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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?

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    $\begingroup$ 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 $(-18,18)$? . $\endgroup$
    – Mirko
    Nov 29, 2016 at 18:26
  • $\begingroup$ @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. $\endgroup$
    – Zelazny
    Nov 29, 2016 at 18:33

2 Answers 2

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Look at the graphs of:

  • $ 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.

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Hint for the first.

$$f(A)=f([0,3)×[-2,2])=(-18,18)$$

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  • $\begingroup$ Absolutely right, I bungled $f[A]$. $\endgroup$
    – Zelazny
    Nov 29, 2016 at 18:47

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