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Given f ′(x) = 4x3 12x2, determine the interval(s) on which f is both increasing and concave up,

User Mkmitchell
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Final answer:

To find where the function f is increasing and concave up, we analyze the first derivative f'(x) to identify intervals of increase, and the second derivative f''(x) to determine the concavity. f(x) is increasing where f'(x) > 0 and concave up where f''(x) > 0, resulting in the function being both increasing and concave up for x > 3.

Step-by-step explanation:

To determine the interval(s) on which the function f is both increasing and concave up, given that f ′(x) = 4x^3 - 12x^2, we need to analyze the first and second derivatives of f(x). The first derivative, f ′(x), helps us identify where the function is increasing or decreasing, while the second derivative, f ′′(x), indicates the concavity of the function.

First, let's find where f(x) is increasing. f(x) is increasing where f ′(x) > 0. For f ′(x) = 4x^3 - 12x^2, we find the critical points by setting the derivative equal to zero: 4x^3 - 12x^2 = 0. Factoring out 4x^2 gives 4x^2(x - 3) = 0, so x = 0 and x = 3 are critical points. The function is increasing in intervals where the first derivative is positive.

Next, we assess concavity with the second derivative, f ′′(x). To find f ′′(x), we differentiate f ′(x) again. Doing so yields: f ′′(x) = 12x^2 - 24x. We set this equal to zero to find potential inflection points: 12x^2 - 24x = 0, and then factor to get 12x(x - 2) = 0. Therefore, the inflection points are at x = 0 and x = 2. The function is concave up where f ′′(x) > 0.

By testing intervals around the critical points and inflection points, we can determine where the function is both increasing and concave up. It turns out that f(x) is both increasing and concave up for x > 3, which is the interval (3, ∞).

User Daniel Baulig
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