Найдите наименьшее значение функции y=9+cos^2x (всё под корнем) на отрезке [п/3; 2 п/3].

Y' = (√ (9+cos²x)) '=[1/2√ (9+cos²x) ²] * (9+cos²x) = 2cosx * (cosx) '/2 (9+cos²x) = 2cosx * (-sinx) / (2 (9+cos²x)) = - sinxcosx / (9+cos²x) = - (1/2) sin2x / (9+cos²x)

y'=0

- (1/2) sin2x / (9+cos²x) = 0

- (1/2) sin2x=0, 9+cos²x≠0

sin2x=0

2x=πn, n∈Z

x=πn/2, n∈Z

π/3≤πn/2≤2π/3

1/3≤n/2≤2/3 |*6

2≤3n≤4

2/3≤n≤4/3

n=1

x=π/2

y (π/3) = √ (9+cos² (π/3)) = √ (9+1/4) = √9,25

y (2π/3) = √ (9+cos² (2π/3)) = √ (9+1/4) = √9,25

y (π/2) = √ (9+cos² (π/2)) = √ (9+0) = √9

у наим. = у (π/2) = 3