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