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....

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- Nov 27th 2010, 03:03 PMmetaname90Stuck 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 PMMondreus
What exactly is the question?

- Nov 27th 2010, 04:58 PMmetaname90
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 PMdwsmith
Establish same common denominators:

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

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

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

$

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

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

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

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

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

check the critical values for extrema ... don't forget the endpoints. - Nov 27th 2010, 06:17 PMmetaname90
Thank you both of you!

- Nov 27th 2010, 06:19 PMdwsmith