Find the function \( F \) that satisfies the following differential equations and initial conditions. \( F^{\prime \prime}(x)=\cos x, F^{\prime}(0)=9, F(\pi)=9 \) \( F(x)=\square \) (Type an exact answer in terms of \( \pi \).)

To solve for the function \( F(x) \) that satisfies \( F^{\prime \prime}(x) = \cos x \), we start by integrating \( F^{\prime \prime}(x) \). 1. First, integrate \( F^{\prime \prime}(x) \): \[ F^{\prime}(x) = \int \cos x \, dx = \sin x + C_1 \] where \( C_1 \) is a constant to be determined. 2. Next, we use the initial condition \( F^{\prime}(0) = 9 \) to find \( C_1 \): \[ F^{\prime}(0) = \sin(0) + C_1 = 0 + C_1 = C_1 \] Thus, \( C_1 = 9 \). Therefore, we have: \[ F^{\prime}(x) = \sin x + 9 \] 3. Now, integrate \( F^{\prime}(x) \): \[ F(x) = \int (\sin x + 9) \, dx = -\cos x + 9x + C_2 \] where \( C_2 \) is another constant to be determined. 4. We use the next condition \( F(\pi) = 9 \) to find \( C_2 \): \[ F(\pi) = -\cos(\pi) + 9\pi + C_2 = -(-1) + 9\pi + C_2 = 1 + 9\pi + C_2 \] We set this equal to 9: \[ 1 + 9\pi + C_2 = 9 \] Solving for \( C_2 \): \[ C_2 = 9 - 1 - 9\pi = 8 - 9\pi \] 5. Finally, substituting \( C_2 \) back into \( F(x) \): \[ F(x) = -\cos x + 9x + (8 - 9\pi) \] Combining the constants gives the final function: \[ F(x) = -\cos x + 9x + 8 - 9\pi \] Thus, the function \( F(x) \) is: \[ \boxed{-\cos x + 9x + 8 - 9\pi} \]