Question

найдите наименьшее значение функции y=6x+2П-5cosx-7 на отрезке [П/3,4]

Answer 1

y=6x+2pi-5cosx-7

y'=6+5cosx

5cosx=-6

cosx=-6/5 - нет решений

f(pi/3) = 6(pi/3)+2pi-5cos(pi/3)-7 = 3.066

f(4) = 6(4)+2pi-5cos(4)-7 = 26.55 

 

Answer 2

$y=6x+2pi-5cosx-7 \y'=6+5cosx \5cosx=-6 \cosx=-6/5 - net korney \f(pi/3) = 6(pi/3)+2pi-5cos(pi/3)-7 = 3.066 \f(4) = 6(4)+2pi-5cos(4)-7 = 26.55$