 Solution

A critical exponent specifies a leading singular power in a response or order parameter as the thermodynamic critical point is approached. Analytic backgrounds and the selected thermodynamic branch must be distinguished from that singular part. Write $r=at$, $a>0$, and first take $u>0$. The equation of state is $h=rM+u{M}^3$.

At zero field on the ordered branch, $M=(-r/u{)}^{1/2}$, giving the order-parameter critical exponent $\beta =1/2$. Differentiating the equation of state gives $\chi =(r+3u{M}^2{)}^{-1}$: above the transition $\chi =1/r$, below it $\chi =1/(2|r|)$, so the magnetic-susceptibility critical exponent is $\gamma =1$. At $r=0$, $M=\mathrm{sgn}(h)(|h|/u{)}^{1/3}$, giving the critical-isotherm exponent $\delta =3$.

The minimized singular free-energy density is $f_s=0$ above and $f_s=-r^2/(4u)$ below. Two temperature derivatives produce a finite specific-heat jump, corresponding to the heat-capacity critical exponent $\alpha =0$. The inverse quadratic fluctuation kernel is $K{p}^2+r+3u{M}^2$, so $\xi =\sqrt{K/(r+3u{M}^2)}$ gives $\nu =1/2$; the critical propagator proportional to $p^{-2}$ gives $\eta =0$. ThusThese mean-field critical exponents follow from minimization and linear response. More generally, linearize the renormalization group and differentiate its singular free-energy scaling form to obtain the actual fluctuation-controlled exponents.