## Precalculus (6th Edition)

Published by Pearson

# Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.1 Inverse Functions - 4.1 Exercises: 42

#### Answer

$f$ and $g$ are inverses of each other.

#### Work Step by Step

RECALL: (1) $(f \circ g)(x) = f\left(g(x)\right)$ (2) The function $g(x)$ is the inverse of function of a one-to-one function $(x)$ if for every element of the domain, $(f \circ g)(x) =x$ and $(g \circ f)(x)=x$ Find $(f \circ g)(x)$ by substituting $\frac{1}{3}x-3$ to $x$ in $f(x)$: $(f\circ g)(x) \\=f\left(g(x)\right) \\=3\left(\frac{1}{3}x-3\right)+9 \\=3(\frac{1}{3}x)-3(3)+9 \\=x-9+9 \\=x$ Find $(g \circ f)(x)$ by substituting $3x+9$ to $x$ in $g(x)$: $(g\circ f)(x) \\=g(\left(f(x)\right) \\=\frac{1}{3}\left(3x+9\right)-3 \\=\frac{1}{3}(3x) + \frac{1}{3}(3)-3 \\=x+3-3 \\=x$ Since $(f\circ g)(x) = (g\circ f)(x)=x$, then $f$ and $g$ are inverses of each other.

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