# Find and graph $f^{-1}$

We have $f: \mathbb{R}^2 \to \mathbb{R}, \ f(x,y)=x^2y, \ A = (-3,1] \times [-2,2], \ B = [-1,2)$. We want to find $f[A]$ and find and graph $f^{-1}[B]$.

$f[A] = (-18,18)$ but I get stuck on finding $f^{-1}$. How might I approach this?

• Pick various $b\in B$ and find $f^{-1}(b)$ first. Then take the union over $b\in B$. For example could you describe $f^{-1}(1)$? Why do you use semicolon instead of a comma? I think the standard notation would be $B=[-1,2)$ rather than $B=[-1;2)$. I think your $f[A]$ should be an interval symmetric about the origin, perhaps $(-18,18)$? . Nov 29, 2016 at 18:26
• @Mirko No reason other than that I've seen this notation a couple of times. A quick review reveals that in fact [-1,2) is the standard approach. Nov 29, 2016 at 18:33

• $x^2y = -1$
• $x^2y = 2$
Solve for $y = \dots$, and plot them both. Those are the borders of your interval in the preimage, and it should be quite easy to decide which areas belong to the preimage and which not.
$$f(A)=f([0,3)×[-2,2])=(-18,18)$$
• Absolutely right, I bungled $f[A]$. Nov 29, 2016 at 18:47