You have some sign errors in your exponents. I get
You're flipping things from numerator to denominator more times than is required in the quotient rule.
I came across two functions I am having trouble with and I am not really sure why.
the function is f(t)= 64t/(8+t)^3
I tried the quotient rule which then translates into
64[(1)((8+t)^-3)-3t(8+t)^-4]/[(8+t)^3]^2
from there it goes to crap, the answer that it should be is 128(4-t)/(8+t)^4 which is not the answer I end up with. thank you.
oh wow, haha. didn't even occur to me. gives me 64(8+t-3t)/(8+t)^4 which goes onto be able to be factored into 128(4-t)?(8+t)^4. I see why I was screwing that up.
since that was fairly quick maybe you can help this, I also need to solve for f'(t) = 0 because I am doing maximum and minimum values.
I know that it gives t=4, it wasn't ever shown to me how to solve f'(t) or whatever the vairable maybe = 0
A fraction can only be zero when the numerator is zero. However, if you're trying to maximize and minimize the original function f(t), you'll also need to look at what happens when the derivative doesn't exist (in this case, when t = -8). Are you trying to find just local extrema, or are you trying to find global extrema over some interval?
I'm guessing you meant the closed interval [0,10], not the open interval which you wrote, which is (0,10). Over a closed interval, a continuous function is guaranteed to have global extrema, but all bets are off when the interval is open.
You need to evaluate the function at the critical point (t = 4 is the only critical point inside the interval) and at both endpoints. What do you get for that?
It doesn't matter what a denominator does - it can't make a fraction zero. Only the numerator can make a fraction zero. The only place the numerator of your derivative is zero is at t = 4. Hence, that is a critical point - the only critical point inside your interval.
I agree with your function values. Which one is the greatest? That's your max. Which one's the smallest? That's your min.