Roots of $x\sin(1/x)$ and Inverse Function theorem

Question

I'm trying to do the following inverse value theorem question:

Consider the function $$f(x) = \begin{cases} \frac{x}{2} + x^2 \sin\left( \frac{1}{x} \right) &\text{ if } x\neq0\\ 0 &\text{ if } x=0 \end{cases}$$ Determine on what subsets of $\mathbb{R}$ the function $f(x)$ is invertible.

To do this, we know that the function is locally invertible in the neighborhoods of all points where $\dfrac{df}{dx}\neq 0$. However looking for the points where this fails gets us:

$$\dfrac12 + 2x\sin(1/x)-\cos(1/x) = 0$$

This is an exceptionally nasty function, and I don't know if there's a missing trig identity I could be using or some special way to solve it, but I was wondering if I could try to solve the simpler $$x\sin(1/x)=0$$ and use that same idea to solve the bigger equation. I don't really want to just cheat and use the set-builder notation and essentially say "All the points where this is false work" unless there's no other option.

Regardless, any help would really be appreciated. Thanks!

Answer 1

I think that there is more to aggregate in your reasoning. The theorem says that, for functions of a single variable, if $g$ is a continuously differentiable function with nonzero derivative at the point $a$, then $g$ is injective in a neighborhood of $a$, the inverse is continuously differentiable near $b=g(a)$ and the derivative of the inverse function at b is the reciprocal of the derivative of $g$ at $a$, which is $$ \frac{d(g^{-1})}{dx}(b) = \frac{1}{\frac{dg}{dx}(a)} $$ The function you are using has not a continuous derivative in $\frac{df}{dx}(0)$. If you look at the graphic of the function in here you can see that it tricks you into thinking that $\frac{df}{dx}$ is going close to $\frac{1}{2}$, nevertheless it is not because the derivative is not defined in $x = 0$. The theorem says that you are allowed to apply it ONLY if $f$ is continuously differentiable AND has nonzero derivative, so you cannot make any statement (solely based on inverse function theorem) if the derivative is zero because the requirements were not satisfied firstly. Looking at the graphic, I think that you may have a great guess on the possible values for which the function is invertible.

Answer 2

In fact you need to find the largest value of $$f(x)= \frac{x}{2} + x^2 \sin\left( \frac{1}{x} \right)$$ with a positive derivative that is to say the largest root of $$f'(x)=2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)+\frac{1}{2}=0$$ or the smallest root of $$g(t)=\frac{2 \sin (t)}{t}-\cos (t)+\frac{1}{2}=0$$ By inspection $$g(\pi)=\frac 32 \qquad \text{and} \qquad g(2\pi)=-\frac 12$$

Using a series expansion $$g(t)=-\frac{1}{2}+\frac{t-2 \pi }{\pi }+\left(\frac{1}{2}-\frac{1}{2 \pi ^2}\right) (t-2 \pi )^2+O\left((t-2 \pi )^3\right)$$ Solving the quadratic gives $$t=\frac{\pi (2 \pi -3)}{\pi -1}\qquad \implies\qquad x=\frac{\pi -1}{\pi (2 \pi -3)}$$ which is an overestimate; so, you are safe with this bound.

Otherwise, it is a simple numerical optimization leading to a value of $0.0848882$ when $x=0.141311$.