Найдите наибольшее значение функции y y = 2 cos x + x - П/3 на отрезке [0; П/2]

[tex]y=2cos x+x-\frac{\pi}{3}[/tex]
Ищем критические точки
[tex]y'=-2sin x+1[/tex]
[tex]y'=0[/tex]
[tex]-2sin x+1=0[/tex]
[tex]sin x=\frac{1}{2}[/tex]
[tex]x=(-1)^k*\frac{\pi}{6}+\pi*k[/tex]
k є Z
[tex]0 \leq x \leq \frac{\pi}{2}[/tex]
[tex]x=\frac{\pi}{6}[/tex]
0 __(+)________pi/6_________(-)______pi/2
значит точка [tex]x=\frac{\pi}{6}[/tex] - точка максимума
[tex]y(0)=2cos0+0-\frac{\pi}{3}=2-\frac{\pi}{3}[/tex]
[tex]y(\frac{\pi}{6})=\sqrt{3}-\frac{\pi}{6}[/tex]
[tex]y(\frac{\pi}{2})=\frac{\pi}{6}[/tex]
[tex]y_{max}=y(\frac{\pi}{6})=\sqrt{3}-\frac{\pi}{6}[/tex]

[tex]y=\cos x+x-\frac{\pi}{3},\ x\in[0;\frac{\pi}{2}];\\ y'=-\sin x+1;\\ y'=0;\\ 1-\sin x=0;==>\ \sin x=1==>\ x=\frac{\pi}{2}+2\pi n, n\in Z\\ y(0)=\cos0+0-\frac{\pi}{3}=1-\frac{\pi}{3}=\frac{3-\pi}{3}<0;\\ y(\frac{\pi}{2})=\cos\frac{\pi}{2}+\frac{\pi}{2}-\frac{\pi}{3}=0+\pi(\frac{1}{2}-\frac{1}{3})=\\ =\pi(\frac{3-2}{2\cdot3})=\pi\frac{1}{6}=\frac{\pi}{6}>0;\\ y_{max}=\frac{\pi}{6} [/tex]