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 48: 18

Answer

(a) The graph is shown in the image below. It is symmetric about the origin. (b) $f$ is its own inverse. To prove that, find the inverse of function $f(x)=\frac{1}{x}$ like normal.
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Work Step by Step

(a) The graph of function $f(x)=\frac{1}{x}$ is shown in the image below. The graph, as we can see, is symmetric about the origin. (b) To see if the function $f(x)=\frac{1}{x}$ is its own inverse or not, we would find the inverse of $f(x)=\frac{1}{x}$ - First, solve for $x$ in terms of $f(x)$: $$f(x)=\frac{1}{x}$$ $$x=\frac{1}{f(x)}$$ - Now, interchange $x$ and $f(x)$: $$f^{-1}(x)=\frac{1}{x}$$ As you can see, both the function and its inverse are the same. Therefore, $f$ is its own inverse.
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