Given function $f(x) = \sqrt{x^2 + 2}$, find out, what are values of $f^{-1}(A)$, where $A = (1, 2]$.
At first, I have to find inverse function, which is $f^{-1}(x) = \pm\sqrt{x^2 -2}$. Now solving inequality $1 < f^{-1}(x) < 2$ I have to get an interval in domain, which corresponds to interval $(1, 2]$ in codomain. From graph, it has to be $[-\sqrt{2}, \sqrt{2}] \to (1, 2]$, but inequality yields me wrong values. What am I doing wrong here?
Where you write "solving inequality" is not that clear what you do and tried. The following is an idea:
$$x\in f^{-1}(A)\iff 1<\sqrt{x^2+2}\le2\,,\,\,\text{and now you've to solve these two inequalities}$$
...and observe the expression within the square root is always positive (in fact, always equal or greater than $\;2\;$)
$$\sqrt{x^2+2}>1\implies x^2+2>1\implies x^2>-1\implies x\in\Bbb R$$
Take care: in general, we'd have to do in the first step $\;|x^2+1|>1\;$ , but as noted above this is unnecessary in this case.
$$\sqrt{x^2+2}\le2\implies x^2+2\le4\implies\ldots\text{etc. (complete here)}$$