from f"(x)=5x+2
f ' (x) = (5/2)x^2 + 2x + c
but f ' (-3) = 3
3 = (5/2)(9) + 6 + c
c = ...
find c and
now you have f ' (x)
integrate once more, adding another constant
and since you have (-3,6) on it, you can find that constant as well
Given f"(x)=5x+2 and f'(-3)=3 and f(-3)=-6
find f'(x)=
find f(3)=
2 answers
So once I solve for c, how does that translate to f'(x)?