Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.3 Inverse, Exponential, and Logarithmic Functions - 1.3 Exercises - Page 36: 33

Answer

$f(x)=\sqrt{x}$ $f^{-1}(x)=x^{2}$
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Work Step by Step

$f(x)=\sqrt{x}$, for $x\ge0$ Substitute $f(x)$ by $y$: $y=\sqrt{x}$ Solve the equation for $x$. Do so by squaring both sides and rearranging the equation: $y^{2}=(\sqrt{x})^{2}$ $x=y^{2}$ Interchange $x$ and $y$: $y=x^{2}$ Substitute $y$ by $f^{-1}(x)$: $f^{-1}(x)=x^{2}$ The graph of both the original function and its inverse is shown in the answer section.
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