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Let$$f(x)= \begin{cases} \frac{\sin(x)}{x} & x \ne 0 \\ 1 & x=0 \end{cases} $$

What's the number of extreme points for this function in the interval $(-2\pi; 2\pi)$ and which are they?

$f'(x)=\frac{x\cos(x)-\sin(x)}{x^2}$

It says that the function has three extreme points but I can only find this one:

$x\cos(x)-\sin(x)=0\iff x=\tan(x)\Rightarrow x=0$

Any ideas on how to find the other two ?

Thanks in advance.

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  • $\begingroup$ draw graphs of $y=x$ and $y=\tan x$ to see the other points. They can be found using numerical methods. $\endgroup$ – Vasya Jul 19 at 11:45
  • $\begingroup$ wolframalpha.com/input/… notice from the graph that extreme points for $\frac{\sin(x)}x$ correspond to intersections of the diagonal $y=x$ with the many pieces of $\tan(x)$. $\endgroup$ – Mirko Jul 19 at 13:21
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You have to be careful with $\tan \, x$ because it is undefined at $\pm \frac {\pi} 2$ and $\pm \frac {3\pi} 2$. Consider the interval $(\frac {\pi} 2, \frac {3\pi} 2)$. Show that in this interval $\tan \, x -x$ takes all values from $-\infty$ to $\infty$. [$\lim_{x \to \pi /2+} \tan \, x -x=-\infty $ and $\lim_{x \to 3\pi /2-} \tan \, x -x=\infty $]. Hence it vanishes at some point in that interval. Similarly there is solution on the negative side.

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