Find the range of $f(x)=x/(3x^2−3x)$ 
Find the range of $f(x)=x/(3x^2−3x)$

From my computations, the range I get is all reals except for -1/3. However, after I looking at the graph of the function on google, it looks as if the range is all reals except for 0. I've been working on this for hours but I don't know what I did wrong.
 A: So we have the function $f(x)=\frac{x}{3x^2-3x}=\frac{1}{3x-3}$. Now set this equal to $y$/ That is,
$$\frac{1}{3x-3}=y$$ This will boil down to $$x=\frac{1}{3y}+{1}$$ You can see that this function is defined for all value except $0$. So the range of the function will be $\Bbb{R}-\{0\}$
A: Assuming $x\ne 0$ where $f(x)$ isn't defined we get
$$\frac{x}{3x^2-3x}=\frac{1}{3}\frac{1}{x-1}$$
the latest function is defined everywhere but at $x=1$.
Thus, the graph of $f(x)$ is a shifted version of $\frac{1}{x}$ ( just around $x=1$ instead of $x=0$) and also with a hole at $x=0$.
This determine the range.
A: When you knock out the $x$ from top and bottom, you essentially have a hyperbola with the x-axis as the horizontal asymptote. Therefore $y=0$ is out of your range. Next, there is a hole in the graph for $x=0$. The hole is located at $(0,-1/3)$ So this y-value also needs to be taken out of the range. So there are two values that are not in the range. Can you set up your range now?
A: You are probably looking for the horizontal asymptote of your function where $x \to \infty$:
$$
\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x}{3x^2-3x} = \lim_{x \to \infty} \frac{1}{3(x-1)} = 0
$$
That's how you get that the range is all $x \neq 0$.
