## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 12 - Exponential Functions and Logarithmic Functions - 12.1 Composite Functions and Inverse Functions - 12.1 Exercise Set - Page 788: 60

#### Answer

$a.\quad 40,42,46,50$ $b.\quad$Yes, $\quad f^{-1}(x)=x-32$ $c.\quad 8,10,14,18$

#### Work Step by Step

$a.\quad$ $f(8)=8+32=40$ $f(10)=10+32=42$ $f(14)=14+32=46$ $f(18)=18+32=50$ $b.\quad$ $f(x)=x+32$ is a linear function, not constant, so its graph is an oblique line that passes the horizontal line test. It is one-to-one and has an inverse. To find a formula for the inverse, 1. Replace $f(x)$ with $y.$ $y=x+32$ 2. Interchange $x$ and $y$. (This gives the inverse function.) $x=y+32$ 3. Solve for $y.$ $x-32=y$ 4. Replace $y$ with $f^{-1}(x)$ . (This is inverse function notation.) $f^{-1}(x)=x-32$ $c.\quad$ $f^{-1}(40)=40-32=8$ $f^{-1}(42)=42-32=10$ $f^{-1}(46)=46-32=14$ $f^{-1}(50)= 50-32=18$

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