How do I find the critical points for the absolute value of sine x?

[John O' Meara]

i am trying to solve the following, and the derivative of the |sin x| is required. Find the critical points and which ones are stationary points for |sin x|?
To find the points we must solve $$f'(x)=0$$
Since f'(x)=|cos x|, and solving for this gives $$\frac{\pi}{2}, \frac{3\pi}{2}$$ etc.
That is $$n\pi + \frac{\pi}{2}$$ as the total number of critial points, but using a graphing calculator I see $$n\frac{\pi}{2}$$ critical points. So I was thinking maybe f'(x)=|cos x| is not correct. I did a sign analysis and only got all +s, suggesting that (according to the calculus of it) f does not have any relative extremum! I am studing this on my own, so please help, Thanks.

 

[disregardthat]

Think about the alternative definition of $$|x|$$, namely $$\sqrt{x^2}$$.

 

[tiny-tim]

Hi John!

Since f'(x)=|cos x|

Then f'(x) would always be positive, but looking at the graph of f(x), that clearly isn't true.

 

[sutupidmath]

f(x)=|sinx| is differentiable at every point in R\{k*pi}. In general: |x|=x if x>0 and -x if x<0. Do the same for |sinx|.

 

[John O' Meara]

Hi,
I was thinking alright of (sin(x))^2, its derivative is 2sinx*cosx and it gives you the correct number of critical points for |sin x|, but when I use the definition of the derivative to find f'(x) I get |cos x|. Is there any way to get $$\sqrt{(\sin x )^2}$$ from the definition of the derivative? Thanks.

 

[disregardthat]

Just differentiate as you would with $$\sqrt{f(x)}$$. Don't cancel out, that is; $$\sqrt{y^2}$$ to $$y$$, at any point.

 

[sutupidmath]

Hi,
I was thinking alright of (sin(x))^2, its derivative is 2sinx*cosx and it gives you the correct number of critical points for |sin x|, but when I use the definition of the derivative to find f'(x) I get |cos x|. Is there any way to get $$\sqrt{(\sin x )^2}$$ from the definition of the derivative? Thanks.

Perform the steps that i suggested, and then apply the def. of the derivative on each interval separately. Because, like i mentoned |sinx|, as you will find out, is not differentiable at any multiple of pi.

i.e. |sinx|={ sinx if sinx>0 and -sinx if sinx<0}

so basically, find the values of x for which sinx>o,and the values of x for which sinx<0, at this point you know that |sinx| is differentiable whenever sinx<0 and sinx>0, all you need to do is determine what happens when sinx=0, namely at those points x. and you can use the def. of the derivative in terms of sequences or however you have learned it to show that whenever x=k*pi, |sinx| is not differentiable.