= Solution
For $s>0$, the denominator $s(s+1)$ is positive, so every value of $f$ is positive. Conversely, for every $r>0$, the positive number $s=(\sqrt{1+4/r}-1)/2$ satisfies $f(s)=r$. Thus its <image of a function>, or range, is
$$
\boxed{f((0,\infty))=(0,\infty).}
$$
One can also see the full range from <continuity> and <monotonicity>: $f$ is a <continuous function> with $f'(s)=-(2s+1)/(s^2+s)^2<0$, and its endpoint <limits> are $+\infty$ as $s\to0^+$ and $0$ as $s\to\infty$. The endpoints themselves are not attained.
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