Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Review - Exercises - Page 68: 2

Answer

(a) 3 (b) $g$ is one-to-one because for any two x-coordinates $x_1$ and $x_2$, $g(x_1) \neq g(x_2)$ (c) 0.2 (d) $[-1,3.5]$ (e) We can see a sketch of the graph of $g^{-1}$ below.

Work Step by Step

(a) On the graph, we can see that $g(2) = 3$ (b) $g$ is one-to-one because for any two x-coordinates $x_1$ and $x_2$, $g(x_1) \neq g(x_2)$. Note that $g$ is an increasing function. Since the function is increasing, for any two x-coordinates $x_1$ and $x_2$, $g(x_1) \neq g(x_2)$. (c) $g(0.2) = 2$ Then: $g^{-1}(2) = 0.2$ (d) The domain of $g^{-1}$ is the range of $g$. The range of $g$ is $[-1,3.5]$ The domain of $g^{-1}$ is $[-1,3.5]$ (e) We can find some coordinate pairs in the graph of $g$: $(-2,-1)$ $(0,1)$ $(1,2.6)$ $(2,3)$ $(4,3.5)$ In the graph of $g^{-1}$, these coordinate pairs are reversed: $(-1,-2)$ $(1,0)$ $(2.6,1)$ $(3,2)$ $(3.5,4)$ We can see a sketch of the graph of $g^{-1}$ below.
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