Range $R=[0,1]$ of function with domain $(0,1)$ Is there a function g continuous in the domain $(0,1)$ with range $R=[0,1]$.
Explain the answer.
This is a first semester calculus question, therefore I am curious about the depth someone has to reach to prove this one.
I believe that simple stating that for any function f that in domain $(0,1)$  $$\lim_{x\to z^-} f(x)= \lim_{x\to z^+} f(x) =f(z)$$
is true $\forall z\in (0,1)$ because the Range has no discontinuities is not enough.

The simplest continuous function with bounded range familiar to all students of Calc 1 is $f(x) = \sin x$. Its domain is the line and its range is $[-1,1]$.  The function $g(x) = \frac 12 (\sin x + 1)$ is continuous, its domain is the line, and its range is $[0,1]$.
To get a continuous function on $(0,1)$ whose range is $[0,1]$ just boost the periods up a bit. Try $g(x) = \frac 12 (\sin 100x + 1)$, for instance.
