Question

F(x):3x³ - 2

a) Determine the intervals on which f is concave up and concave down.
f is concave up on:
f is concave down on:

b) Based on your answer to part (a), determine the inflection points of f. Each point should be entered as an ordered pair
(that is, in the form (x, y)).


c) Find the critical numbers of f and use the Second Derivative Test, when possible, to determine the relative extrema. List
only the x-coordinates.
Relative maxima at:
Relative minima at:

d) Find the x-value(s) where f'(x) has a relative maximum or minimum.
f' has relative maxima at:
f' has relative minima at:

Answer 1

a) To find the intervals on which f is concave up and concave down, we need to find the second derivative of f:

f(x) = 3x³ - 2

f'(x) = 9x²

f''(x) = 18x

f''(x) is positive when x > 0, so f is concave up on (0, ∞).

f''(x) is negative when x < 0, so f is concave down on (-∞, 0).

b) To find the inflection points of f, we need to find the x-values where f''(x) changes sign:

f''(x) changes sign at x = 0, so the inflection point is (0, -2).

c) To find the critical numbers of f, we need to find the values of x where f'(x) = 0 or f'(x) does not exist:

f'(x) = 9x² = 0

x = 0

The Second Derivative Test tells us that if f''(x) > 0, then f has a relative minimum at x, and if f''(x) < 0, then f has a relative maximum at x. Since f''(x) is positive when x > 0 and negative when x < 0, there is no relative maximum or minimum for f.

d) To find the x-values where f'(x) has a relative maximum or minimum, we need to find the critical points of f':

f'(x) = 9x²

f''(x) = 18x

f''(x) = 0 when x = 0, so x = 0 is a critical point of f'.

Since f''(x) is positive when x > 0 and negative when x < 0, f'(x) has a relative minimum at x = 0.