Question

The inverse of the function
f(x)=(x+1)^2+2 if x ≥ 0 is

Answer 1

If $f^{-1}(x)$ is the inverse of $f(x)$, then

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

We're given a domain for $f(x)$ of $x\ge0$, so $f\left(f^{-1}(x)\right) = x$ is valid only for $f^{-1}(x)\ge0$.

Now,

$f\left(f^{-1}(x)\right) = \left(f^{-1}(x) + 1\right)^2 + 2 = x$

Solve for the inverse :

$\left(f^{-1}(x) + 1\right)^2 = x - 2 \\\\ \sqrt{\left(f^{-1}(x)+1\right)^2} = \sqrt{x-2} \\\\ \left|f^{-1}(x) + 1\right| = \sqrt{x-2}$

Since $f^{-1}(x)\ge0 \implies f^{-1}(x)+1 \ge0$, by definition of absolute value we have

$\left|f^{-1}(x)+1\right| = f^{-1}(x) + 1$

Then we end up with

$f^{-1}(x) + 1 = \sqrt{x-2} \\\\ \boxed{f^{-1}(x) = \sqrt{x-2}-1}$