SOLUTION: Find the maximum and minimum values of each given function and state the corresponding values of x . ( 0 <= x < 2pi ) (1) {{{y=cos(x)+2cos(x/2)}}}

Take the derivative on "x": y'(x) = -sin(x) + 2*(-sin(x/2)*(1/2) = - sin(x) - sin(x/2).

Equate it to zero. You will get an equation

-sin(x) - sin(x/2) = 0,   or

2*sin(x/2)*cos(x/2) + sin(x/2) = 0,   or   ( after factoring )

sin(x/2)*(2*cos(x/2) + 1) = 0.

It deploys in two independent equations


1.  sin(x/2) = 0  --->  x/2 = 0, ,   --->  x = 0  and  x =   ( the only roots in the segment [,] ).


2.  2*cos(x/2) + 1 = 0  --->  cos(x/2) =   ---> x/2 = ,   --->  x =   ( the only root in the segment [,] ).
 

Answer:

  a)  Maximum at x = 0:                        3.


  b)  Minimum at x = :     = -1.5.


  c)  Local maximum at x = : 1 + 2*(-1) = -1.

Plot y = cos(x)+2cos(x/2)