= Solution
The positive-domain graph starts with limiting value $f(0^+)=1$ and limiting slope $-1$, then decreases towards the horizontal asymptote $f=0$, with $f(q)\sim1/(2q)$ for large $q$. In fact direct <differentiation> gives
$$
f'(q)=-1+\left(1+\frac1q\right)e^{-1/q}<0,\qquad f''(q)=\frac{e^{-1/q}}{q^3}>0.
$$
The first inequality follows from $e^s>1+s$ for $s>0$. Thus the graph is strictly decreasing and convex upward, with $0<f(q)<1$ and no <stationary points>. The integral representation $f(q)=\int_0^1(1-e^{-s/q})\,ds$ also shows its positivity.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1-q12-mass-factor.png]
{title=The spherical <mass> factor $f(q)=1-q+qe^{-1/q}$ and its two limiting approximations}
These properties belong to the reusable <exponential spherical mass factor>.
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