Question

Найдите наибольшее и наименьшее значения функции f(x)=cos^2x-cos x на отрезке[2;п/4]

Answer 1

y' = -2cosx*sinx + cosx = 0

cosx(-2sinx + 1) = 0

cosx = 0   x = π/2 + πk, k ∈ Z

-2sinx + 1 = 0  sinx = 1/2  x = (-1)^k * π/6 + πk, k ∈ Z

Найдем значения x, принадлежащие промежутку [π/3;π]

x = π/2 + πk

при k = 0  x = π/2  

x = (-1)^k * π/6 + πk

при k = 1; x = 5π/6

Проверим значния ф-ии в точках π/3; π/2; 5π/6 и π

y(π/3) = cos^2(π/3) + sin(π/3) = 1/4 + √3/2 = (2√3 + 1)/4 ≈ 1,11

y(π/2) = 0 + 1 = 1

y(5π/6) = 3/4 + 1/2 = 5/4 = 1,25

y(π) = 1 + 0 = 1

yнаиб = 1,25

yнаим = 1