Let x&y be real numbers such that : $\displaystyle -2 \leq x \leq 1$ and $\displaystyle 1 \leq 3x-x^2+2y^2 \leq 22 $ Show that : $\displaystyle -4 \leq y \leq 4$
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I would start by letting $\displaystyle f(x)=3x-x^{2},$ and finding its extremes on the interval $\displaystyle [-2,1].$ What do you get?
Finding extremes is a method in calculus. This is pre-calculus subforum.
-2<x<1 -6<3x<3 0<x^2<1 -6<3x-x^2<2
Originally Posted by TWiX Finding extremes is a method in calculus. You have a parabola, so you can do it with basic algebra.
Originally Posted by Also sprach Zarathustra -2<x<1 -6<3x<3 0<x^2<1 -6<3x-x^2<2 I think there is an error here. The correct inequality is $\displaystyle -10\le 3x-x^{2}\le 2.$ Do you see how to get this?
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