# Inequalities. (1)

• May 26th 2011, 01:12 PM
TWiX
Inequalities. (1)
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 : $\displaystyle -4 \leq y \leq 4$
• May 26th 2011, 01:17 PM
Ackbeet
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?
• May 26th 2011, 01:26 PM
TWiX
Finding extremes is a method in calculus.
This is pre-calculus subforum.
• May 26th 2011, 01:29 PM
Also sprach Zarathustra
-2<x<1

-6<3x<3

0<x^2<1

-6<3x-x^2<2
• May 26th 2011, 03:46 PM
LoblawsLawBlog
Quote:

Originally Posted by TWiX
Finding extremes is a method in calculus.

You have a parabola, so you can do it with basic algebra.
• May 26th 2011, 04:42 PM
Ackbeet
Quote:

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?