Antiderivatives Calculus

Question

Given

f ''(x) = 3x - 3 and f '(-3) = 1 and f(-3) = 1

find f '(x) =

and find f(2) =

Answer 1

∫f''(x)dx=f'(x)+C so

f (x)=∫3x-3dx=(3x²)/2-3x+C

f'(-3)=(3(-3)²/2)-3(-3)+C=1

f'(-3)=-27/2+9+C=1

Solving for C yields C=43/2

f'(x)=3x²/2-3x-43/2

∫f'(x)dx=f(x)

f(x)=∫3x²/2-3x-43/2 dx = x³/2-3x²/2-43x/2+K

f(-3)=(-3)³/2-3(-3)²/2-43(-3)/2+K=1 Solving for K yields K=-73/2

f(x)=(x³-3x^2-43x-73)/2

f(2)=(8-12+86-73)/2=-163/2=-81.5