University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 174: 5

Answer

See below for detailed explanations and answers.
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Work Step by Step

a) $f(x)=x^3$ and $g(x)=\sqrt[3]x$ To show that $f$ and $g$ are inverses of one another, we plug the formula for $f(x)$ for every $x$ in $g(x)$ and vice versa: - Plug $f(x)$ for every $x$ in $g(x)$: $$(gof)(x)=g(f(x))=\sqrt[3]{x^3}=x$$ - Plug $g(x)$ for every $x$ in $f(x)$: $$(fog)(x)=f(g(x))=(\sqrt[3]x)^3=x$$ Since they both result in $x$, we conclude that $f$ and $g$ are inverses of one another. b) The graphs of $f$ and $g$ are shown below. c) First, we need to find $f'$ and $g'$: $$f'(x)=3x^2$$ $$g'(x)=(\sqrt[3]x)'=(x^{1/3})'=\frac{1}{3}x^{-2/3}$$ - At the point $(1,1)$, we have: $$f'(1)=3\times1^2=3$$ The tangent line to $f(x)$ at $(1,1)$ is $$y-1=3(x-1)=3x-3$$ $$y=3x-2$$ $$g'(1)=\frac{1}{3}\times1^{-2/3}=\frac{1}{3}$$ The tangent line to $g(x)$ at $(1,1)$ is $$y-1=\frac{1}{3}(x-1)=\frac{1}{3}x-\frac{1}{3}$$ $$y=\frac{1}{3}x+\frac{2}{3}$$ - At the point $(-1,-1)$, we have: $$f'(-1)=3\times(-1)^2=3$$ The tangent line to $f(x)$ at $(-1,-1)$ is $$y+1=3(x+1)=3x+3$$ $$y=3x+2$$ $$g'(-1)=\frac{1}{3}\times(-1)^{-2/3}=\frac{1}{3}\times1=\frac{1}{3}$$ The tangent line to $g(x)$ at $(-1,-1)$ is $$y+1=\frac{1}{3}(x+1)=\frac{1}{3}x+\frac{1}{3}$$ $$y=\frac{1}{3}x-\frac{2}{3}$$ d) At the origin, we have: $$f'(0)=3\times0^2=0$$ The tangent line to $f(x)$ at the origin is $$y-0=0(x-0)=0$$ $$y=0$$ $$g'(0)=\frac{1}{3}\times0^{-2/3}=\frac{1}{3\times0^{2/3}}=\frac{1}{0}=\infty$$ This means $g(x)$ has a vertical tangent here. Since this point is the origin, that vertical tangent is $x=0$.
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