Question

For the following function, find

(a) the critical numbers,
(b) the open intervals where the function is increasing, and
(c) the open intervals where it is decreasing f(x)={4} / {3} x^{3}-12 x^{2}

Answer 1

To analyze the function f(x) = 4/3x^3 - 12x^2, we take the derivative, set it to zero to find critical numbers, and test intervals to determine where the function is increasing or decreasing. Critical numbers are x = 0 and x = 6. The function is increasing on the open interval (0, 6) and decreasing on the open intervals (−∞, 0) and (6, ∞).

Step-by-step explanation:

To find the critical numbers, open intervals where the function is increasing, and the open intervals where it is decreasing for the function f(x) = \(\frac{4}{3}x^3 - 12x^2\), we follow these steps:

1. Differentiate f(x) to get f'(x), the derivative.
2. Set f'(x) equal to zero and solve for x to find the critical numbers.
3. Determine the sign of f'(x) on either side of each critical number to identify the intervals where f(x) is increasing or decreasing.

The derivative f'(x) is f'(x) = 4x^2 - 24x. Setting the derivative equal to zero gives us 4x(x - 6) = 0. This yields the critical numbers x = 0 and x = 6.

Next, we test values of x to determine the sign of f'(x) in the intervals (−∞, 0), (0, 6), and (6, ∞) to find where the function is increasing and decreasing. The function f(x) is increasing on the open interval (0, 6) as f'(x) is positive, and it is decreasing on the open intervals (−∞, 0) and (6, ∞) as f'(x) is negative in these intervals.