<s>[INST] Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$ Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$
$$f(x) = \frac{\sin x}{x}, x \not= 0 $$
$$f(x) = 1, x = 0 $$
I calculated its derivative.
$$f'(x) = \frac{x\cos x-\sin x}{x^2}, x \not= 0$$
$$f'(x) = 0, x = 0$$
I can see that $f'(x)$ has a solution for x = 0.
I am not sure how to solve for $x\not= 0$ though. I have tried this:
$$ \frac{x\cos x - \sin x}{x^2} = 0 $$
Multiply both sides by $x^2$
$$ x\cos x - \sin x = 0 $$
Add $\sin x$ to both sides
$$ x\cos x = \sin x$$
Now divide both sides by $\cos x$
$$ x = \tan x $$
Did I do anything wrong? Can this equation be figured out without using any kind of graphing tool?
 [/INST] What you did is very correct and there is no explicit solution to the equation.
Discarding the trivial case $x=0$, it is better to consider that you look for the zero's of
$$f(x)=x \cos (x)-\sin (x)$$ Now, consider a Taylor expansion close to $\frac {3 \pi} 2$ (remember the symmetry) to get
$$f(x)=1+\frac{3}{2} \pi  \left(x-\frac{3 \pi }{2}\right)+\frac{1}{2} \left(x-\frac{3 \pi
   }{2}\right)^2-\frac{1}{4} \pi  \left(x-\frac{3 \pi }{2}\right)^3+O\left(\left(x-\frac{3 \pi   }{2}\right)^4\right)$$ and use series reversion to get
$$x=\frac{3 \pi }{2}-\frac{2}{3 \pi }-\frac{16}{81 \pi ^3}-\frac{16}{243 \pi ^5}- \cdots$$ This truncated expansion would give $x=4.4936$ while the "exact" solution would be $4.4934$.
Just compute the value of the function at this point.
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