Find original function given second derivative

Hi I need some guided help on this :
If f "(x) = 5x^3 - 3x^2, f ' (0) = 4 and f ' (0) = 4 and f(0) = -4 . Find f(x)
I appreciate any help :)

if f"(x) = 5x^3 - 3x^2, then f'(x) = (5/4)x^4 - (3/3)x^3 + C = (5/4)x^4 - x^3 + C by integration

Since f'(0) = 4
(5/4)0^4 - 0^3 + C = 4
C = 4

f'(x) = (5/4)x^4 - x^3 + 4

Integrating one more time yields
f(x) = (5/20)x^5 - (1/4)x^4 + 4x + C

Since f(0) = -4
-4 = (5/20)0^5 - (1/4)0^4 + 4(0) + C
C = -4

f(x) = (5/20)x^5 - (1/4)x^4 + 4x -4

f'(0)=4, so when you antidifferentiate (get the antiderivative of) f''(x), set it equal to 4 and substitute all the x for 0.
f'(x)=(5/4)(x^4)-x^3+C (C is unknown constant that is zeroed when deriving)
f'(0)=(5/4)(0^4)-0^3+C=4
f'(0)=0+C=4

When f'(0)=4, C=4
This means that f'(x)=(5/4)(x^4)-x^3+4

Now that you know the equation of the first derivative, do the same to get the original equation