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