## Thomas' Calculus 13th Edition

$$f^{-1}(x)=\sqrt{x-1}$$
Given: $y=x^{2}+1$ We solve for $x$: $\Rightarrow x^{2}=y-1$ $\Rightarrow x=\sqrt{y-1}$ Interchanging $x$ and $y$, we get: $y=\sqrt{x-1}=f^{-1}(x)$