(a) The critical numbers happen when x = 3 or x = -1/2
(b) f is decreasing on (-∞, -1/2), increasing on (-1/2, 3), and increasing on (3, ∞).
(c) f has a local minimum value of -22 at x = 3, and a local maximum value of 25.5 at x = -1/2.
(d) f is concave downward on (-∞, 1/4) and concave upward on (1/4, ∞).
(e) The inflection point of f is at x = 1/4.
(a) To find the critical numbers of f, we need to find the values of x where the derivative of f equals zero or does not exist.
f'(x) = 12x² - 6x - 18 = 6(2x² - x - 3) = 6(x - 3)(2x + 1)
Setting f'(x) equal to zero, we get:
6(x - 3)(2x + 1) = 0
x = 3 or x = -1/2
These are the critical numbers of f.
(b) To find the intervals where f is increasing and decreasing, we need to examine the sign of the derivative f'(x) in the intervals determined by the critical numbers.
When x < -1/2, f'(x) < 0, so f is decreasing on the interval (-∞, -1/2).
When -1/2 < x < 3, f'(x) > 0, so f is increasing on the interval (-1/2, 3).
When x > 3, f'(x) > 0, so f is increasing on the interval (3, ∞).
(c) To find the local minimum and maximum values of f, we need to examine the critical numbers and the end points of the intervals.
f(3) = 4(3)³ - 3(3)² - 18(3) + 5 = -22
f(-1/2) = 4(-1/2)³ - 3(-1/2)² - 18(-1/2) + 5 = 25.5
Thus, f has a local minimum value of -22 at x = 3, and a local maximum value of 25.5 at x = -1/2.
(d) To find the intervals where f is concave upward and concave downward, we need to examine the sign of the second derivative f''(x).
f''(x) = 24x - 6 = 6(4x - 1)
When x < 1/4, f''(x) < 0, so f is concave downward on the interval (-∞, 1/4).
1/4 < x, f''(x) > 0, so f is concave upward on the interval (1/4, ∞).
(e) To find the inflection points of f, we need to examine the points where the concavity changes.
The concavity changes at x = 1/4, which is the only inflection point o