Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 10D / c

No such continuous function exists. The extreme value theorem makes every continuous function on the closed interval $[0,1]$ bounded and gives it a finite maximum. Its image therefore cannot equal the unbounded real interval $[0,\infty )$.

The continuous functionhas the required image. Its values are nonnegative, and for every $y\ge 0$ the point $x=1/(y+1)\in (0,1]$ satisfies $f(x)=y$. Thus the image is exactly $[0,\infty )$. The missing left endpoint permits unboundedness without a failure of continuity on the domain.

A suitable continuous function isFor $t_k=\pi /2+2\pi k$ and $u_k=3\pi /2+2\pi k$,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 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.