Question

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

Answer 1

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.