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: 25


$f^{-1}(x)=\sqrt[5]x$ - Domain: $(-\infty,\infty)$ - Range: $(-\infty,\infty)$

Work Step by Step

$$y=f(x)=x^{5}\hspace{1cm}$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=x^{5}$$ $$x=\sqrt[5]{y}$$ 2) Interchange $x$ and $y$: $$y=\sqrt[5]x$$ Therefore, $$f^{-1}(x)=\sqrt[5]x$$ - 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, $\sqrt[5]x$ also ranges from $-\infty$ to $\infty$. So the range of $f^{-1}$ is $(-\infty,\infty)$ *Check: $f(f^{-1}(x))=(\sqrt[5]x)^5=(x^{1/5})^5=x^{\frac{1}{5}\times5}=x^1=x$ $f^{-1}(f(x))=\sqrt[5]{x^5}=(x^5)^{1/5}=x^{5\times\frac{1}{5}}=x^1=x$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$
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