Question

Let f be a twice-differentiable function with derivative given by f ′(x) = −4x³ + 12x².

Part A: Find the x-coordinate of any possible critical points of f. Show your work.
Part B: Find the x-coordinate of any possible inflection points of f. Show your work.
Part C: Use the Second Derivative Test to determine any relative extrema and inflection points. Justify your answers.
Part D: If f has only one critical point on the interval [2, 4], what is true about the function f
on the interval [2, 4]? Justify your answers.

Answer 1

For part A, the critical points of f(x) are x = 0 and x = 3. For part B, the inflection points of f(x) are x = 0 and x = 2. For part C, x = 0 and x = 3 are inflection points. And for part D, if f has only one critical point on the interval [2, 4], it means that f does not have any relative extrema on the interval.

Step-by-step explanation:

Part A:

To find the critical points of f(x), we need to find the values of x where f'(x) is equal to 0 or undefined. Since f'(x) = -4x^3 + 12x^2, we set -4x^3 + 12x^2 = 0 and solve for x. Factoring out x^2, we get x^2(-4x + 12) = 0. This gives us two critical points: x = 0 and x = 3.

Part B:

To find the inflection points of f(x), we need to find the values of x where the concavity of the function changes. Since concavity is determined by the second derivative, we need to find the values of x where f''(x) = 0 or undefined. We differentiate f'(x) with respect to x to find f''(x). f''(x) = -12x^2 + 24x. Setting this equal to 0 and solving for x gives us two candidates for inflection points: x = 0 and x = 2.

Part C:

To determine the relative extrema and inflection points, we use the Second Derivative Test. For the critical point x = 0, we evaluate f''(0) = -12(0)^2 + 24(0) = 0. Since f''(0) = 0, the Second Derivative Test is inconclusive and we need to consider additional information. We evaluate f''(x) for values less than and greater than 0 to determine the concavity. For x = -1, f''(-1) = -12(-1)^2 + 24(-1) = 12 - 24 = -12, indicating concavity downwards. For x = 1, f''(1) = -12(1)^2 + 24(1) = 12, indicating concavity upwards. Therefore, x = 0 is an inflection point. For the critical point x = 3, we evaluate f''(3) = -12(3)^2 + 24(3) = 0. Since f''(3) = 0, the Second Derivative Test is inconclusive and we need to consider additional information. We evaluate f''(x) for values less than and greater than 3 to determine the concavity. For x = 2, f''(2) = -12(2)^2 + 24(2) = 0, indicating no change in concavity. Therefore, x = 3 is also an inflection point.

Part D:

If f has only one critical point on the interval [2, 4], it means that the critical point is x = 3. Since x = 3 is an inflection point, it means that f does not have any relative extrema on the interval [2, 4]. The function f may be increasing or decreasing on the interval depending on the values of f(x) for x < 3 and x > 3.