1. First Derivative: - Compute \( f'(x) \): \[ f'(x) = 8x - \cos x - 2\sin 2x. \] 2. Critical Points: - Set \( f'(x) = 0 \): \[ 8x - \cos x - 2\sin 2x = 0. \] - This equation has exactly one solution because the quadratic term \( 8x \) dominates. 3. Second Derivative: - Compute \( f''(x) \): \[ f''(x) = 8 + \sin x - 4\cos 2x. \] - Since \( f''(x) > 0 \), \( f(x) \) has a local minimum at the critical point
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is:
A cylindrical tank of radius 10 cm is being filled with sugar at the rate of 100Ο cm3/s. The rate at which the height of the sugar inside the tank is increasing is: