Given f"(x)=5x+2 and f'(-3)=3 and f(-3)=-6

find f'(x)=
find f(3)=

2 answers

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
So once I solve for c, how does that translate to f'(x)?