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Given that f ′′(x) = cos(x),f ′ (π/2)=7 and f(π/2)=14 find: f'(x) =
f(x) =


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Final answer:

To find f'(x) and f(x), we integrate f''(x)=cos(x) using the given initial conditions. The first derivative is f'(x)=sin(x)+7, and the original function is f(x)=-cos(x)+7x+14-7(π/2).

Step-by-step explanation:

The question is about finding the expressions for the first derivative f'(x) and the original function f(x) given the second derivative f''(x) = cos(x), along with initial conditions at x = π/2: f'(π/2) = 7 and f(π/2) = 14.

To find f'(x), we integrate the second derivative f''(x):

∫cos(x) dx = sin(x) + C

Using the initial condition f'(π/2) = 7, we solve for C:

sin(π/2) + C = 7

C = 7, since sin(π/2) = 1.

Therefore, the first derivative is f'(x) = sin(x) + 7.

Next, we repeat this process to find f(x). We integrate f'(x):

∫(sin(x) + 7) dx = -cos(x) + 7x + D

Using the initial condition f(π/2) = 14, we solve for D:

-cos(π/2) + 7(π/2) + D = 14

D = 14 - 7(π/2), since cos(π/2) = 0.

Therefore, the original function is f(x) = -cos(x) + 7x + 14 - 7(π/2).

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