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marisabel16
@marisabel16
July 2022
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найдите критические точки функции f(x)=x^3-3x
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v4volande
Ответ: экстремумы (критические точки): (1;-2) и (-1;2)
2 votes
Thanks 1
pro100vesna
F'(x)=3x²-3=0
x=⁺₋1
-1+3=2
1-3=-2
критич точки -2 2
0 votes
Thanks 0
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Answers & Comments
Ответ: экстремумы (критические точки): (1;-2) и (-1;2)
x=⁺₋1
-1+3=2
1-3=-2
критич точки -2 2