Question

Найдите наибольшее и наименьшее значение функции y=f(x) на данном промежутке: f(x)=√3x+cos2x , [-П/2;П/2]

Answer 1

F(-π/2) = √3 × (-π/2) + cos2(π/2) = -1.7 × 3.14/2 + cosπ = -3.669
f(π/2) = √3 × π/2 + cos2(π/2) = 1.669
f '(x) = √3 - sin2x × 2 = 0
√3 - 2sin2x = 0
-2sin2x = -√3 | ÷ (-2)
sin2x = √3/2
2x = (-1)ⁿ × arcsin√3/2 + πn | ÷2
x = (-1)ⁿ × π/6 + πn/2

n = 0
x = (-1)⁰ × π/6 + π×0/2 = π/6
n = 1
x = (-1)¹ × π/6 + π×1/2 = π/3

f(π/6) = √3 × π/6 + cos2(π/6) = 1.7 × 3.14/6 + 1/2 = 1.389
f(π/3) = √3 × π/3 + cos2(π/3) = 1.7 × 3.14/3 - 1/2 = 1.279

f(max) = 1.389
f(min) = 1.279