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Gudi10
@Gudi10
July 2022
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найти тангенс угла наклона касательной к графику функции y=2x-x^3 в точке х0=-1
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ShirokovP
Verified answer
Tga = f ' (x0)
f ' (x) = (2x - x^3) ' = 2 - 3x^2
f ' (-1) = 2 - 3*(-1)^2 = - 3 + 2 = - 1
Ответ
tga = - 1
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Answers & Comments
Verified answer
Tga = f ' (x0)f ' (x) = (2x - x^3) ' = 2 - 3x^2
f ' (-1) = 2 - 3*(-1)^2 = - 3 + 2 = - 1
Ответ
tga = - 1