Attachment 1238

Printable View

- Nov 15th 2006, 04:58 PMMr_GreenMath question
- Nov 15th 2006, 05:33 PMQuick
are you the same mr. green that posted this?

Because you certainly have asked a lot of questions - Nov 15th 2006, 05:36 PMMr_Green
- Nov 15th 2006, 06:08 PMMr_Green
a and b will be decimals rite?

- Nov 16th 2006, 03:39 AMtopsquark
You want an interval such that $\displaystyle \left | f(x) - 6 \right | \leq 2$.

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

So

$\displaystyle x^2 - 2x - 3 \leq 2$

and

$\displaystyle x^2 - 2x - 3 \geq -2$

For the first inequality solve:

$\displaystyle x^2 - 2x - 3 = 2$

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

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

For the second inequality solve:

$\displaystyle x^2 - 2x - 3 = -2$

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

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

Putting the two together gives an interval for x:

$\displaystyle 1 + \sqrt{2} \leq x \leq 1 + \sqrt{6}$

-Dan