Question

Let $f\left(x\right)=\{\begin{array}{cc}x& -1\le x\le 1\\ x^2& 1<x\le 2\end{array}$, the range of $h^{-1}\left(x\right)$ where $h\left(x\right)=fof\left(x\right)$ is

Answer 1

range of $h^{-}\left(x\right)$ is domain of $h\left(x\right)$

for $-1\le x\le 1$ $f\left(x\right)\in [-1,1]$ and for $1<x\le 2$ $f\left(x\right)\in (1,4]$

$\because 2$ to $4$ should not be domain domain $⟹x^2=2⟹x=\sqrt{2}$

the domain of $h\left(x\right)\in [-1,\sqrt{2}]$