Find the Inverse function of f. $f(x)=1+\sqrt{1+x}$

Question

I found the Inverse of the function, $f^{-1}(x)= x^2-2x$. The back of my pre-cal book gives me the inverse of the function and the domain.

What I don't understand is, how the domain comes to be $x \geq 1$.

Can someone show me the steps onto how to get the domain?

Answer 1

Hint Domain of $f^{-1}$= Range of $f$.

Note that the reason why we need this restriction is that $y=1+\sqrt{1+x} \geq 1$. Without this restriction, the function $f(x)=x^2-2x$ is not one to one, thus cannot be the inverse of something ;)

Answer 2

If $f(x) = 1+ \sqrt{1+x}$, then presumably the domain is $x \ge -1$, or $[-1,\infty)$. The range can be seen to be $[1,\infty)$.

Here is a plot of $f$: