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dmitrija7531
@dmitrija7531
July 2022
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Найти наибольшее значение функции
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valshev
F'(x)=3*x^2+12*x=0 => 3*x*(x+4)=0 => x1=0, x2=-4 - не входит в промежуток
f(0)=0 - минимум
f(-3)=27 -максимум
f(1)=7
1 votes
Thanks 1
dmitrija7531
как ты в f(1)=7 получил?
valshev
1^3+6*1^2=7
dmitrija7531
спасибо
valshev
:)
dmitrija7531
наибольший будет 27?:)
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Answers & Comments
f(0)=0 - минимум
f(-3)=27 -максимум
f(1)=7