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Sunny56
@Sunny56
June 2022
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Дана функция f(x) = 2/3cos(3x-П/6).Составьте уравнение касательной к графику функции в точке с абсциссой x=П/3
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mikael2
Y=2/3 cos(3x-π/6) x0=π/3 y(x0)=2/3cos(π-π/6)=2/3 cos(5π/6)=-2/3*(√3/2)=
=-√3/3
y'=-2/3*sin(3x-π/6)*3=-2sin(3x-π/6)
y'(π/3)=-2sin(5π/6)=-2*1/2=-1
у-е касательной -1(x-π/3)+√3/3
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Answers & Comments
=-√3/3
y'=-2/3*sin(3x-π/6)*3=-2sin(3x-π/6)
y'(π/3)=-2sin(5π/6)=-2*1/2=-1
у-е касательной -1(x-π/3)+√3/3