Question

найти наиб. и наим. значения функции
y=cosx-√3 sinx на отрезке [-П;0]

Answer 1

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