Let 𝑓(𝑥)=𝑥3−3𝑥2−9𝑥+25.Determine the absolute maximum and absolute minimum values of 𝑓 on the interval [0,4].

To find the absolute maximum and minimum values of the function on the interval [0,4], we need to follow these steps:

1. Find the derivative of the function f(x) = x^3 - 3x^2 - 9x + 25. The derivative, f'(x), is 3x^2 - 6x - 9.

2. Set the derivative equal to zero and solve for x to find critical points. So, 3x^2 - 6x - 9 = 0. Solving this quadratic equation gives us x = -1 and x = 3. However, -1 is not in the interval [0,4], so we only consider x = 3.

3. Evaluate the function at the endpoints of the interval and at the critical point. So, we find f(0), f(3), and f(4).

   f(0) = (0)^3 - 3*(0)^2 - 9*(0) + 25 = 25 f(3) = (3)^3 - 3*(3)^2 - 9*(3) + 25 = -2 f(4) = (4)^3 - 3*(4)^2 - 9*(4) + 25 = 1

4. The absolute maximum value of the function on the interval [0,4] is the largest of these values, which is 25 at x = 0. The absolute minimum value is the smallest of these values, which is -2 at x = 3.