Inequalities. (1)

**TWiX** · May 26th 2011, 01:12 PM

Let x&y be real numbers such that : $-2 \leq x \leq 1$ and $1 \leq 3x-x^2+2y^2 \leq 22$

Show that : $-4 \leq y \leq 4$

**Ackbeet** · May 26th 2011, 01:17 PM

I would start by letting $f(x)=3x-x^{2},$ and finding its extremes on the interval $[-2,1].$ What do you get?

**TWiX** · May 26th 2011, 01:26 PM

Finding extremes is a method in calculus.
This is pre-calculus subforum.

**Also sprach Zarathustra** · May 26th 2011, 01:29 PM

-2<x<1

-6<3x<3

0<x^2<1

-6<3x-x^2<2

**LoblawsLawBlog** · May 26th 2011, 03:46 PM

Originally Posted by TWiX

Finding extremes is a method in calculus.

You have a parabola, so you can do it with basic algebra.

**Ackbeet** · May 26th 2011, 04:42 PM

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

$-10\le 3x-x^{2}\le 2.$

Do you see how to get this?

Thread: Inequalities. (1)