Type:
Solution Manual
Calculus for Business, Resource: Economics, Life Sciences, and Social Sciences Edition:
13th Edition
Author(s):
Raymond Barnett Michael Ziegler Karl Byleen
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