 Solution

Put $t=T-T_c$, so $a_4\sim t$. At zero field the stationary equation factors asFor $t>0$, $m=0$ is the unique minimum. For $t<0$, it is unstable and the two minima areThe order parameter therefore tends continuously to zero, and its order-parameter critical exponent is $\boxed{\beta =1/2}$.

At either ordered minimum the singular free-energy density iswhereas it is zero for $t>0$. Two temperature derivatives give a singular heat capacity proportional to $|t|$ below the transition and zero above it, so the heat-capacity critical exponent is $\boxed{\alpha =-1}$.

The inverse magnetic susceptibility at a stable minimum is the curvature ${\chi }^{-1}={\partial }^{2}f/\partial m^2$. Below $T_c$,and hence ${\chi }_{-}\sim |t{|}^{-2}$ and $\boxed{{\gamma }_{-}=2}$. Above $T_c$, however, the curvature at $m=0$ vanishes for every $t>0$. Indeed, at small field $B=4a_4{m}^3+O(m^5)$, so $m\sim (B/(4a_4){)}^{1/3}$ and the linear susceptibility is already infinite away from the critical point. Consequently the usual magnetic-susceptibility critical exponent ${\gamma }_{+}$ is not defined for this exceptional free energy; assigning it a finite value would incorrectly assume a quadratic term.

At $T=T_c$, the equation of state is $B=6a_6{m}^5$, so $m\sim |B{|}^{1/5}$ and the critical-isotherm exponent is $\boxed{\delta =5}$. Thus the transition is continuous, although its missing quadratic term makes the high-temperature linear response singular throughout that phase.