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: 20



Work Step by Step

$$y=f(x)=x^2\hspace{1cm}x\le0$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y = x^2$$ Since our domain for $x$ is $(-\infty,0]$, we take the negative value of $x$ here, which means $$x=-\sqrt{y}$$ 2) Interchange $x$ and $y$: $$y=-\sqrt{x}$$ Therefore, the inverse of function $y=f(x)=x^2$, $x\le0$ is the function $y=-\sqrt{x}$. In other words, $$f^{-1}(x)=-\sqrt{x}$$
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