# Math question

• Nov 15th 2006, 05:58 PM
Mr_Green
Math question
• Nov 15th 2006, 06:33 PM
Quick
Quote:

Originally Posted by Mr_Green

are you the same mr. green that posted this?

Because you certainly have asked a lot of questions
• Nov 15th 2006, 06:36 PM
Mr_Green
Quote:

Originally Posted by Quick
are you the same mr. green that posted this?

Because you certainly have asked a lot of questions

yes in deed
• Nov 15th 2006, 07:08 PM
Mr_Green
a and b will be decimals rite?
• Nov 16th 2006, 04:39 AM
topsquark
Quote:

Originally Posted by Mr_Green

You want an interval such that $\left | f(x) - 6 \right | \leq 2$.

So you want $\left | x^2 - 2x - 3 \right | \leq 2$

So
$x^2 - 2x - 3 \leq 2$
and
$x^2 - 2x - 3 \geq -2$

For the first inequality solve:
$x^2 - 2x - 3 = 2$

$x^2 - 2x - 5 = 0$

So $x = 1 \pm \sqrt{6}$ by the quadratic formula. We want the "+" solution as the "-" solution is less than 0. So we want $x \leq 1 + \sqrt{6}$ to solve the inequality.

For the second inequality solve:
$x^2 - 2x - 3 = -2$

$x^2 - 2x - 1 = 0$

So $x = 1 \pm \sqrt{2}$ by the quadratic formula. We again want the "+" solution since the "-" solution is less than 0. So we want $x \geq 1 + \sqrt{2}$.

Putting the two together gives an interval for x:
$1 + \sqrt{2} \leq x \leq 1 + \sqrt{6}$

-Dan