Question

f (x)=cos2xcosx-sinx в точке x=П/4 помогите

Answer 1

y' = -2cosx*sinx + cosx = 0cosx(-2sinx + 1) = 0cosx = 0   x = π/2 + πk, k ∈ Z-2sinx + 1 = 0  sinx = 1/2  x = (-1)^k * π/6 + πk, k ∈ ZНайдем значения x, принадлежащие промежутку [π/3;π]x = π/2 + πkпри k = 0  x = π/2  x = (-1)^k * π/6 + πkпри k = 1; x = 5π/6Проверим значния ф-ии в точках π/3; π/2; 5π/6 и πy(π/3) = cos^2(π/3) + sin(π/3) = 1/4 + √3/2 = (2√3 + 1)/4 ≈ 1,11y(π/2) = 0 + 1 = 1y(5π/6) = 3/4 + 1/2 = 5/4 = 1,25y(π) = 1 + 0 = 1yнаиб = 1,25yнаим = 1