# What's the number of extreme points for this function

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 ?

• draw graphs of $y=x$ and $y=\tan x$ to see the other points. They can be found using numerical methods. – Vasya Jul 19 at 11:45
• 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 $\tan(x)$. – Mirko Jul 19 at 13:21
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.