Question

Задание во вложении.

Answer 1

1) y=x^5-x^3-2x+3; [-2;2]
На концах отрезка
y(-2)=-32-(-8)-2(-2)+3=-32+8+4+3=-17
y(2)=32-8-2*2+3=32-8-4+3=23
Экстремумы
y' =5x^4-3x^2-2=(x^2-1)(5x^2+2)=0
x1=-1; y(-1)=-1-(-1)-2(-1)+3=-1+1+2+3=5
x2=1; y(1)=1-1-2+3=1
Минимум y(-2)=-17; максимум y(2)=23

2) y=sin x*cos^2(x/2); [0;Π/2]
На концах отрезка
y(0)=sin 0*cos^2 0=0
y(Π/2)=sin(Π/2)*cos^2(Π/4)=1*1/2=1/2
Экстремумы
y' = cos x*cos^2(x/2) + sin x*2cos(x/2)*(-sin(x/2))*1/2 =
= cos x*cos^2(x/2) - sin x*cos(x/2)*sin(x/2) = 0
cos(x/2)*(cos x*cos(x/2) - sin x*sin(x/2)) = 0
cos(x/2)=0; x/2=Π/2+Π*k; x=Π+2Π*k
На отрезке [0;Π/2] этих корней нет.
cos x*cos(x/2)-sin x*sin(x/2) = cos(3x/2) = 0
3x/2=Π/2+Π*n; x=Π/3+2Π/3*n
На отрезке [0;Π/2] корень
x=Π/3; y(Π/3)=sin(Π/3)*cos^2(Π/6)=√3/2*3/4=3√3/8~0,65
Минимум y(0)=0; максимум y(Π/3)=3√3/8