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nalobina3
@nalobina3
July 2022
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Напишите уравнение касательной к графику функции у= 3sinx-12x в точке с абциссой x0 = -π/2
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maryfine
Y'=3cosx-12
y'(x0)=3cos(-π/2)-12=0-12=-12
y(x0)=3sin(-π/2)-12*(-π/2)=-3+6π
уравнение касательной:
y=y'(x0)(x-x0)+y(x0)
y=-12*(x+π/2)+6π-3=-12x-6π+6π-3=-12x-3
Ответ: y=-12π-3
2 votes
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Answers & Comments
y'(x0)=3cos(-π/2)-12=0-12=-12
y(x0)=3sin(-π/2)-12*(-π/2)=-3+6π
уравнение касательной:
y=y'(x0)(x-x0)+y(x0)
y=-12*(x+π/2)+6π-3=-12x-6π+6π-3=-12x-3
Ответ: y=-12π-3