Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.1 Inverse Functions - 6.1 Exercises - Page 406: 20


(a) Because it passes the horizontal line test. (b) The range is the segment $[-3,3]$ while the domain is the segment $[-1,3]$. (c) $f^{-1}(2)=0$ (d) $f^{-1}(0)\approx1.7$

Work Step by Step

(a) Any horizontal line intercepts the graph of the function at most once. This means that every value of $y$ from the range corresponds to only one value of $x$ i.e. the function is one-to-one. (b) From the graph we read that $x$ i.e. the argument takes values from $[-3,3]$ so this segment is the domain. On the other hand, the function takes values from $[-1,3]$ so this is the range. All of this is for $f$. Knowing that for $f^{-1}$ the domain and the range are reversed with respect to $f$we have that Domain: $[-1,3]$ Range: $[-3,3]$ (c) We see that the graph passes through the point $(0,2)$ so $f(0)=2$, meaning that $f^{-1}(2)=0$. (d) We see that the graph intercepts the $x$ around one third of a unit short of $x=-2$ which we estimate to be $x_0\approx1.7$. Because $f(x_0)=0$ we have that $f^{-1}(0)=x_0\approx1.7$.
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