Sets and inverse functions I am working on an assignment and I want to know if my thinking is right for my answer. i have been giving the following functions and sets:
$$f(x)=x^2$$
I found the inverse to be.
$$f^{-1}(x)=\sqrt{x}$$
All $x\in \mathbb{R}$
$$T_1=[-2,1]$$
$$T_2=(-1,2)$$
I have to determine the following four sets.
$$f^{-1}(T_1\cup T_2)$$
$$f^{-1}(T_1)\cup f^{-1}(T_2)$$
$$f^{-1}(T_1\cap T_2)$$
$$f^{-1}(T_1) \cap f^{-1}(T_2)$$
I cannot see how I can give a resulting set because with the radical causes part of these sets become undefined and if that is true none of these are functions.
 A: Remember the definition of $f^{-1}(S)$, where $S$ is a set: this is the set of all $x$ in $\Bbb{R}$ such that $f(x)$ is in $S$. So, for example, $f^{-1}(T_1 \cup T_2)$ is the set of all $x \in \Bbb{R}$ such that $f(x) \in T_1 \cup T_2 = [-2,2)$. We can find out what this set is without referring to an explicit inverse function for $f$. The idea is to use what we know about $f$. We know that $x^2$ is nonnegative for all $x$, so really we are looking for all $x$ such that $x^2 \in [0,2)$. Certainly any number from $0$ to $\sqrt{2}$ (not including $\sqrt{2}$) will do. Notice that if $x$ is in the set, then so is $-x$ since $x^2 = (-x)^2$. So we deduce that $f^{-1}(T_1 \cup T_2) = (-\sqrt{2},\sqrt{2})$. The rest work out similarly.
A: The function $ f: R \rightarrow R $ given by $ f(x) = x^2 $ is not bijective so she does not admit inverse. You can only set up an inverse redefine  $f $ so as to make it a bijection. For example, $ f: [0, + \infty) \rightarrow [0, + \infty) $, $ f(x) = x^2 $ is bijective, and therefore it is possible to inverse. In this case, there is $ f^{-1}: [0, + \infty)  \rightarrow [0, + \infty) $ and  is given by $f^{-1} (y) = \sqrt{y} $ . 
What do you want this exercise, as I understand it, is to obtain pre-images of highlighted sets (I think there was confusion between pre-image and inverse function, because the notations are the same, but the meaning is quite different) . 
In this case, given $ f: A \rightarrow B $ and $ Y \subset B $, find all $ x \in A $ that have image in $ Y $, i.e. find $ f ^ {-1} (Y) $ (pre-image of $Y$ by $ f $). So, in your case,  $ f^{-1} (T_1) = [-1,1] $ and $ f ^ {-1} (T_2) = [- \sqrt {2}, \sqrt {2}] $, for example.
Note: It should be noted a detail: let $ f: A \rightarrow B $ and $ Y \subset B $, even if $ Y \neq \emptyset $ may happen that $ f^{-1} (Y) = \emptyset $. For example, given $ f: (0, + \infty) \rightarrow \mathbb {R} $ by $ f (x) = x^2 $ and $ Y = (-1,0) \subset \mathbb {R} $. In this case $ f^{-1}(Y) = \emptyset$ because there is no $ x \in (0,+ \infty)$ such that $ f (x) \in (-1,0) = Y $.
Hope this can help you.
