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

Question

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

Christopher Agresta · Accepted 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.

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.

Evelyn Agard · Answer

undefined To find the critical points of $ f(x) = -\frac{\sin(x)}{2 + \cos(x)} $, follow these steps:

1. Find the first derivative of the function, $ f'(x) $.
2. Set $ f'(x) $ equal to zero and solve for $ x $.
3. Check the second derivative $ f''(x) $ at the critical points to determine concavity.
4. Identify local maxima and minima based on the sign changes of $ f'(x) $ and the concavity of the function.

Now, let's go through each step:

1. Find the first derivative:
$$ f(x) = -\frac{\sin(x)}{2 + \cos(x)} $$
$$ f'(x) = \frac{\cos(x)(2 + \cos(x)) - (-\sin(x))(-\sin(x))}{(2 + \cos(x))^2} $$

2. Set $ f'(x) $ equal to zero and solve for $ x $:
$$ \frac{\cos(x)(2 + \cos(x)) + \sin^2(x)}{(2 + \cos(x))^2} = 0 $$
$$ \cos(x)(2 + \cos(x)) + \sin^2(x) = 0 $$

3. Check the second derivative $ f''(x) $ at the critical points to determine concavity:
$$ f''(x) = \frac{2\sin(x)(2 + \cos(x)) - 2\cos^2(x) - 2\sin^3(x)}{(2 + \cos(x))^3} $$

4. Identify local maxima and minima based on the sign changes of $ f'(x) $ and the concavity of the function.