Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.1 Composite Functions - 4.1 Assess Your Understanding - Page 280: 47

Answer

Refer to the answer below.

Work Step by Step

The functions $f(x)$ and $g(x)$ are known as inverse functions. Thus, we simplify $(f\circ g) (x)=(g\circ f) (x)$ or, $f[g(x)]=g[f(x)]$ as follows: $(f\circ g) x =f[g(x)]=f(\dfrac{1}{2}x)=f (\dfrac{x}{2})=2(0.5x)=(2)(\dfrac{x}{2}) =x$ and $(g\circ f) x = g[f(x)]=f(\dfrac{1}{2}x)=g(2x)= [\dfrac{1}{2}](2x)=x$
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