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} = √(x-2) \\\\ \left|f^(-1)(x) + 1\right| = √(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 = √(x-2) \\\\ \boxed{f^(-1)(x) = √(x-2)-1}`