Calculus - Curve Sketching

Question

Use the derivative to find the exact critical points and intervals of increase and decrease for

f (x) = 4x³ – 21x² – 24x + 5.



Answer 1

f'(x) = 12x² - 42x - 24 = 6(2x² - 7x - 4) = 6(2x + 1)(x - 4)

0s are at -1/2 and 4

f''(x) = 24x - 42 = 24(x - 7/4)

f''(x) < 0 at -1/2 and f''(x) > 0 at 4

Thus, there is a relative maximum at -1/2 and a relative minimum at 4;

f(-1/2) = 4(-1/2)³ - 21(-1/2)² -24(-1/2) + 5 = 11 1/4

f(4) = 4(4)³ - 21(4)² -24(4) + 5 = -171

The function is increasing on (-∞, -1/2) and (4, ∞) and is decreasing on (-1/2, 4)