How to prove that these two functions do not intersect? I have this function $\;g(x)-x\;$ on the domain $\;0&lt;x&lt;\frac{1}{3}\;$, where
$g(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi  x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi  x)+1}}{2 \sqrt{2} \pi  \cosh (\pi  x)}+\frac{18 \sqrt{2} x^2 \left(9 \pi  x^3+4 \tanh (\pi  x)\right)}{2 \sqrt{2} \pi }.$
I need to prove that $\;g(x)\;$ does not intersect $\;x\;$ over the mentioned domain. Calculating the first derivative does not work since it makes the problem more complicated. The plot of the functions is attached.
Any hint or suggestion is welcome.
Partial answer
Denote
$$g_1(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi  x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi  x)+1}}{2 \sqrt{2} \pi  \cosh (\pi  x)}$$ and
$$g_2(x) = \frac{18 \sqrt{2} x^2 \left(9 \pi  x^3+4 \tanh (\pi  x)\right)}{2 \sqrt{2} \pi }$$
$g_1, g_2$ are positive maps and we have $g = g_1+g_2$.
$g_2(x)/x$ is increasing on $(0, 1/3)$ as a product of positive increasing maps. Moreover $g_2(0.06) / 0.06 &gt;1.0$. Hence we have the partial result that $g(x) - x$ is positive on $(0.06, 1/3)$. Remains to prove the result on $(0, 0.06]$.
$$ g(0)=0, g'(0)=1 $$
$$t(x)= x, t(0)=0,t'(0)=1$$
so there is a tangential contact for the two curves with two coinciding intersections as seen in your graph. It is confirmed that at the double root there is common slope$=1$