## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 49: 19

#### Answer

$$f^{-1}(x)=\sqrt{x-1}$$

#### Work Step by Step

$$y=f(x)=x^2+1\hspace{1cm}x\ge0$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y = x^2+1$$ $$x^2=y-1$$ Since our domain for $x$ is $[0,\infty)$, we take the positive value of $x$ here, which means $$x=\sqrt{y-1}$$ 2) Interchange $x$ and $y$: $$y=\sqrt{x-1}$$ Therefore, the inverse of function $f(x)=x^2+1$, $x\ge0$ is the function $y=\sqrt{x-1}$. In other words, $$f^{-1}(x)=\sqrt{x-1}$$

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