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kisakamiren
@kisakamiren
July 2022
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найдите наибольшее значение функции y=x^3-12x+7 на отрезке [-3;0]
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sedinalana
Verified answer
Y`=3x²-12=0
x²=4
x=-2∈[-3;0]
x=2∉[-3;0]
y(-3)=-27+36+7=16
y(-2)=-8+24+7=23 наиб
y(0)=7
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Answers & Comments
Verified answer
Y`=3x²-12=0x²=4
x=-2∈[-3;0]
x=2∉[-3;0]
y(-3)=-27+36+7=16
y(-2)=-8+24+7=23 наиб
y(0)=7