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?
Best Answer
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.)