Finding Max and Mins Part III

[Hckyplayer8]

Determine the absolute max and min values of f(x) = sin x over 2 + cos x on the interval [0,2pi] and the x values where they occur.

The first step is to find the derivative which will involve the quotient rule. So (cos x)(2+cos x) - (-sin x)(sin x) over (2+cos x)².

I got the top simplified to (cos x)(2 + cos x) + (sin x)² which checks out when I run it through symbolab. After that our processes diverge as I want to distribute the (cos x). Symbolab instead changes the sin x² into 1-cos x²...

 

[Harry_the_cat]

Just checking that \(\displaystyle f(x) =\frac{sin x}{2+cos x}\) rather than \(\displaystyle \frac{sin x}{2} + cos x\).

Please make sure you insert brackets if you are going to write an equation in a line

 

[Jagulba]

So simplified would be
2cosx +( cosx)^2+( sinx)^2 which is 2cosx+1
It should be easy to find the root now, just remember to check the sign of derivative around the root to see wether the point is max min or inflection point

 

[Hckyplayer8]

Just checking that \(\displaystyle f(x) =\frac{sin x}{2+cos x}\) rather than \(\displaystyle \frac{sin x}{2} + cos x\).

Please make sure you insert brackets if you are going to write an equation in a line

The problem's function is \(\displaystyle f(x) =\frac{sin x}{2+cos x}\)

Sorry about that.

 

[Hckyplayer8]

So simplified would be
2cosx +( cosx)^2+( sinx)^2 which is 2cosx+1
It should be easy to find the root now, just remember to check the sign of derivative around the root to see wether the point is max min or inflection point

I'm with you on the distribution of cos x across 2+cos x². I'm not tracking when it comes to converting sin x² to 2 cos x +1.

 

[Jagulba]

Oh i assumed you are looking for x in Real numbers and not complex.
If my assumption was correct, in Real numbers (2+cosx)>=1 is always true so you don’t need to check if your root make zero over zero case. So you only need to check where the numerator becomes zero.
Now we are only looking at (cos x)(2+cos x) - (-sin x)(sin x)
2cosx+cosx^2-(-sinx^2)= 2cosx+ cosx^2+sinx^2= 2cosx+ (cosx^2+sinx^2)= 2cosx + 1

 

[Hckyplayer8]

Alright. So the next step is to take the derivative and set it to 0. So set [1+2 cos x] over [(2+cos x)²] =0

Which means 1+ 2 cos x = 0 and cos x is = -1/2

Using the unit circle, cos -1/2 is equal to where x = 2pi/3 and x=4pi/3.

f(2pi/3) = [sin 2pi/3] over [2+cos 2pi/3] = [radical 3 over 2] over [2+(-1/2)] = radical 3 over 3
f(4pi/3) = [sin 4pi/3] over [ 2 + cos 4pi/3] = [- radical 3 over 2] over [2+(-1/2)] = - radical 3 over 3

If my interpretation is correct, this problem results in a sine wave. So am I correct to say that there are multiple spots for max and mins?

 

[Jagulba]

Very well done.
You also could use the derivative to see which one is max or min.
You need to check the derivative sign (positive or negative) near those points. If before the point is positive and after the point is negative then that point is maximum for example.
To answer your question i should say no, because the question asked you in a finite window (0,2pi), you would have one max and one min. If there is no window or constraint then the answer would be yes.

 

[Hckyplayer8]

Thanks for the help!

 

[Steven G]

Alright. So the next step is to take the derivative and set it to 0. So set [1+2 cos x] over [(2+cos x)²] =0

You should also check to see where the derivative is undefined as those values are also critical points. As Jagulba pointed out the denominator is always at least 1 so can't be 0.