= Solution
For $b<0$ at $T=T_c$, the quartic term lowers the <free energy> near $m=0$, while the positive sextic term stabilizes it at larger $|m|$. The nonzero stationary points have
$$
\boxed{m^2=-\frac bc,\qquad f(m)=\frac{b^3}{6c^2}<0.}
$$
They are <global minima>. The stationary point $m=0$ has zero quadratic curvature but is a flat local maximum because its leading nonzero term is negative. Thus the system is already ordered at $T_c$, with a finite <magnetization>.
More explicitly, a continuous onset from $m=0$ could occur only where $f''(0)=2r$ vanishes, namely at $r=0$. But the global minima there are the finite values above. For sufficiently large positive $r$, only $m=0$ is stable. Therefore the equilibrium minimum must change at some positive $r$ by crossing between a zero and a finite nonzero <order parameter>. This is a <first-order phase transition>. Its actual temperature lies above $T_c$ and is calculated next; $T_c$ itself is the disordered <spinodal point>, not the coexistence temperature.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2019-q4-landau-phase-diagram.png]
{title=Free-energy coexistence and the continuous and first-order Landau transition lines}
{description=Original plots. The left panel uses $b=-1,c=1$ at $r=0$, at coexistence $r=3/16$, and above coexistence $r=0.24$. The right panel uses $a=c=1$ and shows the equilibrium phase boundaries in the $(T-T_c,b)$ plane. Dashed gray curves are spinodal limits, not additional equilibrium transitions.}
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