Inequality problem $$|x|^{x^2-x-2}<1$$
I tried to do this problem by taking different cases like first taking $x$ and then $-x$.
For $+x$, the exponent should be less than $1$ so that the whole thing becomes a fraction hence lesser than $1$.
For 2nd cases, I'm confused what cases to take.
Help is appreciated!

$$f(x)=|x|^{x^2-x-2}=|x|^{(x+1)(x-2)}$$ It follows that $f(x)=1$ when $x=-1,1,2$ You can verify that $-1$ correspond to a minimun, that a second minimun is between $1$ and $2$ and there is not another one. Hence, by continuity, you have as answer the open interval $$1\lt x\lt2$$
