On the request of OP I am converting my comment into an answer.
The result in question is a non-trivial property of continuous functions and it is a consequence of completeness of real numbers. The completeness of real numbers is the only property of real numbers that distinguishes it from the rational numbers and it can be expressed in many equivalent forms and perhaps the most popular one is as follows:
Completeness of Real Numbers: Let $A$ be a non-empty subset of the set of real numbers $\mathbb{R} $ which is bounded above (ie there is a specific real number $K$ such that no member of $A$ exceeds $K$). Then there exists a unique real number $M$ with the following properties:
- No member of $A$ exceeds $M$.
- Every real number less than $M$ is exceeded by at least one member of $A$.
Numbers of the form $K$ are said to be an upper bound of set $A$ and it should be obvious that if $K$ is an upper bound of $A$ then any number greater than $K$ is also an upper bound of $A$. The number $M$ in the above result is called the least upper bound or supremum of $A$ and we write $M=\sup A$.
The terms bounded below, lower bound, greatest lower bound or infimum can be defined similarly and it can be proved easily (using the above mentioned result) that if a non-empty subset of the set of real numbers is bounded below then it has a greatest lower bound or infimum.
Let's assume the following property of continuous functions (which is a consequence of the completeness of real numbers as will be shown later):
Theorem 1: If a function $f: [a, b] \to\mathbb{R} $ is continuous continuous on $[a, b] $ then it is bounded on $[a, b] $.
Next we come to the Extreme Value Theorem which is the result in question:
Extreme Value Theorem: If a function $f:[a, b] \to\mathbb{R} $ is continuous on $[a, b] $ then there are two numbers $c, d$ in interval $[a, b] $ such that $f(c) \leq f(x) \leq f(d) $ for all $x\in[a, b] $ ie $f(c) $ is the (absolute) minimum value of $f$ on $[a, b] $ and $f(d) $ is the (absolute) maximum value of $f$ on $[a, b]$.
This theorem is an easy consequence of theorem 1 mentioned above. Since $f$ is continuous on $[a, b] $ it follows by theorem 1 that it is bounded on $[a, b] $. Let $M, m$ be the supremum and infimum (respectively) of the values of $f$ on $[a, b] $ (these numbers $m, M$ exist via completeness property of real numbers). We will show that there is a point $d\in[a, b] $ such that $f(d) =M$.
Let's assume on the contrary that $f(x) \neq M$ for all $x\in[a, b] $. Then the function $g$ defined by $g(x) =1/(M-f(x))$ is continuous on $[a, b] $ and by theorem $1$ it is bounded on $[a, b] $. But since $M$ is the supremum of values of $f$ there are values of $f$ which can be as near to $M$ as we please so that $g(x) =1/(M-f(x))$ is unbounded. This contradiction proves that $M=f(d) $ for some $d\in[a, b] $. Similarly we can show that $m=f(c) $ for some $c\in[a, b] $.
Theorem 1 stated earlier can be proved easily using completeness of real numbers. Since $f$ is continuous at $a$ there is a $c\in(a, b) $ such that $|f(x) - f(a) |<1$ for all $x\in[a, c] $. Thus $f$ is bounded on $[a, c] $. Consider the following set: $$A=\{x\mid x\in[a, b], f\text{ is bounded on }[a,x] \} $$ It is clear from the above argument that $[a, c] \subseteq A\subseteq [a, b] $ so that $A$ is non-empty and bounded above. By completeness property of real numbers $M=\sup A$ exists and clearly $M\in(a, b] $. We will show that $M\in A$ and $M=b$.
Since $f$ is continuous on $M$, it follows that there is a number $h>0$ such that $[M-h, M]\subseteq [a, b] $ and $f$ is bounded on $[M-h, M]$. Since $M=\sup A$, there is a member $x\in A$ such that $x\in(M-h, M] $ and therefore $f$ is bounded on $[a, x] $. It is now clear that $f$ is bounded on $[a, M] $ so that $M\in A$.
Next let us assume on the contrary that $M<b$. Then as before there is a number $h>0$ such that $[M-h, M+h] \subseteq [a, b] $ and $f$ is bounded on $[M-h, M+h] $. And as before there is an $x\in A$ such that $x\in(M-h, M] $ so that $f$ is bounded on $[a, x] $. It follows that $f$ is bounded on $[a, M+h] $ and thus $M+h\in A$. This is a contradiction as $M=\sup A$. Thus $M=b$ and therefore $f$ is bounded on $[a, b] $.
Both theorem 1 and Extreme Value Theorem can be proved independently using various formulations of completeness property and one should try to prove these results using all the different forms of completenes. This helps in understanding the completeness property as well the properties of continuous functions on closed intervals.
Best Answer
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.