Find the smallest and largest value of the function y = x²-4x + 3 on the segment [1,3].

Let’s find the derivative of the function.

f (x) = x² – 4x + 3.

f` (x) = 2x – 4.

Let us equate the derivative to zero.

2x – 4 = 0.

2x = 4;

x = 2.

Determine the signs of the derivative at each interval.

(-∞; 2) let x = 0; 2 * 0 – 4 = -4, derivative (), the function decreases.

(2; + ∞) let x = 3; 2 * 3 – 4 = 2, derivative (+), the function is increasing.

Point 2 is the minimum point of the function (included in the interval [1; 3]).

The maximum point will be 3, as the function grows from 2 to 3.

Let’s calculate the smallest value of the function:

x = 2; y = 2² – 4 * 2 + 3 = 4 – 8 + 3 = -1.

Let’s calculate the largest value of the function:

x = 3; y = 3² – 4 * 3 + 3 = 9 – 12 + 3 = 0.

Answer: the smallest value of the function in the interval [1; 3] is -1 and the largest value is 0.