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

First you need to find the critical numbers and see if they are in your interval (0,9). To find the critical numbers, derive the equation and set it equal to 0.

$\displaystyle V'(x)=-3x^2-12x+15$

$\displaystyle 0=x^2+4x-5=(x+5)(x-1)$

$\displaystyle x=1$ $\displaystyle x=-5$

So you have a critical value at x=1. To find the maximum, you must plug in the endpoints of your interval (i.e. x=0 and x=9) and the critical number within the interval (i.e. x=1) into the original equation and find which one is the highest. That is your maximum.

$\displaystyle V(0)=-15$

$\displaystyle V(1)=-1-6+15-15=-7$

$\displaystyle V(9)=-9^3+6*9^2+15*9-15=-729+486+135-15=-123$

Well, that's what I got and it doesn't make sense. Can anyone else help? I am pretty sure that I have the right method of getting the maximum. I just don't understand how the answer could be negative.