Sorry. If f'(x) is a constant, f"(x) = 0
Fix the typo and try again.
f"(x)=10, f'(x)=3 f(0)=1/3 find f(1)
2 answers
I will assume you meant f'(0) = 3
in that case,
f"(x) = 10
f'(x) = 10x+c
3 = 10*0+c -----> c = 3
f'(x) = 10x+3
f(x) = 5x^2+3x+c
1/3 = 5*0+3*0+c
c = 1/3
f(x) = 5x^2 + 3x + 1/3
I figure you can now find f(1).
in that case,
f"(x) = 10
f'(x) = 10x+c
3 = 10*0+c -----> c = 3
f'(x) = 10x+3
f(x) = 5x^2+3x+c
1/3 = 5*0+3*0+c
c = 1/3
f(x) = 5x^2 + 3x + 1/3
I figure you can now find f(1).
1 answer