Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.1 - Inverse Functions and Their Derivatives - Exercises 7.1 - Page 373: 29


$ f^{-1}(x)= \sqrt{\frac{1}{x}}$ Domain of $ f^{-1}$ is $(0,\infty)$ Range of $ f^{-1}$ is $(0,\infty)$

Work Step by Step

Given: $ f(x) =\frac{1}{x^{2}}$ Find the inverse: $y=\frac{1}{x^{2}}$ $x^{2} =\frac{1}{y}$ $x=\sqrt{\frac{1}{y}}$ Switch $x$ and $y$: $y= \sqrt{\frac{1}{x}} =f^{-1}(x) $ We see that: Domain of $ f^{-1}$ is $(0,\infty)$ and Range of $ f^{-1}$ is $(0,\infty)$ Confirm the inverse: \begin{aligned} f\left(f^{-1}(x)\right) &=f\left(\frac{1}{\sqrt{x}}\right) \\ &=\frac{1}{\left(\frac{1}{\sqrt{x}}\right)^{2}} \\ &=\frac{1}{1/x} \\ &=x \\f^{-1}(f(x)) &=f^{-1}\left(\frac{1}{x^{2}}\right) \\ &=\frac{1}{\sqrt{\frac{1}{x^{2}}}} \\ &=\frac{1}{\frac{1}{x}} \\ &=x \end{aligned}
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