Find the smallest and largest value of the function y = x³-3x² + 3x + 2 on the segment [-2; 2].

1. Let’s find the first derivative of the function:

y ‘= (x ^ 3 – 3x ^ 2 + 3x + 2)’ = 3x ^ 2 – 6x + 3.

2. Let us equate this derivative to zero and find the zeros of the function:

3x ^ 2 – 6x + 3 = 0.

Divide the equation by 3:

x ^ 2 – 2x + 1 = 0;

D = b ^ 2 – 4ac = 4 – 4 = 0.

D = 0, the equation has one root.

x = -b / 2a = 2/2 = 1.

3. Find the value of the function at this point and at the ends of the given segment [-2; 2]:

y (1) = 1 ^ 3 – 3 * 1 ^ 2 + 3 * 1 + 2 = 1 – 3 + 3 + 2 = 3;

y (-2) = (-2) ^ 3 – 3 * (-2) ^ 2 + 3 * (-2) + 2 = -8 – 12 – 6 + 2 = -24;

y (2) = 2 ^ 3 – 3 * 2 ^ 2 + 3 * 2 + 2 = 8 – 12 + 6 + 2 = 4.

Answer: fmax = 4, fmin = -24.