Let 𝑓(𝑥)=𝑥3−3𝑥2+2𝑥f(x)=x 3 −3x 2 +2x. Find the maximum and minimum values of the function 𝑓(𝑥)f(x) on the interval [0,3][0,3].

To find the maximum and minimum values of the function f(x) = x^3 - 3x^2 + 2x on the interval [0,3], we need to follow these steps:

1. Find the derivative of the function, f'(x), which represents the slope of the function at any point. The derivative of f(x) = x^3 - 3x^2 + 2x is f'(x) = 3x^2 - 6x + 2.

2. Set the derivative equal to zero and solve for x to find critical points. This gives us the equation 3x^2 - 6x + 2 = 0. Solving this quadratic equation gives us x = 1/3 and x = 2.

3. Evaluate the function at the critical points and the endpoints of the interval. This gives us f(0) = 0, f(1/3) = 1/27 - 1/3 + 2/3 = 1/3, f(2) = 8 - 12 + 4 = 0, and f(3) = 27 - 27 + 6 = 6.

4. The maximum value of the function on the interval [0,3] is 6, which occurs at x = 3. The minimum value of the function on the interval [0,3] is 0, which occurs at x = 0 and x = 2.