## Precalculus (6th Edition)

Published by Pearson

# Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.1 Inverse Functions - 4.1 Exercises - Page 417: 40

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#### Work Step by Step

See Inverse Function, p.409. Let $f$ be a one-to-one function. Then $g$ is the inverse function of $f$ if $(f\circ g)(x)=x$ for every $x$ in the domain of $g$, and $(g\circ f)(x)=x$ for every $x$ in the domain of $f$. The condition that $f$ is one-to-one in the definition of inverse function is essential. ---------------------- g(f(1))=g(1)=1 g(f(3))=g(3)=3 g(f(5))=g(5)=5 f(g(1))=g(1)=1 f(g(3))=g(3)=3 f(g(5))=g(5)=5 So, $(f\circ g)(x)=x$ for every $x$ in the domain of $g$, and $(g\circ f)(x)=x$ for every $x$ in the domain of $f$, ... f and g are inverses.

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