# Given function and intervals, determine the set

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?

• 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.
– user457568
Nov 3, 2018 at 15:30

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.)

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.