What are the critical values, if any, of # f(x)=cscxtanx-sqrt(xcosx) in [0,2pi]#?

1 Answer
May 22, 2017

Answer:

#x in {0, 0.455, 2pi}#

Explanation:

Critical values are points where #f'(x)=0# or #f'(x)# is undefined, BUT #f(x)# is defined.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

First of all, #cscxtanx=1/sinx*sinx/cosx=1/cosx=secx#, so we can substitute #secx# into the equation.

#f(x)=secx-sqrt(xcosx)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
First, we must find points where #f'(x)=0#

#f'(x) = secxtanx-(cosx-xsinx)/(2sqrt(xcosx))=0#

Graphing this function shows that the only solution on the interval #[0,2pi]# is #0.455#.

Next, we find points where #f'(x)# is undefined. By using a calculator, or by manipulating the equation, we can see that #f'(x)# is not defined on the ENTIRE interval #[pi/2,(3pi)/2]#. This would normally mean that #x=pi/2# and #x=(3pi)/2# are critical points, but #f(pi/2)#and#f((3pi)/2)#are also not defined, since #secx# has asymptotes at those points.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Therefore, our critical points are the endpoints #x=0# and #x=2pi# as well as the point #x=0.455#.