Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.1 - Finding Limits Using Tables and Graphs - Exercise Set - Page 1140: 57

Answer

$$f^{-1}(x)= \sqrt[3]{{x}+2}$$ $$\lim_{x \to 6 }f^{-1}(x)=\sqrt[3]{{6}+2}=2$$
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Work Step by Step

$$y=x^3-2 \quad \to \quad x=\sqrt[3]{{y}+2} \\ \Rightarrow \quad f^{-1}(x)=\sqrt[3]{{x}+2}$$ To find $\lim_{x\to 6 }f^{-1}(x)$, examine the graph of $f^{-1}$ near $x=6$. As $x$ gets closer to $6$, the values of $f^{-1}(x)$ get closer to $\sqrt[3]{{6}+2}$. We conclude from the graph that$$\lim_{x \to 6 }f^{-1}(x)=\sqrt[3]{{6}+2}=2.$$
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