1) y = sin(2x)
y' = 2cos(2x) = 0,
2x = π/2 + πk
x = π/4 + πk/2
π/12 ≤ π/4 + πk/2 ≤ π/2
π/12 - π/4 ≤ πk/2 ≤ π/2 - π/4
-π/3 ≤ πk ≤ π/2
-1/3 ≤ k ≤ 1/2
k=0
x=π/2
y(π/2) = sin(π) = 0 - наименьшее значение на отрезке
y(π/12) = sin(π/6) = 0.5 - наибольшее значение на отрезке
2) y=(x+1)/(x^2 + 2x + 2)
y' = (x^2 + 2x + 2 - (x+1)(2x+2))/(x^2 + 2x + 2)^2 = (x^2 + 2x + 2 - 2x^2 - 4x - 2)/(x^2 + 2x + 2)^2 = -(x^2 + 2x)/(x^2 + 2x + 2)^2 = 0
x^2 + 2x = x*(x+2) = 0
x=0, x=-2
y(0) = 1/2 = 0.5 - наибольшее значение
y(-2) = -1/2 = -0.5 - наименьшее значение
y(1) = 2/5 = 0.4