= Solution
For a <fixed point> $g^*$ with $\beta(g^*)=0$, put $g=g^*+h$. To first order in the displacement,
$$
\frac{dh}{d\log\mu}=\beta'(g^*)h,\qquad
h(\mu)=h(\mu_s)\left(\frac\mu{\mu_s}\right)^{\beta'(g^*)}.
$$
Hence $\beta'(g^*)<0$ means ultraviolet attraction, since $h\to0$ as $\mu\to\infty$; $\beta'(g^*)>0$ means infrared attraction, since $h\to0$ as $\mu\to0$. If the derivative vanishes, the linear test is inconclusive and higher terms determine the direction of flow.
For $\delta=4-D>0$ and $b=3/(16\pi^2)$, the retained <beta function> is $-\delta g+bg^2$. Its two zeros and slopes are
$$
\begin{array}{c|c}
g&\beta'(g)\\\hline
0&-\delta\\
\delta/b&+\delta
\end{array}.
$$
Thus the positive interacting <Wilson-Fisher fixed point> is
$$
\boxed{g^*=(4-D)\frac{16\pi^2}{3},\qquad \beta'(g^*)=4-D>0,}
$$
and is infrared attractive in the coupling direction. Couplings between zero and $g^*$ increase towards $g^*$ when the scale decreases; those just above $g^*$ decrease towards it. The Gaussian <fixed point> $g=0$ is ultraviolet attractive for $D<4$.
The conclusion is local and perturbative. For $4-D\ll1$, $bg^*=4-D$ is small, making the expansion controlled; extrapolation far below four dimensions requires more information. It is also a statement about the displayed coupling flow, not stability against every possible deformation of the full theory. In particular, the <mass term> is a relevant perturbation of the critical theory and must be tuned to reach the massless critical surface. Calling this fixed point infrared stable must not erase that additional relevant direction.
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