# Stuck on simple calculus problem...

• Nov 27th 2010, 03:03 PM
metaname90
Stuck on simple calculus problem...
Let

Defined in the interval 1<(or equal to) x <(or equal to) 9

I've found the derivative:

But don't know what to do next....
• Nov 27th 2010, 03:17 PM
Mondreus
What exactly is the question?
• Nov 27th 2010, 04:58 PM
metaname90
Oh sorry. The maximum and minimum value (both the x position and the actual value, though obviously finding one it will be easy to find the other). That's why i took the derivative but I don't know how I would set that to zero or what...
• Nov 27th 2010, 05:02 PM
dwsmith
Establish same common denominators:

$\displaystyle \frac{(15-x)(2x-6)-2(x^2-6x+25)}{2\sqrt{x^2-6x+25}}=0\rightarrow (15-x)(2x-6)-2(x^2-6x+25)=0$

and

whenever the denominator is zero you will have an asymptote unless it factors out and then you have a whole in the graph.

Your critical values to test are when the numerator and denominator are 0.
• Nov 27th 2010, 05:25 PM
skeeter
$\displaystyle \frac{(15-x)(x-3)}{\sqrt{x^2-6x+25}} - \sqrt{x^2-6x+25} = 0$

$\displaystyle \frac{(15-x)(x-3)}{\sqrt{x^2-6x+25}} = \sqrt{x^2-6x+25}
$

$(15-x)(x-3) = x^2-6x+25$

$-x^2+18x-45 = x^2-6x+25$

$0 = 2x^2-24x+70$

$0 = x^2-12x+35$

$0 = (x-7)(x-5)$

check the critical values for extrema ... don't forget the endpoints.
• Nov 27th 2010, 06:17 PM
metaname90
Thank you both of you!
• Nov 27th 2010, 06:19 PM
dwsmith
Quote:

Originally Posted by skeeter
$\displaystyle \frac{(15-x)(x-3)}{\sqrt{x^2-6x+25}} - \sqrt{x^2-6x+25} = 0$

$\displaystyle \frac{(15-x)(x-3)}{\sqrt{x^2-6x+25}} = \sqrt{x^2-6x+25}
$

$(15-x)(x-3) = x^2-6x+25$

$-x^2+18x-45 = x^2-6x+25$

$0 = 2x^2-24x+70$

$0 = x^2-12x+35$

$0 = (x-7)(x-5)$

check the critical values for extrema ... don't forget the endpoints.

Oops, I see you factored it out