## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 0 - Before Calculus - 0.2 New Functions From Old - Exercises Set 0.2 - Page 25: 42

#### Answer

(a) $$g(x)=\frac{1}{1-x} \qquad h(x)=x^2$$ (b) $$g(x)=|x| \qquad h(x)= 5+2x$$

#### Work Step by Step

(a) By considering the functions $$g(x)=\frac{1}{1-x} \qquad h(x)=x^2$$ we have $$f(x)=(g \circ h)(x)=g(h(x))=\frac{1}{1-x^2}.$$ (b) By considering the functions $$g(x)=|x| \qquad h(x)= 5+2x$$ we have $$f(x)=(g \circ h)(x)=g(h(x))=|5+2x|.$$

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