The graph of f ′ (x), the derivative of f of x, is continuous for all x and consists of five line segments as shown below. Given
f (–3) = 6, find the absolute maximum value of f (x) over the interval [–3, 0]. Graph of line segments increasing from x equals negative 4 to x equals negative 3, decreasing from x equals negative 3 to x equals 0, increasing from x equals 0 to x equals 3, constant from x equals 2 to x equals 4 and decreases from x equals 4 to x equals 5. x intercepts at x equals negative 4, x equals 0, x equals 5.
The other answer is correct until solving for C. f(-3) = 6, not -6. So C = 21/2. This makes f(0) = 10.5. Local Max should occur at x intercepts on the f'(x) graph.