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Anonimi16
@Anonimi16
August 2021
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Записать уравнение касательной к графику функции ...(смотреть вложение)
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sedinalana
Y=f(x0)+f`(x0)(x-x0)
f(3)=1/3*27-9+5=5
f`(x)=x²-2x
f`(3)=9-6=3
y=5+3(x-3)=5+3x-15=3x-10
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Answers & Comments
f(3)=1/3*27-9+5=5
f`(x)=x²-2x
f`(3)=9-6=3
y=5+3(x-3)=5+3x-15=3x-10