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

1 Answer
Sep 14, 2016

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)=0to 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)
Sincef(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.