![\displaystyle f(x)=x+2\cos x\\\\f'(x)=1-2\sin x=0\\\\2\sin x=1\\\\\sin x=\frac{1}2\\\\\left[\begin{array}{ccc}\displaystyle x=\frac{\pi}{6}+2\pi n;\quad n\in Z\\\\\displaystyle x=\frac{5\pi}6+2\pi n;\quad n \in Z\end{array}\right \\\\\\\underline{...\quad-\quad\quad\frac{\pi}6\quad\quad+\quad\quad\frac{5\pi}6\quad\quad-\quad\quad\frac{13\pi}6\quad\quad+\quad\quad\frac{17\pi}6\quad\quad-\quad...}](https://tex.z-dn.net/?f=%5Cdisplaystyle+f%28x%29%3Dx%2B2%5Ccos+x%5C%5C%5C%5Cf%27%28x%29%3D1-2%5Csin+x%3D0%5C%5C%5C%5C2%5Csin+x%3D1%5C%5C%5C%5C%5Csin+x%3D%5Cfrac%7B1%7D2%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cdisplaystyle+x%3D%5Cfrac%7B%5Cpi%7D%7B6%7D%2B2%5Cpi+n%3B%5Cquad+n%5Cin+Z%5C%5C%5C%5C%5Cdisplaystyle+x%3D%5Cfrac%7B5%5Cpi%7D6%2B2%5Cpi+n%3B%5Cquad+n+%5Cin+Z%5Cend%7Barray%7D%5Cright+%5C%5C%5C%5C%5C%5C%5Cunderline%7B...%5Cquad-%5Cquad%5Cquad%5Cfrac%7B%5Cpi%7D6%5Cquad%5Cquad%2B%5Cquad%5Cquad%5Cfrac%7B5%5Cpi%7D6%5Cquad%5Cquad-%5Cquad%5Cquad%5Cfrac%7B13%5Cpi%7D6%5Cquad%5Cquad%2B%5Cquad%5Cquad%5Cfrac%7B17%5Cpi%7D6%5Cquad%5Cquad-%5Cquad...%7D)
Точки минимума (знак меняется с - на +): ![\displaystyle \boxed{x=\frac{\pi}6+2\pi n;\quad n\in Z}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cboxed%7Bx%3D%5Cfrac%7B%5Cpi%7D6%2B2%5Cpi+n%3B%5Cquad+n%5Cin+Z%7D)
Точки максимума (знак меняется с + на -): ![\displaystyle \boxed{x=\frac{5\pi}6+2\pi n;\quad n\in Z}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cboxed%7Bx%3D%5Cfrac%7B5%5Cpi%7D6%2B2%5Cpi+n%3B%5Cquad+n%5Cin+Z%7D)
Ответ:
розпишем 4=2^2, далее 2^2•2^(n+1)=
2^(2+n+1)= 2^(3+n)
<span>ОДЗ: 6-x^2>0 => x^2<6 => 0<x<sqrt(6)
-x^2-5x+6=0
D=25+24=49
x(1)=-6 - Лишний корень
х(2)=1 - Входит в ОДЗ.
Ответ: х=1
log_2(6-1</span>²<span>)=log_2(5*1) log_2(5)=log_2(5)</span>≈2.32 Но, по-моему значение логарифма не надо, да? Только значение х.