Question

Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 3 at x = −3 and a local minimum value of 0 at x = 1.

Answer 1

Answer:

f(x) = (3x³ + 9x² - 27x + 15)/32

Step-by-step explanation:

f'(x) = 3ax² + 2bx + c

Maximum or minimum: f'(x) = 0

Maximum: f'(-3) = 0 and f(-3) = 3

Minimum: f'(1) = 0 and f(1) = 0

f'(-3) = 0 leads 27a - 6b + c = 0       (1)

f'(1) = 0 leads 3a + 2b + c = 0           (2)

f(-3) = 3 leads -27a + 9b -3c + d = 3    (3)

f(1) = 0 leads a + b + c + d = 0        (4)

(1) - (2):   24a - 8b = 0, or 3a - b = 0, b = 3a

(3) - (4)   -28a + 8b - 4c = 3    (5)

(1) * 4 + (5):  80a - 16b = 3,

use b = 3a, get 80a - 16*3a = 3, 32a = 3, a = 3/32, b = 3a = 9/32

From (2): c = -3a - 2b = -9/32 - 18/32 = -27/32

From (4): d = -a - b - c = -3/32 - 9/32 + 27/32 = 15/32

So

f(x) = (3x³ + 9x² - 27x + 15)/32