Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$

Question

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?

Answer 1

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.

Answer 2

Let $g(x)= x \cos x -\sin x.$

We have:

1. $g(0)=0.$

2. $g(2 \pi)=2 \pi$ and $g(\pi)= - \pi$. By the intermediate value theorem, $g$ has a zero in $( \pi, 2 \pi).$

3. $g(-2 \pi)=-2 \pi$ and $g(- \pi)= \pi$. By the intermediate value theorem, $g$ has a zero in $( -2\pi, -\pi).$

So far, we have that $g$ has three zeros in $( -2 \pi, 2 \pi).$

Now it is your turn to show that $g$ has exactly three zeros in $( -2 \pi, 2 \pi).$

(Use $g'$ and look in which subintervalls of $( -2 \pi, 2 \pi)$ the function $g$ is increasing / decreasing.)

Answer 3

Notice that $$ f'(x)=\frac{x\cos(x)-\sin(x)}{x^2} $$ Therefore, $f(x)=0\implies|f'(x)|=\frac1{|x|}\ne0$.

$f$ has simple roots at $-2\pi$, $-\pi$, $\pi$, and $2\pi$. In between each pair of adjacent roots, $f$ has at least one local extremum. Thus, there are at least $3$ local extrema in $[-2\pi,2\pi]$.

$f'(x)=0\implies\tan(x)=x$ which happens once on $\left[-\frac{3\pi}2,-\pi\right]$, at $0$, and once on $\left[\pi,\frac{3\pi}2\right]$. Thus there are at most $3$ roots.