How can I solve this...any help? To save us time I just screen shot it...

Question

anonymous · Answer

attachment

anonymous · Answer

Which part are you having trouble with? Have you found the first and second derivatives of the function?

anonymous · Answer

Nope...I just took the derivate of the function....which is -6x^2 + 42x-36 

but i didn't understand how to find or what A and B equals to

accessdenied · Answer

find the x-values where the slope is 0, which will be the places where we can have a maximum or a minimum.

anonymous · Answer

You want to solve the equation $-6x^2+42x-36=0$. There should be two solutions, which they're calling A and B.

anonymous · Answer

Ah ok...i will try

anonymous · Answer

yup i got the first two A=1 and B= 6

anonymous · Answer

Right on. Now take the second derivative for the next part.

anonymous · Answer

so -12x+42=0 again?

anonymous · Answer

Actually, you're looking for the values of the second derivative at A and B. Have you learned about the second derivative test for local extrema?

anonymous · Answer

oh ok...not really...

anonymous · Answer

Okay, it's actually pretty easy. If $f'(A)=0$ and $f''(A)\ge0$, then it's a local minimum. If $f'(A)=0$ and $f''(A)\le0$, then it's a local maximum. Pretty simple :)

dumbcow · Answer

here is a shortcut, since it is a negative cubic function
|dw:1335768644832:dw|
so the min x-value comes before max x-value
Given: A<B
f'(A) = min
f'(B) = max