Easiest to make change of variable:
$y = (x-4) \implies \frac{dy}{dx} = 1.$
$$f'(y) = \frac{12y}{3y^2 + 1}.$$
It is immediate that at $y = 0, f'(y) = 0$
and $f''(y) > 0.$
This verifies that $f(y)$ has a minimum at $y = 0$.
Further, it is straight forward that
$$f(y) = \int f'(y)dy = \int \frac{12y}{3y^2 + 1}
= 2 \log (3y^2 + 1) + C.$$
This means that when $y = 0, 5 = f(y) = C.$
Thus, $f(y) = 2 \log (3y^2 + 1) + 5.$
This immediately allows the conclusion that $f(y)$ has
no extreme value
for any finite value of $y$, except at $y = 0$.
Alternatively, you can reason that $f'(y) = 0$
when and only when $y = 0$.
Edit
The following analysis resulted from back and forth comments with Mick, following this answer. I think that this analysis deserves to be included in the answer.
$f(y)$ is continuous throughout $\mathbb{R}.$
$f''(y)$ and $f'(y)$ are both well defined throughout $\mathbb{R}.$
$f''(0) > 0, f'(0) = 0,$ and $\forall y\neq 0, f'(y) \neq 0.$
Therefore, $\forall y < 0, f'(y) < 0.$
Similarly, $\forall y > 0, f'(y) > 0.$
Therefore $f(y)$ must be strictly increasing on $(0,\infty)$
and $f(y)$ must be strictly decreasing on $(-\infty, 0).$
Therefore $y$ has a global minimum at $y = 0.$
Anyway...
As $y \to \infty, \log(3y^2 + 1) \to \infty.$
Therefore, as $y \to \infty ~f$ is unbounded.
Best Answer
Use AM-GM: $$x^4+\frac1{x^2}=x^4+\frac1{2x^2}+\frac1{2x^2}\ge 3\sqrt[3]{\frac1{4}},$$ equality occurs when $x^4=\frac1{2x^2}=\frac1{2x^2} \Rightarrow x=\pm\frac1{\sqrt[6]{2}}$.