Find the largest value of the function f (x) = – x ^ 3 + 3x ^ 2 + 9x – 29 on the segment [-1; 4]

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

y = -3x ^ 2 + 6x + 9.

2. Let us equate this derivative to zero:

-3x ^ 2 + 6x + 9 = 0.

D = b ^ 2 – 4ac = 36 + 4 * 3 * 9 = 144.

x1 = (-b + √D) / 2a = (-6 + 12) / (- 6) = 6 / (- 6) = -1;

x2 = (-b – √D) / 2a = (-6 – 12) / (- 6) = -18 / (- 6) = 3.

3. Find the value of the function at critical points and at the ends of the segment [-1; 4]:

y (-1) = – (- 1) ^ 3 + 3 (-1) ^ 2 + 9 * (-1) – 29 = 1 + 3 – 9 – 29 = -34;

y (3) = -3 ^ 3 + 3 * 3 ^ 2 + 9 * 3 – 29 = -27 + 27 + 27 – 29 = -2;

y (4) = -4 ^ 3 + 3 * 4 ^ 2 + 9 * 4 – 29 = -64 + 48 + 36 – 29 = -9.

Answer: fmax = -2.