Declaring function range, what else can be excluded than values where it's undefined? I've seen many "complicated" functions' range's merely declared as $(-\infty, \infty)$ or perhaps $(-\infty, \infty) \setminus \{0\}$.
Is that everything that can be said?
Take for example the function:
$$\frac{x^2+1}{2x}$$
It's undefined at $0$, but can we say (or how can we say) that the range is $(-\infty, \infty) \setminus \{0\}$, because $x^2+1≥2x$ (which says that the function isn't bounded) and it doesn't seem like one can specify the range more specifically?
Do we need to take limits or something? Use continuity? Injectivity?
Or how does one specify the "exact" range of e.g. the above function?
 A: Clearly $\dfrac{x^2+1}{2x}$ can never be $0$ so the range cannot be $(-\infty,+\infty)$.  We have
$$
\frac{x^2+1}{2x} = \frac x 2 + \frac 1 {2x}.
$$
Notice that $\dfrac x 2$ goes up to $\infty$ as $x\to+\infty$ and $\dfrac 1 {2x}$ goes up to $+\infty$ as $x\downarrow0$, and the sum is positive everywhere in between, so the range of the restriction of this function to $x>0$ must be $[a,\infty)$ where $a$ is some positive number.  The difficulty of finding that number is typical of the reasons why it is often more work to find the range than the domain.
The algorithmic way of finding the number $a$ that is taught in the conventional curriculum requires calculus, and it doesn't seem unusual to see otherwise intelligent people rashly asserting that you need caluclus in order to do that.  Calculus, however, is sufficient but not necessary for the task.
$$
\overbrace{\frac x 2 + \frac 1 {2x} = \left( \frac x 2  - 1 + \frac 1 {2x} \right) + 1}^\text{This is a sort of completion of the square.} = \left( \sqrt{\frac x 2} - \frac 1 {\sqrt{2x}} \right)^2 + 1
$$
The square after the $\text{“}=\text{''}$ is never negative and is $0$ only when $x=1$.  Therefore the value of this expression is everywhere $\ge$ the value of the $1$ that follows the $\text{“}+\text{''}$.  Hence the range of the restriction of this function to $x>0$ is $[1,\infty)$.
One can do a similar thing when $x<0$.
