.90 is interest. b. 𝑃 = 2,000𝑒 0.31𝑡 = 2,000(𝑒 0.31) 𝑡 ≈ 2,000(1.3634) 𝑡, which corresponds to an annually compounded interest rate of 36.34%

25. Substitute 𝐴 = 700, 𝑃 = 500, and 𝑟 = 0.10 in the continuous compounding formula: 700 = 500𝑒 0.10𝑡. Solve for 𝑡: 𝑒 0.10𝑡 = 700∕500 = 7∕5, 0.10𝑡 = ln(7∕5), 𝑡 = 10 ln(7∕5) ≈ 3.36 years. 26. Substitute 𝐴 = 700, 𝑃 = 500, and 𝑟 = 0.15 in the continuous compounding formula: 700 = 500𝑒 0.15𝑡. Solve for 𝑡: 𝑒 0.15𝑡 = 700∕500 = 7∕5, 0.15𝑡 = ln(7∕5), 𝑡 = ln(7∕5)∕0.15 ≈ 2.24 years. 27. Substitute 𝐴 = 3𝑃 and 𝑟 = 0.10 in the continuous compounding formula: 3𝑃 = 𝑃 𝑒 0.10𝑡. Solve for 𝑡: 𝑒 0.10𝑡 = (3𝑃 )∕𝑃 = 3, 0.10𝑡 = ln 3, 𝑡 = 10 ln 3 ≈ 11 years. 28. Substitute 𝐴 = 1,000,000, 𝑃 = 1,000 and 𝑟 = 0.10 in the continuous compounding formula: 1,000,000 = 1,000𝑒 0.10𝑡. Solve for 𝑡: 𝑒 0.10𝑡 = 1,000,000∕1,000 = 1,000, 0.10𝑡 = ln 1,000, 𝑡 = 10 ln 1,000 ≈ 69 years. 29. The bank can afford to pay its customers 𝑟 𝑛𝑡 𝑃 (1 + ) = 𝑃 (1.036) 𝑡 𝑛