= Solution
A suitable <continuous function> is
$$
\boxed{f(x)=\frac{\sin(1/x)}x\qquad(0<x\leq1).}
$$
For $t_k=\pi/2+2\pi k$ and $u_k=3\pi/2+2\pi k$,
$$
f(1/t_k)=t_k\longrightarrow+\infty,\qquad
f(1/u_k)=-u_k\longrightarrow-\infty.
$$
All these arguments lie in $(0,1]$. Its image is a <real interval> by the <continuous image of a real interval> result, and it is unbounded both above and below. Given any real $y$, choose two of the displayed points with function values on opposite sides of $y$, then apply the <intermediate value theorem> on the segment joining them. Hence \b[the image is all of $\mathbb R$]. A monotone example would not suffice here; the oscillation gives both directions of unboundedness near the same missing endpoint.
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