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Возможно ли найти эту книгу в электроном виде?
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Ответы на вопрос:
y=8x-4tgx-2π+2
y' = 8 - 4/cos²x;
y' = 0; 8 - 4/cos²x = 0;
4/cos²x = 8;
cos²x = 4/8;
cos²x = 1/4;
2cos²x = 1/2;
1 + cos2x = 1/2;
cos2x = 1/2 - 1;
cos2x = -1/2;
2x = ±arccos( -1/2) + 2πn;
2x = ±( 2π/3) + 2πn;
x = ±( π/3) + πn;
критические точки, принадлежащие данному отрезку: ± π/3.
y(-π/3) = 8(-π/3)-4tg(-π/3)-2π+2 = -8π/3 + 4√3 - 2π + 2 = 4√3 - 14π/3 + 2
y(π/3) = 8π/3-4tg(π/3)-2π+2 = 8π/3 - 4√3 - 2π + 2 = -4√3 + 2π/3 + 2
miny(x) = y(-π/3) = 4√3 - 14π/3 + 2
[-π/3; π/3]
miny(x) = y(π/3) = -4√3 + 2π/3 + 2
[-π/3; π/3]
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