Question

Помогите пожалуйста решить!

Answer 1

4.3.36
y`=14-14√2cosx=0
cosx=1/√2
x=+-π/4+2πn,n∈z
x=π/4∈[0;π/2]
y(0)=-28-3,5π≈-39
y(π/4)=-28-3,5π+3,5π-14=-42 наим
y(π/2)=-28-3,5π+7π-14√2=-28+11-19,6=-37,6-наиб
4.3.38
y`=-27√3/2+27sinx=0
sinx=√3/2
x=(-1)^n*π/3+πn,n∈z
x=π/3∈[0;π/2]
y(0)=-22+9√3π/2-27=-49+24=-25 наиб
y(π/3)=-22+9√3π/2-9√3π/2-27*1/2=-22-13,5=-35,5 наим
y(π/2)=-22+9√3π/2-27√3π/4=-22-12=-34