Suppose that f(x)=x3?6x2?3. (A) List all the critical values of f(x). Note: If t

Question

Suppose that f(x)=x3?6x2?3. (A) List all the critical values of f(x). Note: If there are no critical values, enter 'NONE'. (B) Use interval notation to indicate where f(x) is increasing. Note: Use 'INF' for ?, '-INF' for ??, and use 'U' for the union symbol. Increasing: (C) Use interval notation to indicate where f(x) is decreasing. Decreasing: (D) List the x values of all local maxima of f(x). If there are no local maxima, enter 'NONE'. x values of local maxima = (E) List the x values of all local minima of f(x). If there are no local minima, enter 'NONE'. x values of local minima = (F) Use interval notation to indicate where f(x) is concave up. Concave up: (G) Use interval notation to indicate where f(x) is concave down. Concave down: (H) List the x values of all the inflection points of f. If there are no inflection points, enter 'NONE'. x values of inflection points = (I) Use all of the preceding information to sketch a graph of f. When you're finished, enter a "1" in the box below.

Explanation / Answer

f(x) = x^3 - 6x^2 - 3 f'(x) = 3x^2 - 12x Set this equal to 0. 3x^2 - 12x = 0 3x(x - 4) = 0 x = 0, x = 4 are the two critical points. consider the intervals: (-inf, 0), (0, 4), (4, inf) f(x) is increasing where f'(x) > 0. f(x) is increasing on (-inf, 0)u(4, inf) f(x) is decreasing where f'(x) < 0 f(x) is decreasing on (0, 4) x = 0 is a local max. x = 4 is a local min. f"(x) = 6x - 12 set equal to 0. 6x - 12 = 0 x = 2 consiser (-inf, 2) and (2, inf) f(x) is concave up where f"(x) > 0. f(x) is concave up on (2, inf) f(x) is concave down where f"(x) < 0. f(x) is concave down on (-inf, 2) x = 2 is thus an inflection point.

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