How do you find the critical points for #f(x)= -(sinx)/ (2+cosx) # and the local max and min?

1 Answer
Sep 14, 2016

Answer:

The critical points are at:
#((2pi)/3,sqrt(3)/3)#is a minimum point

#((4(pi)/3),sqrt(3)/3)# is the maximum point.

Explanation:

To find the critical points we have to find #f'(x)#
then solve for #f'(x)=0#

#f'(x)=-((sinx)'(2+cosx)-(2+cosx)'sinx)/(2+cosx)^2#

#f'(x)=-(cosx(2+cosx)-(-sinx)sinx)/(2+cosx)^2#

#f'(x)=-(2cosx+cos^2(x)+sin^2(x))/(2+cosx)^2#

Since #cos^2(x)+sin^2(x)=1# we have:
#f'(x)=-(2cosx+1)/(2+cosx)^2#

Let us dolce for #f'(x)=0#to find the critical points:

#f'(x)=0#
#rArr-(2cosx+1)/(2+cosx)^2=0#
#rArr-(2cosx+1)=0#
#rArr(2cosx+1)=0#
#rArr2cosx=-1#
#rArrcosx=-1/2#
#cos(pi-(pi/3))=-1/2#

or
#cos(pi+(pi/3))=-1/2#

Therefore,
#x=pi-(pi/3)=(2pi)/3#
or #x=pi+(pi/3)=(4pi)/3#

Let's compute #f((2pi)/3)=-sin((2pi)/3)/(2+cos((2pi)/3)#

#f((2pi)/3)=-(sqrt(3)/2)/(2-1/2)#
#f((2pi)/3)=-(sqrt(3)/2)/(3/2)#
#f((2pi)/3)=-(sqrt(3)/3)#
Since#f(x)# is decreasing on #(0,(2pi)/3)#
Then#(((2pi)/3),-sqrt(3)/3)# is minimum point

Since then the function increases till #x=(4(pi)/3)# then the point
#((4(pi)/3),sqrt(3)/3)# is the maximum point.