Domain and range of $g(x) = \frac{3x^2+4}{x-2}$ 
$$g(x) = \frac{3x^2+4}{x-2}$$

for the domain, I find, $\{x|x\ne 2\}$'
and for the range
$y = \dfrac{3x^2+4}{x-2}$ and when $x=2-\epsilon$, $y \to \dfrac{3(2-\epsilon)^2 + 4}{(2-\epsilon)-2} = \dfrac{K}{-\epsilon} \to -\infty$
When $x = 2+\epsilon$, $y \to \dfrac{3(2+\epsilon)^2+4}{(2+\epsilon)-2} = \dfrac{K}{\epsilon} \to \infty$
Therefore, $\text{ran}(g(x)) = (-\infty, \infty)$ 
have I done this right or am I missing something? For instance, I thought about the case when $x \to \infty$, but this looks like $y \to 1$ in this case, nothing remarkable. I don't even know if that is what we say when we have an $\frac{\infty^2}{\infty}$....Do we just say that is one? Or do we say that is $\infty$ still?

The domain is $$\mathbb R\setminus\{2\}.$$
Now, let $x>2$.
Thus, by AM-GM
$$\frac{3x^2+4}{x-2}=\frac{3(x-2)^2+12(x-2)+16}{x-2}=3(x-2)+\frac{16}{x-2}+12\geq$$
$$\geq2\sqrt{3(x-2)\cdot\frac{16}{x-2}}+12=12+8\sqrt{3}.$$
The equality occurs for $3(x-2)=\frac{16}{x-2}$ or $x=2+\frac{4}{\sqrt3}.$
Also, let $x<2$.
Thus, by AM-GM again we obtain:
$$\frac{3x^2+4}{x-2}=\frac{3(x-2)^2+12(x-2)+16}{x-2}=3(x-2)+\frac{16}{x-2}+12=$$
$$=12-\left(3(2-x)+\frac{16}{2-x}\right)\leq12-2\sqrt{3(2-x)\cdot\frac{16}{2-x}}=12-8\sqrt{3}.$$
The equality occurs for $3(2-x)=\frac{16}{2-x}$ or $x=2-\frac{4}{\sqrt3}$ and since $g$ is a continuous function on any interval of the domain and
$$\lim_{x\rightarrow+\infty}g(x)=+\infty$$ and
$$\lim_{x\rightarrow-\infty}g(x)=-\infty,$$ we got the following range:
$$\left(-\infty,12-8\sqrt3\right]\cup\left[12+8\sqrt3,+\infty\right).$$
Because $$3x^2+4=3x^2-12x+12+12x-24+16=$$
$$=3(x^2-4x+4)+12(x-2)+16=3(x-2)^2+12(x-2)+16.$$
I used AM-GM:

For non-negatives $a$ and $b$ we have:
  $$\frac{a+b}{2}\geq\sqrt{ab}.$$
  The equality occurs for $a=b$.

For $x>2$ we have $a=3(x-2)$ and $b=\frac{16}{x-2}.$
