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yuraseven77
@yuraseven77
July 2022
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Знайдіть екстремуми функції у = х ^3 – 6х^ 2 .
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sergeygutsko
Y' = 3x² – 12x.
3x² – 12x = 0 при x = 0 и x = 4.
y(0) = 0.
y(4) = –32.
Ответ: max = 0, min = –32.
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Answers & Comments
3x² – 12x = 0 при x = 0 и x = 4.
y(0) = 0.
y(4) = –32.
Ответ: max = 0, min = –32.