Question 1047187
.
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)}}}
~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
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, {{{pi}}}, {{{2pi}}}  --->  x = 0  and  x = {{{2pi}}}  ( the only roots in the segment [{{{0}}},{{{2pi}}}] ).


2.  2*cos(x/2) + 1 = 0  --->  cos(x/2) = {{{-1/2}}}  ---> x/2 = {{{2pi/3}}}, {{{4pi/3}}}  --->  x = {{{4pi/3}}}  ( the only root in the segment [{{{0}}},{{{2pi}}}] ).
 

<U>Answer</U>:

  a)  Maximum at x = 0:                        3.


  b)  Minimum at x = {{{4pi/3}}}:  {{{-1/2 + 2*(-1/2)}}}   = -1.5.


  c)  Local maximum at x = {{{2pi}}}: 1 + 2*(-1) = -1.
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  <TD>

{{{graph( 330, 330, -1.5, 14.5, -4.5, 4.5,
          cos(x)+2*cos(x/2)
)}}}


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

  </TD>
  </TR>
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