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
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
To find the function f(x), we need to integrate the second derivative of f(x). The integral of 10x with respect to x is (5/6)x^2, and the integral of 2sin(x) with respect to x is -2cos(x).
Thus, the indefinite integral of f(x) = 10x + 2sin(x) would be (5/6)x^2 - 2cos(x) + C, where C is the constant of integration.
To determine the constant C, we need to use the given information f(0) = 4. Substituting x = 0 into the equation f(x) = (5/6)x^2 - 2cos(x) + C, we have:
4 = (5/6)(0)^2 - 2cos(0) + C
4 = 0 - 2(1) + C
4 = -2 + C
C = 6
Now, we can rewrite the function f(x) without the constant C:
f(x) = (5/6)x^2 - 2cos(x) + 6
Therefore, the function f(x) satisfying f''(x) = 10x + 2sin(x), f(0) = 4, and f'(0) = 4 is:
f(x) = (5/6)x^2 - 2cos(x) + 6