Number of roots of $f(x)=x^2-2^{x-\frac{1}{x}}$

Question

How many roots does $f(x)=x^2-2^{x-\frac{1}{x}}$ have in $(0,1]$?
Since this function is continuous, I plugged in a couple of values and looked at the signs of the function's values and I concluded that there is a root between $\frac{1}{10}$ and $\frac{1}{5}$.$1$ is also a root, so I believe there are two roots in $(0,1]$, but how to rigorously prove this?

Answer 1

Instead of finding the roots $x^2 = 2^{x - 1/x}$, let's consider the roots of $f(x) = 2 \ln x - (x - 1/x) \ln 2$. We have $$f'(x) = -\frac {\ln 2} {x^2} \left( x^2 - \frac {2 x} {\ln 2} + 1 \right).$$ $f'(x)$ has two positive roots the product of which is $1$, therefore it has exactly one root $x_0$ on $(0, 1)$.

This means that $f$ decreases from $\infty$ to $f(x_0)$ on $(0, x_0)$ and increases from $f(x_0)$ to $f(1) = 0$ on $(x_0, 1)$. Since $f(1) = 0$, we have $f(x_0) < 0$. Therefore $f$ has exactly two roots on $(0, 1]$.

Answer 2

Compute $$f(0.2),f(0.4)$$ and $$f(1)$$