Answer
$f$ is restricted to $[0,\infty)$.
$f^{-1}(x)=\sqrt{x-1}$ for $x\in[1,\infty)$
Work Step by Step
We are given the function:
$f(x)=x^2+1$
Graph the function and use the Horizontal Line Test to check if it has an inverse. We notice that a horizontal line intersects the graph at more than one point, so the function does not have an inverse on the domain $(-\infty,\infty)$.
We restrict the domain of $f$ to the interval $[0,\infty)$, on which the function has an inverse.
Determine the inverse:
$y=x^2+1$
$x=y^2+1$
$y^2=x-1$
$y=\sqrt{x-1}$ as the domain of $f$ equals the range of $f^{-1}$, and the range of $f$ equals the domain of $f^{-1}$.
