Math question

**Quick** · November 15th 2006, 05:33 PM

Originally Posted by Mr_Green

$Click image for larger version. Name: amath23.jpg Views: 19 Size: 11.9 KB ID: 1238$

are you the same mr. green that posted this?

Because you certainly have asked a lot of questions

**Mr_Green** · November 15th 2006, 05:36 PM

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

**Mr_Green** · November 15th 2006, 06:08 PM

a and b will be decimals rite?

**topsquark** · November 16th 2006, 03:39 AM

Originally Posted by Mr_Green

$Click image for larger version. Name: amath23.jpg Views: 19 Size: 11.9 KB ID: 1238$

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

Thread: Math question

LinkBack

Thread Tools

Display

Math question

Similar Math Help Forum Discussions

Math question

2 math question help

tricky math math question

Cr. Math and a dumb math question

math question help

Search Tags