Find f. f '''(x) = cos x, f(0) = 1, f '(0) = 7, f ''(0) = 8

Question

 I need to find the orginal function of the question

Answer 1

The solution is found by repeated integration. Integrating both sides of the equation f'''(x) = cos x yields f''(x) = sin x + A, where A is a constant. The initial condition f''(0) = 8 implies A = 8. So f''(x) = sin x + 8. Integrating again gives f'(x) = -cos x + 8x + B, where B is a constant. The initial condition f'(0) = 7 implies B = 8. Thus f'(x) = -cos x + 8x + 8. Integrate this once more and use the initial condition f(0) = 1 to obtain the solution f(x) = -sin x + 4x^2 + 8x + 1.

Answer 2

We can use  f'''(x)=cosx to find f"(x) by taking the anti-derivative.

 

f"(x) = sin(x) + C

 

where C is an arbitrary constant.

 

 

Then we can find f'(x) by taking the anti-derivative of f"(x).

 

f'(x) = -cos(x) + Cx

 

 

Anti-derivative this to get

 

f(x) = -sin(x) + (C/2)x²

 

 

Now, we use the initial value conditions f'(0)=7  and   f"(x)=8    to solve for C.

 

-cos(0) + C(0) = 7

 

sin(0) + C = 8

 

 

Use this system of equations to solve for C.  But I think you made an error with the inputs of the initial value conditions.