найдите величайшее значение функции y=8x-4tgx-2pi+2 на отрезке [-pi/3;pi/3]

y=8x-4tgx-2+2

y' = 8 - 4/cosx;

y' = 0; 8 - 4/cosx = 0;

4/cosx = 8;

cosx = 4/8;

cosx = 1/4;

2cosx = 1/2;

1 + cos2x = 1/2;

cos2x = 1/2 - 1;

cos2x = -1/2;

2x = arccos( -1/2) + 2n;

2x = ( 2/3) + 2n;

x = ( /3) + n;

Критичные точки, принадлежащие данному отрезку: /3.

y(-/3) = 8(-/3)-4tg(-/3)-2+2 = -8/3 + 43 - 2 + 2 = 43 - 14/3 + 2

y(/3) = 8/3-4tg(/3)-2+2 = 8/3 - 43 - 2 + 2 = -43 + 2/3 + 2

miny(x) = y(-/3) = 43 - 14/3 + 2

[-/3; /3]

miny(x) = y(/3) = -43 + 2/3 + 2

[-/3; /3]