- #1
John O' Meara
- 330
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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 [tex] f'(x)=0[/tex]
Since f'(x)=|cos x|, and solving for this gives [tex] \frac{\pi}{2}, \frac{3\pi}{2} [/tex] etc.
That is [tex] n\pi + \frac{\pi}{2} [/tex] as the total number of critial points, but using a graphing calculator I see [tex] n\frac{\pi}{2} [/tex] 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.
To find the points we must solve [tex] f'(x)=0[/tex]
Since f'(x)=|cos x|, and solving for this gives [tex] \frac{\pi}{2}, \frac{3\pi}{2} [/tex] etc.
That is [tex] n\pi + \frac{\pi}{2} [/tex] as the total number of critial points, but using a graphing calculator I see [tex] n\frac{\pi}{2} [/tex] 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.