University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 49: 27


$$f^{-1}(x)=\sqrt[3]{x-1}$$ - Domain: $(-\infty,\infty)$ - Range: $(-\infty,\infty)$

Work Step by Step

$$y=f(x)=x^{3}+1\hspace{1cm}$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=x^3+1$$ $$x^3=y-1$$ $$x=\sqrt[3]{y-1}$$ 2) Interchange $x$ and $y$: $$y=\sqrt[3]{x-1}$$ Therefore, $$f^{-1}(x)=\sqrt[3]{x-1}$$ - Domain: $x$ is defined in $R$. So the domain of $f^{-1}$ is $(-\infty,\infty)$. - Range: $x$ ranges from $-\infty$ to $\infty$. All the same, $\sqrt[3]{x-1}$ also ranges from $-\infty$ to $\infty$. So the range of $f^{-1}$ is $(-\infty,\infty)$ *Check: $f(f^{-1}(x))=(\sqrt[3]{x-1})^3+1=(x-1)+1=x$ $f^{-1}(f(x))=\sqrt[3]{(x^3+1)-1}=\sqrt[3]{x^3}=x$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$
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