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rachampt76
02.03.2022 •
Mathematics
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.
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Ответ:
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
Ответ: