 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, isOne can also see the full range from continuity and monotonicity: $f$ is a continuous function with $f^{\prime }(s)=-(2s+1)/(s^2+s{)}^2<0$, and its endpoint limits are $+\infty$ as $s\to 0^{+}$ and $0$ as $s\to \infty$. The endpoints themselves are not attained.