Given $f(x) = x \sin\frac1x$, find roots of $f'(x)$ in the interval $0\le x \le \frac 1{\pi}$. This question is taken from book: Advanced Calculus: An Introduction to Classical Analysis, by Louis Brand. The book is concerned with introductory analysis.

If $f(x) = x \sin\frac1x\;(x\ne 0), f(0) = 0$, does Rolle's theorem guarantee a root of $f'(x)$ in the interval $0\le x \le \frac 1{\pi}$? Show that $f'(x)$ has an infinite number of roots $x_l \gt x_2 \gt x_3\gt \cdots$ in the given interval which may be put in one-to-one correspondence with the roots of $\tan y = y\,$ in the interval $\pi \le y \lt \infty$. Calculate $x_1$ to three decimal places. 

Given $f(x) = x \sin\frac1x(x\ne 0), f(0) = 0$.  
At $x=0, f(0) = 0 \sin(\infty)$, but $\sin(\infty)\in[-1,1]$, which means the range corresponding to $x=0$ is undefined.
But, the value of $f(0)$ is stated to be $0$. This is a point of confusion as how this range point is specified. 
Also at $x =\frac 1{\pi}$, the fn. yields $f(x) = \frac 1{\pi} \sin(\pi) =0.$ 
So, $f(0)= f\left(\frac 1{\pi}\right) = 0$.
Rolle's theorem needs three conditions: 


*

*Let $f(x)$ be continuous on a closed interval $[a, b]$,

*and, $f(x)$ be differentiable on the open interval $(a, b)$.

*If $f(a) = f(b)$, then there is at least one point $c$ in $(a, b)$ where $f'(c) = 0$.


By being a product of polynomial & a trigonometric function, both of which are differentiable & continuous, the product too is.
Hence, all three conditions are satisfied. So, root of $f'(x)$ is guaranteed in the given interval $\big[0,\frac{1}{\pi}\big]$.
First need calculate $x_1$, so find $f'(x)$. 
It is given by $\sin\left(\frac 1x\right)-\frac 1x \cos \left(\frac 1x\right)$.
$f'(x)=0\implies x\sin\left(\frac 1x\right)=\cos \left(\frac 1x\right)\implies x=\cot\left(\frac 1x\right)$.
Unable to solve further.
I hope that the solution of the above equation can help with the rest two questions, although have doubts for each as stated below:


*

*$f'(x)$ has an infinite number of roots $x_l \gt x_2 \gt x_3\gt \cdots$ in the interval $0 \le x \le \frac 1{\pi}$.
Unable to understand how it is possible to have the given scenario of infinite roots in a given order.

*These roots may be put in one-to-one correspondence with the roots of $\tan y = y\,$ in the interval $\pi \le y \lt \infty$.
Here, the two equations whose roots are to be paired are:
$x = \cot\left(\frac 1x\right)$ and $ y = \tan(y)$ with connection not visible.
Edit The book states the answer for $x_1=0.2225$. Still have no clue about attaining it. 
 A: Partial answer. $f'(x)=\sin(\frac 1 x)-\frac 1 x \cos(\frac 1 x)$ so $\tan (\frac 1 x)=\frac 1 x$. So the correspondence between roots of $\tan \, y=y$ and solutions of the given equation is obvious. Also $tan\, y -y$ changes sign in every interval of the type $(2n\pi-\pi/2,2n\pi+\pi /2)$ so it has  a root in that interval. It follows that the given equation has infintely many roots. 
A: For $x>0$,$$f'(x)=\sin\left(\frac1x\right)-\frac1x\cos\left(\frac1x\right)=\sin(y)-y\cos(y).$$
Hence the roots of $f'$ are the inverse solutions of $y=\tan(y)$ and $x_1$ corresponds to the smallest $y$ above $\pi$.
As $\dfrac{df'(y)}{dy}=\cos(y)-\cos(y)+y\sin(y)$ is negative between $\pi$ and $2\pi$, and $f'(\pi)$ and $f'(2\pi)$ differ in sign, we can approach the isolated root by the secant method.
The successive approximations are
$$4.1887902\cdots (f>0)\\
4.5312881\cdots (f<0)\\
4.4901885\cdots (f>0)\\
4.4933831\cdots (f>0)\\
4.4934095\cdots (f<0)\\
$$
and
$$\frac1{4.4934095\cdots}=0.2225\cdots.$$
A: Guide:


*

*The part that you mention at $x=0$, $f(0)=0\cdot \sin (\infty)$ makes no sense. It has been stated that the rule $f(x)=x\sin \frac1x$ only holds if $x\ne 0$. Also, $\frac10$ is undefined.

*You haven't argued that it is continuous at $0$. You have to show that $\lim_{x \to 0^+}f(x)=f(0)$.

*To show that there are countably infinite solutions, after you partition the domain to countably many partitions, you want to check that each partition has at least $1$ and also finitely many solutions.

*Check that $\tan y - y$ increases on $(\pi, \frac{3\pi}2)$ and there is a unique root in that interval.

*Hence we know that $x_1 \in \left(\frac{2}{3\pi} ,\frac1{\pi}\right)$. Now, you can perform a binary search on that interval to narrow down $x_1$ up to $3$ decimal places.
