What is the range of the function $\frac{3}{2-x^2}$ I'm so, so very confused about finding the ranges of real functions, no concept in Mathematics has yet confused me more than this, please tell me what's wrong in my solution for finding the range of the function : $\dfrac{3}{2-x^2}$
Here's how I do it and get a partial answer, please check it out...
$x^2 \geq 0$

$-x^2 \leq 0$

$2 - x^2 \leq 2$

$\dfrac {1}{2 - x^2} \geq \dfrac{1}{2}$

So, $\dfrac {3}{2 - x^2} \geq \dfrac{3}{2}$

So, $f(x) \geq \dfrac{3}{2}$
By this, $Range(f) = [\dfrac{3}{2}, ∞)$
But as per my textbook, the answer is $(-∞,0)∪[\dfrac {3}{2},∞)$, which is (obviously) correct
What my main question here is : How can I add the proof of the negative values in the range in my proof? 
I would be very, very grateful to you if you help (no exaggeration, I would be so very thankful cause this topic is frustrating me)
Also, this is a general question : Am I the only one so confused about finding domains and ranges? I mean did you, when you began, also face problems with this concept?
Thanks

We have: $$y=f(x)=\frac{3}{2-x^2} ; x \ne \pm\sqrt{2}$$
Write $x$ as a function of $y$ as: $$x=\sqrt{2-\frac{3}{y}}$$
Now we have to find the set of values of $y$ for which $x$ is $\mathbf {real}$.
For this, $2-{3\over y} \ge 0$. This can be acquired in two ways: 
$\mathbf {Case\ 1:}$ Substituting $x=0$ into the second equation we get $y={3\over 2}$. Hence talking of positive $y$, it can go from $3\over 2$ all the way upto $\infty$, this is because no matter how large the value of $y$ you take, $x$ will always be a little less than $\sqrt 2$, which is safe. $y$ cannot be less than $3\over 2$ because that would mean $x$ is imaginary. $\mathbf {So},\ \mathbf {y\in [{3\over 2}, \infty)}$
$\mathbf {Case\ 2}:$ We note that from the above step, we get $x\in [0,\sqrt 2)$, when $y\in [{3\over 2},\infty)$. But a little observation shows us that if $y\lt 0$, the whole expression $(2-{3\over y})$ can and will take every value $\mathbf {between}$ $2$ and $\infty$, as $y$ goes from $-\infty$ to $0$ (but never equal to $0$), thus giving $x\in (\sqrt 2,\infty)$.
Thus we have $range(f)= (-\infty,0)\cup[{3\over 2}, \infty)$ that gives us real $x$. 
There is no need to care about the other possble values of $x$ as our main focus is $range(f)$ i.e. we need values of $y$ as expressed in the second equation.
