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

Question

Emilia Aguilar · Accepted Answer

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

Paisley Johnson · Answer

undefined To find the critical values of $ f(x) = \csc(x) \tan(x) - \sqrt{x \cos(x)} $ in the interval $[0, 2\pi]$, we need to find the values of $ x $ where the derivative of $ f(x) $ is equal to zero or undefined.

The derivative of $ f(x) $ with respect to $ x $ is:

$ f'(x) = -\csc(x) \cot(x) - \frac{\sin(x) + x\sin(x) - 2\cos(x)}{2\sqrt{x \cos(x)}} $

To find where $ f'(x) $ is undefined, we look for values of $ x $ where $ x \cos(x) = 0 $. This occurs at $ x = 0, \pi, 2\pi $.

To find where $ f'(x) $ is zero, we solve the equation:

$ -\csc(x) \cot(x) - \frac{\sin(x) + x\sin(x) - 2\cos(x)}{2\sqrt{x \cos(x)}} = 0 $

This equation has no simple algebraic solution. We would typically use numerical methods to approximate the solutions. These critical values would be within the interval $[0, 2\pi]$.

So, the critical values of $ f(x) $ in the interval $[0, 2\pi]$ are $ x = 0, \pi, 2\pi $, and possibly other values found through numerical methods.