I solved F(x) to be equal to $(x^3/3) -4a^2x$
The questions's information is as follows: "For a quadratic function f(x) we define a function as follows"
$$F(x) = \int_0^x f(t)dt$$
Assume that "a" is a positive number and the function F(x) has extreme values at x = -2a,2a. "
a is a positive number. the function F(x) has extreme values at x = -2a and x = 2a.
The local maxmimum value of function F(x)/F'($0$) is: ??
I tried taking the $g(x)f'(x)-g'(x)f(x)/g(x)^2$ thingy but either my math was weird or it just doesn't work with this problem. I ended up getting $(-x^2/4a^2) + 1$ for the first derivative, then solved it for x = $4a^2$. I Then checked to see if that point was a maximum with the second derivative, and it was.
When I plugged $4a^2$ into the original function (F(x)/F'($0$)), I got $-16a^4/3$ + $4a^2$.
I also took this one a step further and solved for a, but t that didn't give me anything. a = $\sqrt(3/4)$
The answer is 4a/3, but I have no idea how to even get there.
Your evaluation of F(x) looks about fine to me. Now you must note that F’(0) is just a number, it is not dependent on x, so you needn’t consider it while performing the derivative (as you have done it seems, from your mention of the quotient rule). So basically you need to find the local maxima /minima (which we’ll come to in a bit)of F(x) and divide it by F’(0).
we already know that F(x) has extreme values at 2a and -2a. So check by plugging in these values into F(x). You will find that F(2a)<F(-2a). But$ F’(0)=-4a^2$ which is negative so $\frac{F(2a)}{F’(0)}\gt \frac{F(-2a)}{F’(0)}$ so $\frac{F(2a)}{F’(0)}$ is the value of the local maxima which is $\frac{-1}{4a^2}$ $(\frac{8a^3}{3}-8a^3)$=$\frac{4a}{3}$.