University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.2 - Combining Functions; Shifting and Scaling Graphs - Exercises - Page 18: 6


$a. -\frac{1}{3}$ $b. 2$ $c. \frac{-x}{x+1}$ $d. \frac{1}{x}$ $e. 0$ $f. \frac{3}{4}$ $g. x-2$ $h. \frac{x+1}{x+2}$

Work Step by Step

$f(x)=x-1$ $g(x)=\frac{1}{x+1}$ $a.\quad f(g(\frac{1}{2}))=?$ $f(g(\frac{1}{2}))=\frac{-\frac{1}{2}}{\frac{1}{2}+1}$ $=-\frac{1}{3}$ $b.\quad g(f(\frac{1}{2}))=?$ $g(f(\frac{1}{2}))=\frac{\ 1\ }{\frac{1}{2}}$ $=2$ $c.\quad f(g(x))=?$ $f(g(x))=\frac{1}{x+1}-1=\frac{1-x-1}{x+1}$ $=\frac{-x}{x+1}$ $d.\quad g(f(x))=?$ $g(f(x))=\frac{1}{(x-1)+1}=\frac{1}{x}$ $e.\quad f(f(2))=?$ $f(f(2))=2-2=0$ $f.\quad g(g(2))=?$ $g(g(2))=\frac{3}{4}$ $g.\quad f(f(x))=?$ $f(f(x))=(x-1)-1=x-2$ $h.\quad g(g(x))=?$ $g(g(x))=\frac{1}{(\frac{1}{x+1})+1}$ $=\frac{x+1}{x+2}$
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