Find the largest and smallest value of the function y = 4-3x + 2x ^ 2 on the segment [0; 3]

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

y ‘= (4 – 3x + 2x ^ 2)’ = -3 + 4x.

2. Let us equate this derivative to zero and find the critical points:

-3 + 4x = 0;

4x = 3;

x = 3: 4;

x = 3/4.

3. Find the value of the function at this point and at the ends of the given segment [0; 3]:

y (0) = 4 – 3 * 0 + 2 * 0 = 4;

y (3/4) = 4 – 3 * 3/4 + 2 * 9/16 = 4 – 9/4 + 9/8 = 4 – 18/8 + 9/8 = 4 – 9/8 = 4 – 1 1/8 = 4 – 1 – 1/8 = 3 – 1/8 = 2 + 1 – 1/8 = 2 + 8/8 – 1/8 = 2 7/8.

y (3) = 4 – 3 * 3 + 2 * 9 = 4 – 9 + 18 = 4 + 9 = 13.

Answer: fmax = 13, fmin = 2 7/8.