Finding a function that fits the "lowest points" of another one I came up with this problem, which I cannot solve myself.
Consider the function:
$\displaystyle f(x) = x^{\ln(|\pi \cos x ^ 2| + |\pi \tan x ^ 2|)}$, which has singularities at $\sqrt{\pi}\sqrt{n + \dfrac{1}{2}}$, with $n \in \mathbb{Z}$.
Looking at its graph:

we can see it is globally increasing:

I was wondering if there exists a function $g(x)$, such that $f(x) - g(x) \ge 0, \forall x \in \mathbb{R^{+}}$ and that best fits the "lowest points" of $f(x)$.
Sorry for the inaccurate terminology but I really don't know how to express this concept mathematically. Here is, for example, $g(x) = x ^ {1.14}$ (in red):

Actually $g(x)$ is not correct because for small values of $x$ it is greater than $f(x)$.

Is it possible to find such a $g(x)$, given that the "nearest" is $g(x)$ to $f(x)$'s "lowest points" the better it is? Again, sorry for my terminology, I hope you could point me in the right direction.  
Thanks,  

As $a^{\ln b}=\exp(\ln a\cdot\ln b)=b^{\ln a}$ the function $f$ can be written in the following way:
$$f(x)=\bigl(\pi|\cos(x^2)|+\pi|\tan(x^2)|\bigr)^{\ln x}\ .$$
Now the auxiliary function
$$\phi:\quad{\mathbb R}\to[0,\infty],\qquad t\mapsto \pi(|\cos(t)|+|\tan(t)|)$$
is periodic with period $\pi$ and assumes its minimum $\pi$ at the points $t_n=n\pi$. The function $$\psi(x):=\phi(x^2)=\pi|\cos(x^2)|+\pi|\tan(x^2)|\bigr)$$ assumes the same values as $\phi$; in particular it is $\geq\pi$ for all $x\geq0$ and  $=\pi$ at the points $x_n:=\sqrt{n\pi}$ $\ (n\geq0)$. Therefore
$$f(x)=\bigl(\psi(x)\bigr)^{\ln x}\geq \pi^{\ln x}=x^{\ln\pi}\qquad(x\geq1)$$
and $=x^{\ln\pi}$ at the $x_n>1$. For $0<x<1$ the inequality is the other way around because $y\mapsto q^y$ is decreasing when $0<q<1$.
