Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.2 - Composite and Inverse Functions - Exercise Set - Page 686: 13

Answer

(a.)$(f\circ g)(x) =x$. (b.)$(g\circ f)(x) =x$. (c.)$(f\circ g)(2) =2$.

Work Step by Step

The given functions are $f(x)=\frac{1}{x}$ and $g(x)=\frac{1}{x}$. (a) $(f\circ g)(x) = f(g(x))$ Replace $x$ with $g(x)$ in the function $f(x)$. $f(g(x))=\frac{1}{g(x)}$ Plug value of $g(x)$ in the right hand side. $f(g(x))=\frac{1}{\frac{1}{x}}$ $f(g(x))=x$. Hence, $(f\circ g)(x) =x$. (b) $(g\circ f)(x) = g(f(x))$ Replace $x$ with $f(x)$ in the function $g(x)$. $g(f(x))=\frac{1}{f(x)}$ Plug value of $f(x)$ in the right hand side. $g(f(x))=\frac{1}{\frac{1}{x}}$ $g(f(x))=x$. Hence, $(g\circ f)(x) =x$. (c) Replace $x$ with $2$ in the part (a) solution. $(f\circ g)(2) =2$.
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