if f''(x) = 10x+2sinx
then f'(x) = 5x^2 - 2cosx + C
but f'(0) = 4, so
4 = 0 -2cos0 + C
4 = -2 + C
C = 6
so f'(x) = 5x2 - 2cosx + 6
then f(x) = (5/3)x^3 - 2sinx + 6x + k
given f(0)=4 , so
4 = 0 - 0 + 0 + k
k = 4
then f(x) = (5/3)x^3 - 2sinx + 6x + 4
I don't understand the instruction not to include the constant,
if you don't have it you wouldn't get f(0) = 4
Consider the function f(x) whose second derivative is f(x)=10x+2sin(x). If f(0)=4 and f'(0)=4, what is f(x)? Please do not include the constant (+C) in your answer.
I got f(x) to be (10/6)x^3-2sin(x), but don't really know what to do with the f(0)=4
1 answer
3 answers