Range $R=[0,1]$ of function with domain $(0,1)$

Question

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.

Answer 1

To be honest the phrasing of the question makes very little sense. Let me try to answer what I think you are asking:

1) Yes there exists such function. Take for example $f(x)=\frac{1}{2}(1+\sin(2πx))$.

2) Continuity of $g$ doesn't guarantee that a function has that property. Indeed $g(x)=1$ doesn't have that property.

Answer 2

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.