Let f(x) = (x + 1)² – 1, x ≥ –1 then the set {x : f(x) = f ^–1(x)} is equal to

Let y = (x + 1)² – 1, x ≥ – 1

$\Rightarrow$(x + 1)² = y + 1

$\Rightarrow$x + 1 = y + 1 as x ≥ – 1

$\Rightarrow$$x = – 1 +\sqrt{y+1} , y \ge – 1$

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

So            f (x) = f ^–1(x)

$\Rightarrow {(x + 1)}^2 – 1 = – 1 + \sqrt{x+1}$

$\Rightarrow \sqrt{x + 1} = 0 or {(x + 1)}^{3/2} = 1$

$\Rightarrow$x = – 1 or x =  0