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ArtemSmulko
@ArtemSmulko
October 2021
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Решите пожалуста А18
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kaef13
F'(x)= 3x^2 - 6x - 9
3x^2 - 6x - 9 = 0 | : 3
x^2 - 2x - 3 = 0
(x-3)(x+1)=0
-2; -1; 0 , тк 3 не принадлежит [-2;0]
f(-2)= -8 - 12 + 18 + 12 = 10
f(-1)= -1 - 3 + 9 + 12= 17
f(0)= 12
Max f(x) на отрезке [-2;0] = 17
Min f(x) на отрезке [-2;0] = 10
Ответ: 1)27
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Answers & Comments
3x^2 - 6x - 9 = 0 | : 3
x^2 - 2x - 3 = 0
(x-3)(x+1)=0
-2; -1; 0 , тк 3 не принадлежит [-2;0]
f(-2)= -8 - 12 + 18 + 12 = 10
f(-1)= -1 - 3 + 9 + 12= 17
f(0)= 12
Max f(x) на отрезке [-2;0] = 17
Min f(x) на отрезке [-2;0] = 10
Ответ: 1)27