Question

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

Answer 1

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

Answer 2

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