Calculus for Business, Economics, Life Sciences, and Social Sciences, 13E By Barnett, Ziegler, Byleen
Email: Richard@qwconsultancy.com
Calculus
Chapter 1
Name ________________________________ Date ______________ Class ____________
Section 1-1 Functions Goal: To evaluate function values and to determine the domain of functions
The domain of the following functions will be the set of real numbers unless it meets one of the following conditions: 1. The function contains a fraction whose denominator has a variable. The domain of such a function is the set of real numbers EXCEPT the values of the variable that make the denominator zero. 2. The function contains an even root (square root , fourth root 4 , etc.). The domain of such a function is limited to values of the variable that make the radicand (the part under the radical) greater than or equal to 0. 1. Evaluate the following function at the specified values of the independent variable and simplify the results.
f ( x) = 4 x - 5 a)
c)
f (1) = 4(1) - 5 f (1) = 4 - 5 f (1) = - 1
b)
f ( x - 1) = 4( x - 1) - 5 f ( x - 1) = 4 x - 4 - 5 f ( x - 1) = 4 x - 9
d)
f (- 3) = 4(- 3) - 5 f (- 3) = - 12 - 5 f (- 3) = - 17
æ1 ö æ1 ö f ç ÷ = 4 ç ÷- 5 è4 ø è4ø æ1 ö f ç ÷= 1- 5 è4 ø æ1 ö f ç ÷= - 4 è4 ø
In problems 2–10 evaluate the given function for f ( x) = x 2 + 1 and g ( x) = x - 4 .
f (3) = (3) 2 + 1 f (3) = 9 + 1 f (3) = 10
2. ( f + g )(3) = f (3) + g (3)
= 10 + (- 1) ( f + g )(3) = 9
1-1 ..
g (3) = 3 - 4 g (3) = - 1
Calculus
Chapter 1
3. ( f - g )(2c) = f (2c) - g (2c)
= 4c 2 + 1 - (2c - 4)
f (2c) = (2c)2 + 1
g (2c) = 2c - 4
f (2c) = 4c 2 + 1
( f - g )(2c) = 4c 2 - 2c + 5
4. ( fg )(- 4) = f (- 4) g (- 4) = (17)(- 8) ( fg )(- 4) = - 136
æf ö f (0) 5. ç ÷(0) = g (0) èg ø 1 -4 æf ö 1 (0) = çè g ÷ 4 ø =
6.
f (- 4) = (- 4) 2 + 1 f (- 4) = 16 + 1 f (- 4) = 17
g (- 4) = - 4 - 4 g (- 4) = - 8
f (0) = (0) 2 + 1 f (0) = 0 + 1 f (0) = 1
g (0) = 0 - 4 g (0) = - 4
2 ×g (- 2) = 2(- 6) 2 ×g (- 2) = - 12
g (- 2) = - 2 - 4 g (- 2) = - 6
7. 3 ×f (4) - 2 ×g (- 1) = 3(17) - 2( - 5) = 51 + 10 3 ×f (4) - 2 ×g (- 1) = 61
8.
f (3) - g (2) 10 - (- 2) = f (1) 2 12 = 2 f (3) - g (2) =6 f (1)
f (3) = (3) 2 + 1 f (3) = 9 + 1 f (3) = 10
1-2 ..
f (4) = (4) 2 + 1 f (4) = 16 + 1 f (4) = 17
g (- 1) = - 1 - 4 g (- 1) = - 5
f (1) = (1) 2 + 1 f (1) = 1 + 1 f (1) = 2
g (2) = 2 - 4 g (2) = - 2
Calculus
9.
10.
Chapter 1
g (- 1 + h) - g (- 1) h - 5 - (- 5) = h h h = h g (- 1 + h) - g (- 1) =1 h
f (2 + h) - f (2) h 2 + 4h + 5 - (5) = h h h 2 + 4h = h
g ( - 1 + h) = - 1 + h - 4 g ( - 1 + h) = h - 5
g (- 1) = - 1 - 4 g (- 1) = - 5
f (2 + h) = (2 + h) 2 + 1
f (2) = (2) 2 + 1 f (2) = 4 + 1 f (2) = 5
f (2 + h) = h 2 + 4h + 4 + 1 f (2 + h) = h 2 + 4h + 5
f (2 + h) - f (2) = h+4 h
In problems 11–18 find the domain of each function. 11. g ( x) =
5 x- 2
The domain is restricted by the denominator. Since it cannot equal zero, the domain is all real numbers except 2. 12.
f ( x) =
2x 3x + 7
The domain is restricted by the denominator. Since it cannot equal zero, the domain 7 is all real numbers except - . 3 13. h(t ) = 4 1 - 2t The domain is restricted by the radicand since it has an even root. Since the radicand 1 must be greater than or equal to zero, the domain is t £ . 2 14. g ( x) = 1 - 2 x 2 There are no restrictions on the domain, therefore the domain is all real numbers. 1-3 ..
Calculus 15.
Chapter 1
f ( x) = 3 x + 4
There are no restrictions on the domain since it has an odd root, therefore the domain is all real numbers. 16. h(w) = w - 3 The domain is restricted by the radicand since it has an even root. Since the radicand must be greater than or equal to zero, the domain is w ³ 3. 17.
f ( x) = 2 x3 + 5 x 2 - x + 17 There are no restrictions on the domain, therefore the domain is all real numbers.
18. g ( x) =
2 x3 5
There are no restrictions on the domain since there is no variable in the denominator, therefore the domain is all real numbers.
1-4 ..