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: 23



Work Step by Step

$$y=f(x)=(x+1)^2\hspace{1cm}x\ge-1$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=(x+1)^2$$ Since $x\ge-1$, it follows that $(x+1)\ge0$. Therefore, $$x+1=\sqrt{y}$$ $$x=\sqrt{y}-1$$ 2) Interchange $x$ and $y$: $$y=\sqrt{x}-1$$ Therefore, the inverse of function $y=f(x)=(x+1)^2$, $x\ge-1$ is the function $y=\sqrt{x}-1$. In other words, $$f^{-1}(x)=\sqrt{x}-1$$
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