How do you construct a function that is continuous over $(0,1)$ whose image is the entire real line? How do you construct a continuous function over the interval $(0,1)$ whose image is the entire real line?
When I first saw this problem, I thought $\frac{1}{x(x-1)}$ might work since it is continuous on $(0,1)$, but when I graphed it, I saw that there is a minimum at $(1/2,4)$, so the image is $[4,\infty)$ and not $(-\infty,\infty)$.
Apparently, one answer to this question is:
$$\frac{2x-1}{x(x-1)}$$
But how is one supposed to arrive at this answer without using a  graphing calculator?

I'd like to provide a different flavor of function. For $x \in (0,1)$
$$f(x) = \tan\bigg(\pi x-\frac{\pi}{2}\bigg)$$

EDIT: How we get to above result?
We start with $\tan(x)$, which is continuous over $x \in (-\pi/2, \pi/2)$ whose image is the entire real line.
We want $x$ to be $(0,1)$ instead. So one easy way is we find a linear transformation (thus is continuous) of $x$ to transform $x$ from $(-\pi/2, \pi/2)$ to $(0,1)$.
Notice that if $\pi x \in (-\pi/2, \pi/2)$, then we have $x \in (-1/2, 1/2)$.
And if $(\pi x - \pi/2) \in (-\pi/2, \pi/2)$, then we have $x\in (0,1)$, and that is what we want.
