Find f.f '''(x) = cos(x),    f(0) = 2,    f '(0) = 6,    f ''(0) = 5f(x) = 6x+1+cos(x)

The problem seems to be a bit mixed up. Let's break it down:

1. You're given that the third derivative of a function, f'''(x), is equal to cos(x).
2. You're also given initial conditions: f(0) = 2, f'(0) = 6, and f''(0) = 5.
3. You're asked to find the function f(x).

To solve this, we need to integrate the third derivative to find the original function, using the initial conditions to find the constants of integration.

Step 1: Integrate f'''(x) = cos(x) to find f''(x). ∫cos(x) dx = sin(x) + C1

Step 2: Use the initial condition f''(0) = 5 to find C1. 5 = sin(0) + C1 So, C1 = 5.

Now we know that f''(x) = sin(x) + 5.

Step 3: Integrate f''(x) = sin(x) + 5 to find f'(x). ∫(sin(x) + 5) dx = -cos(x) + 5x + C2

Step 4: Use the initial condition f'(0) = 6 to find C2. 6 = -cos(0) + 5*0 + C2 So, C2 = 7.

Now we know that f'(x) = -cos(x) + 5x + 7.

Step 5: Integrate f'(x) = -cos(x) + 5x + 7 to find f(x). ∫(-cos(x) + 5x + 7) dx = -sin(x) + (5/2)x^2 + 7x + C3

Step 6: Use the initial condition f(0) = 2 to find C3. 2 = -sin(0) + (5/2)0^2 + 70 + C3 So, C3 = 2.

Finally, we find that f(x) = -sin(x) + (5/2)x^2 + 7x + 2.