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Are there any examples satisfy the continuous function $f:\Bbb R\rightarrow \Bbb R$ such that the image of the closed interval $[0,\infty)$ under $f$ is the open interval $(-1,1)$..like $f([0,\infty))=(-1,1)$?

A picture is also ok.

Christopher Toni
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1 Answers1

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Note that the function

$$g(x) = \frac{2}{\pi} \arctan{x}$$

is monotonically increasing, continuous, and tends to $1$ at infinity (the constant $2/\pi$ is included solely to make the range $[0, 1)$). It follows that something like

$$f(x) = \left(\frac 2 {\pi} \arctan{x}\right) \sin x$$

will have the desired property. It gets arbitrarily close to $\pm 1$, is continuous, but never reaches either.

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    Perhaps more simply, $\frac{x}{x+1} \sin{x}$ works for much the same reasons. – G. H. Faust Jan 24 '14 at 05:15
  • @G.H.Faust Indeed. In fact, if we choose $f$ to be any continuous function that tends to $1$ from below, and $g$ a continuous function such that $g^{-1}(1)$ and $g^{-1}(-1)$ are both unbounded, then $fg$ gives an example. –  Jan 24 '14 at 05:17