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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.

wolfcall
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    The function $f(x)=x^2$ from $\mathbb{R}$ to itself does not have an inverse because it's not injective. The "$f^{-1}$" in the question is a bit of an abuse of notation--$f^{-1}(T_1)$ means "the set of all elements in $\mathbb{R}$ that are mapped into $T_1$ by $f$." So, for example, for $T_1$, which real numbers land inside $[-2,1]$ when you square them? – Nick D. Mar 27 '14 at 01:20

2 Answers2

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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.

Nick
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  • I need this dumbed down a little more. In order to find this I have to take out the values that result in undefined? This is still considered a function regardless of the undefined values? – wolfcall Mar 27 '14 at 03:36
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    Well, suppose $T_3 = [-2,-1]$. Then $f^{-1}(T_3)$ would be empty, since $x^2$ is always nonnegative. There's no problem with $x^2$ being undefined. It's always defined, just never negative when $x$ is a real number. Don't think of $f^{-1}(T_3)$ as $f^{-1}$ applied to $T_3$. Think of it as a set, whose members are the real numbers that $f$ sends to $T_3$. – Nick Mar 27 '14 at 05:18
  • So I should think the preimage is not undefined but does not exist for the first half resulting in an empty set. And for the rest apply the sqrt as normal. – wolfcall Mar 27 '14 at 16:12
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    Yes, and remember that the preimage can contain negative values since $x^2=(-x)^2$. – Nick Mar 27 '14 at 16:54
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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.

ZHN
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