Finding the range of $\frac{x^2}{x^2-9}$ I am a student who is studying about functions (only basic ones) and was practicing till I found this -
Finding the range of 
$$ \frac{x^2}{x^2-9} $$ 
Since I have only learnt the basics , I can only play around with the numbers and not use limits (which was what I found online). 
I was Told to try this method -  to try to assign $y=\frac{x^2}{x^2-9}$ 
This means expressing $x$ in terms of $y$ 
$x = \sqrt { \frac{9y}{y-1} } $
Now I then go on to find the range of this function - 
range =$ y \leq 0  $ or $ y>1$ 
This is simple to do. 
But what I'm confused with is this - 
Since Range is all the possible 'y' values obtainable from the domain, then this confuses me because I express x in terms of y. Finding the x value feels like finding the 'domain' for me . 
I believe I'm having a conceptual problem and I don't understand what does expressing x in terms of y do to help me find the range. does this mean that I can express all other simple functions into x in terms of y to find the range too ? 
Thanks !! 
Note : I'm being taught on How to read the domain And range off a function. So I can't use the graphical method . 
 A: The domain is the set of $x$ values such that the function is defined. Here the expression exists for any $x\notin \{-3,3\}$.
The range is the set of $y$ values for which the equation $y=f(x)$ has solutions.
Indeed, 
$$y=\frac{x^2}{x^2-9}$$ can be written 
$$x^2=\frac{9y}{y-1}$$ and the RHS is only non-negative and defined for $y\notin(0,1]$.
Remember: the domain is for $x$, the range for $y$.
Alternatively, you can study the variations of $f$. Canceling the derivative,
$$f'(x)=-\frac{18x}{(x^2-9)^2}$$ and noting the two vertical asymptotes and the horizontal one, the table of variations is
$$\begin{matrix}x&-\infty&&&-3&&&0&&&3&&&\infty\\\hline f(x)&1&\nearrow&\infty&|&-\infty&\nearrow&0&\searrow&-\infty&|&\infty&\searrow&1\end{matrix}$$
Then all values are reached, except for $(0,1]$.
A: Since you already have $$x = \sqrt { \frac{9y}{y-1} }$$
you can see that $$\sqrt { \frac{9y}{y-1} }$$ exists iff $$\frac{y}{y-1}\ge0~~~ \&~~ y\neq 1 \Longleftrightarrow y(y-1)\ge 0 ~~~ \&~~ y\neq 1 \Longleftrightarrow y\in\Bbb R\setminus(0,1]$$
That is your range is $$\color{blue}{\Bbb R\setminus(0,1] = (-\infty, 0]\cup(1,\infty)}$$
A: The range is the domain of the inverse function (given that it's a bijection).
$f(x)= y$
$f^{-1}  (f(x)) = f^{-1}(y)$
$x=f^{-1}(y)=g(y)$
$x=g(y)$
That's why you're expressing $x$ in terms of $y$. Finding values for the domain of $g$ is equivalent to finding the range of $f$. 

Take a look at $y = x^2+1$
Domain = $\mathbb R$
What about range? We can inspect and see that $y\ge0$
OR
Find the inverse and check it's domain.
$x =\pm \sqrt{y-1}$
$\therefore \space$ Range = $[1,\infty ]$

You can express many functions in terms of others. But sometimes it's really not worth it ($f(x)=x^2+2x+1$) and sometimes it's too time consuming ($f(x)=x^5+x^4+3x^2-10$). And sometimes there's not a closed form ($f(x)=x+e^x$). So, yeah. Sometimes you can. Sometimes you can't.
A: Hint:
What if $x^2=0?$
Else let $y=\dfrac9{x^2-9}+1$
$0<x^2<\infty\iff-9<x^2-9<\infty$
When $0<x^2-9<\infty,0<\dfrac9{x^2-9}<\infty\implies?<y<?$
When $-9<x^2-9<0,0>\dfrac1{x^2-9}>-\dfrac19\iff0>\dfrac9{x^2-9}>-1$
A: A function is a special case of a relation, namely an injective relation (let's call it $r$). That is,
$$r(x,y) \subset \mathbb R^2$$
$y={x^2 \over x^2-9}$ clearly is a relation. You can always substitute two real numbers ($x$ and $y$) and find out if they are related in that way. To check if it is a function, apply the definition of injectivity.
$$\forall x,y_1,y_2:r(x,y_1) \wedge r(x,y_2) \implies y_1=y_2$$
(read: for all $x$ there exists exactly one $y$ so that $x$ and $y$ are related by $r$). Substituting $r$ by the given definition, this gives
$$\forall x,y_1,y_2= \implies {x^2 \over x^2-9}={x^2 \over x^2-9}
$$
This is obviously true, as for any relation of the form $r(x,y): y=...$ or $f(y) = ...$ ("function"). So our $r$ is a function. Let's name it $f$ therefore. Now, the range is all possible $x$, the $domain$ is all possible $y$. You're right that every $x$ can be related (is in relation to) some $y$. Therefore, the $domain$ (in some contexts it is called "support" - you can probably see, why) simply is $\mathbb R$ (the left member of any given pair $(x,y)$). The question is simply about understanding the definition. The $range$ of the function is more interesting, which is why it was asked for.
I'd proceed like this:


*

*The numerator is always positive. Therefore, the sign of the expression is the sign of the denominator. That is: $\forall |x|>\sqrt 9 = 3: f(x)>0$, $\forall |x|<3:f(x)<0)$.

*$|x|$ approaching $3$, the denominator gets arbitrary small, in contrast to the numerator, which stays close to $9$. So $f(x)$ rises indefinitely whenevery this is the case. It also can take any sign.

*There's no $x$ so that $f(x)=0$


So the range is $\mathbb R \setminus \{0\}$. I tried to make a plot of this, however, "gnuplot" decided to pick fancy arbitrary colors (apparently checkered red and blue on the screen, which translated to green in a file, and transparent, which made the plot a little hard to read), so I leave this as an excercise to the reader.
A: A bit different:
$y=\dfrac{x^2}{x^2-9}$,  $x$ real.
Note : $z:= x^2 \ge 0.$
Let's look at:
$F(z):= \dfrac{z}{z-9},$ $ z \ge 0,$ and find its range.
$0 \le z \lt 9:$   $F(z) \in (-\infty, 0].$
$9<z $: $F(z) \in (1,\infty).$
Can you see why?
