# Critical number?

• Dec 9th 2008, 04:29 PM
DHS1
Critical number?
Determine all of the critical numbers of the given function.

I'm like 90% sure that means solve for x, check if what you got for x actually works, and that/those are your critical number(s).

Anyways, here's the problem:

$\displaystyle g(x) = x - \sqrt{14-x} - 2$

Any ideas anyone? I tried solving for x:

$\displaystyle 0 = x-\sqrt{14-x} - 2$

then
$\displaystyle \sqrt{14-x} = x-2$

then squared both sides
$\displaystyle 14-x = x^2-4$

then moved some stuff around
$\displaystyle 18 = x^2+x$

and now I'm stuck there, not even knowing if I'm supposed to be doing this.

HALP
• Dec 9th 2008, 04:46 PM
skeeter
Quote:

Originally Posted by DHS1
Determine all of the critical numbers of the given function.

I'm like 90% sure that means solve for x, check if what you got for x actually works, and that/those are your critical number(s).

Anyways, here's the problem:

$\displaystyle g(x) = x - \sqrt{14-x} - 2$

Any ideas anyone? I tried solving for x:

$\displaystyle 0 = x-\sqrt{14-x} - 2$

then
$\displaystyle \sqrt{14-x} = x-2$

then squared both sides
$\displaystyle 14-x = x^2-4$ ... mistake here, corrected below

$\displaystyle 14 - x = x^2 - 4x + 4$

$\displaystyle 0 = x^2 - 3x - 10$

$\displaystyle 0 = (x - 5)(x + 2)$

$\displaystyle x = 5$ ... $\displaystyle x = -2$

check both solutions in the original equation
• Dec 9th 2008, 05:08 PM
Mentia
also remember that if the factorization isnt obvious that you can always use the quadratic equation.