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nalobina3
@nalobina3
July 2022
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Напишите уравнение касательной к графику функции у= 1/2sin2x-2x в точке с абциссой x0 = π/2
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nafanya2014
Verified answer
Уравнение касательной к кривой у=f(x) в точке x₀:
y-f(x₀)=f`(x₀)·(x-x₀)
f(x₀)=f(π/2)=(1/2)sin(2·π/2)-2·(π/2)=(1/2)·(sinπ)-π=(1/2)·0-π=-π
f`(x)=(1/2)·(cos2x)·(2x)`=cos2x
f`(x₀)=f`(π/2)=cos(2·(π/2))=cosπ=-1
y-(-π)=-1·(x-(π/2))
y=-x-(π/2)
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Answers & Comments
Verified answer
Уравнение касательной к кривой у=f(x) в точке x₀:y-f(x₀)=f`(x₀)·(x-x₀)
f(x₀)=f(π/2)=(1/2)sin(2·π/2)-2·(π/2)=(1/2)·(sinπ)-π=(1/2)·0-π=-π
f`(x)=(1/2)·(cos2x)·(2x)`=cos2x
f`(x₀)=f`(π/2)=cos(2·(π/2))=cosπ=-1
y-(-π)=-1·(x-(π/2))
y=-x-(π/2)