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: 22



Work Step by Step

$$y=f(x)=x^2-2x+1\hspace{1cm}x\ge1$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y = x^2-2x+1$$ $$y=(x-1)^2$$ Since $x\ge1$, it follows that $(x-1)\ge0$. Therefore, $$x-1=\sqrt{y}$$ $$x=1+\sqrt{y}$$ 2) Interchange $x$ and $y$: $$y=1+\sqrt{x}$$ Therefore, the inverse of function $y=f(x)=x^2-2x+1$, $x\ge1$ is the function $y=1+\sqrt{x}$. In other words, $$f^{-1}(x)=1+\sqrt{x}$$
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