How come the preimage of $y=x^2$ given $T = [-2, 1]$ is $-1 \leq x \leq 1$ Shouldn't it be $-\sqrt 2 \leq x \leq 1$?
How do you determine the preimage step by step?
 A: $-2 \leq x^2\leq 1\iff 0\leq x^2\leq 1\iff -1\leq x\leq 1$.
A: You have to consider what the graph of $y=x^2$ looks like. We know $x^2\geq 0$, hence $f^{-1}([-2,1])=f^{-1}([0,1])$, which is now easily checked to be $[-1,1]$.
A: Note, $T = [-2, 1]$ is the given subset of reals into which $f(x) = y = x^2$ maps some values of $x$. 
Since $x^2 \ge 0$ for all $x$, the only subset of $T$ that matters, given the function $y = x^2$, is the interval $[0, 1]$. As you can see in the plot below, the image of $f(x)=y$ is empty below the $x$-axis.
So your task is to determine then pre-image corresponding $f(x) = y = x^2 \geq 0$ when $f(x) = y = x^2 \subseteq [0, 1]$. (See the graph below.)
$\quad\quad\quad\quad\quad\quad\ y = x^2$

As you can see from the graph (pictures help!), $y \in [0, 1] \subset [-2, 1]$ corresponds to $x \in [-1, 1]$. 

A: I see your thinking, but since $x^2>0$ for all real $x$, $T$ can be changed to $\left[0,1\right]$.  (i.e. there are no real $x$ such that $x^2\in \left[-2,0\right)$),  Then, $\sqrt{1}=\pm 1$, so the preimage should be $\left[-1,1\right]$.
