Найти точки экстремумов функции f (x) = 10cosx-5x

f' (x) = - 10sinx-5; f' (x) = 0

sinx=-0.5

x=-π/6+2πk

x=5/6π+2πk

f' (x) = - 10cosx при х=5/6π+2πk

x=x=-π/6+2πk

f' (x) <0 максимум

f (-π / 6+2 π k) = 10*sqrt (3) / 2+5 π / 6+10 π k=5 (sqrt (3) + π / 6+2 π k)

f (5/6 π + 2 π k) = - 10sqrt (3) / 2-25/6 π - 10 π k

f' (x) = - 10sinx-5

f' (x) = 0

sinx=-1/2

x=-П/6+2 Пk

x=5/6 П+2 Пk

f'' (x) = - 10cosx при х=5/6 П+2 Пk сosx0 минимум

x=x=-П/6+2 Пk f'' (x) <0 максимум

f (-П/6+2 Пk) = 10*sqrt (3) / 2+5 П/6+10 Пk=5 (sqrt (3) + П/6+2 Пk)

f (5/6 П+2 Пk) = - 10sqrt (3) / 2-25/6 П-10 Пk