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Function $\mathbb{Q} \rightarrow \mathbb{Q}$ is defined by $f(x)=|x|$. We have the intervals $I_1=[-5, -3), I_2=(2,4]$ and $I_3=[-1,3)$. Determine the sets

  • $f^{-1}(2)=f^{-1}(\left\{2\right\})=...$
  • $f^{-1}(I_1)$

Just some examples so I see how it works correctly because I don't understand what would be the absolute value of the interval $I_1$? Would it just be the length $2$?

Anyway, for the first set we would just have the interval $-2,2$ right?

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    $\begingroup$ Please note that $ 2\sqrt{2} $ belongs to $ I_2 $: the inverse image of a set is usually defined when such set is contained in the codomain. $\endgroup$
    – user457568
    Nov 3, 2018 at 15:30

2 Answers 2

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By definition, for a function $f:A\to B$, and given any $P\subseteq B$ the inverse image of $P$ is defined to be the set $$f^{-1}(P)=\{x\in A\,|\,f(x)\in P\}.$$

In this case where $f(x)=|x|$ and $A=B=\mathbb{Q}$, the inverse image of $P\subseteq\mathbb{Q}$ is the set $$f^{-1}(P)=\{x\in\mathbb{Q}\,|\,|x|\in P\}.$$

The inverse image of $P$ is the set of points $x\in\mathbb{Q}$ such that $|x|\in P$.

For the first case we have $$f^{-1}(\{2\})=\{x\in\mathbb{Q}: |x|=2\}=\{-2,2\}$$ (which is not an interval). To find $f^{-1}(\{2\})$ graphically you could plot the function, find $2$ on the vertical axis, and then draw a horizontal line and see at which $x$-values it intersects the graph. The horizontal line will intersect the graph at $-2$ and $2$.

For the second one, think about what numbers $x\in\mathbb{Q}$ are such that $|x|\in[-5,-3)$. (Again, if it is not obvious, looking at the graph might help.)

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The inverse image $ f^{-1}(T) $ of a function $ f:X\to Y $, $ T\subset Y $ is defined as the set of the elements of the domain $ X $ such that $ f(x)\in T $, a.k.a. $ \left\{x\in X:f(x)\in T\right\} $.

To find $ g^{-1}(a) $ (where $ g:\mathbb{Q}\to\mathbb{R}:x\mapsto\lvert x\rvert $, $ a\in\mathbb{Q} $) means finding all the rational points $ x $ such that $ \lvert x\rvert = a $; if $ a=2\in\mathbb{Q} $, the rationals such that $ \lvert x\rvert = 2$ are simply the elements of $ \{\pm 2\} $.

Then $ f^{-1}(I_1) = \{x\in\mathbb{Q}:\lvert x\rvert\in I_1\} $, i.e. $ -5\leqq\lvert x\rvert \lt -3$. Note that we need to choose only the rational points.

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