So I've been doing this problem for a while now and I keep getting the wrong answer:

Find a cubic function, in the form below, that has a local maximum value of 4 at -4 and a local minimum value of 0 at 2.

$\displaystyle f(x)=ax^3+bx^2+cx+d$

So I just use can obtain 4 equations by subsituting (-4,4) and (2,0) into f(x) and into the derivative of f(x) where $\displaystyle f'(x)=3ax^2+2bx+c=0$:

For f(x): at x=-4, $\displaystyle f(x)=-64a+16b-4c+d-4=0$

at x=2, $\displaystyle f(x)=8a+4b+2c+d=0$

For f'(x): at x=-4, $\displaystyle f'(x)=48a-8b+c=0$

at x=2, $\displaystyle f'(x)=12a+4b+c=0$

So when I solve it I get a= -1/93, b= -1/31, c= 8/31, d= -28/93 but that only solves 3 of the above equations...so I'm doing something wrong. I'll post all my work in the next post since it's a bit long. Also, I don't know how to use matrices so try to refrain from using it. Thanks.