# Problem Statement Given: - \( f''(x) = -4\sin(2x) \) - \( f'() = 5 \) - f() = -2 \) Find: - \( f\left(\frac{\pi}{6}\right) \) --- # Solution ## 1. Integrate \( f''(x) \) to Find \( f'(x) \) \[ f''(x) = -4\sin(2x) \] Integrate once: \[ f'(x) = \int -4\sin(2x) \, dx = -4 \cdot \left( -\frac{1}{2} \cos(2x) \right) + C_1 = 2\cos(2x) + C_1 \] ## 2. Use \( f'() = 5 \) to Solve for \( C_1 \) \[ f'() = 2\cos() + C_1 = 2 \cdot 1 + C_1 = 2 + C_1 = 5 \] \[ C_1 = 3 \] So: \[ f'(x) = 2\cos(2x) + 3 \] ## 3. Integrate \( f'(x) \) to Find \( f(x) \) \[ f(x) = \int [2\cos(2x) + 3] \, dx = 2 \int \cos(2x) dx + \int 3 dx \] \[ = 2 \cdot \frac{1}{2}\sin(2x) + 3x + C_2 \] \[ = \sin(2x) + 3x + C_2 \] ## 4. Use \( f() = -2 \) to Solve for \( C_2 \) \[ f() = \sin() + 3 \cdot + C_2 = + + C_2 = -2 \] \[ C_2 = -2 \] So: \[ f(x) = \sin(2x) + 3x - 2 \] ## 5. Calculate \( f\left(\frac{\pi}{6}\right) \) \[ f\left(\frac{\pi}{6}\right) = \sin\left(2 \cdot \frac{\pi}{6}\right) + 3 \cdot \frac{\pi}{6} - 2 \] \[ = \sin\left(\frac{\pi}{3}\right) + \frac{\pi}{2} - 2 \] \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] \[ f\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{\pi}{2} - 2 \] --- # Final Answer \[ \boxed{ f\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{\pi}{2} - 2 } \]