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NoName9294
@NoName9294
July 2022
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найти наиб. и наим. значения функции
y=cosx-√3 sinx на отрезке [-П;0]
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sedinalana
Verified answer
Y`=-sinx-√3cosx=0
12(1/2sinx+√3/2cosx)=0
sin(x+π/3)=0
x+π/3=πn,n∈z
x=-π/3+πn,n∈z
x=-π/3∈[-π;0]
y(-π)=cos(-π)-√3sin(-π)=-1наим
y(-π/3)=cos(-π/3)-√3sin(-π/3)=1/2-√3*(-√3/2)=1/2+3/2=2 наиб
y(0)=cos0-√3sin0=1
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Answers & Comments
Verified answer
Y`=-sinx-√3cosx=012(1/2sinx+√3/2cosx)=0
sin(x+π/3)=0
x+π/3=πn,n∈z
x=-π/3+πn,n∈z
x=-π/3∈[-π;0]
y(-π)=cos(-π)-√3sin(-π)=-1наим
y(-π/3)=cos(-π/3)-√3sin(-π/3)=1/2-√3*(-√3/2)=1/2+3/2=2 наиб
y(0)=cos0-√3sin0=1