= Solution
The stationary points of the <Landau free energy> obey
$$
f'(m)=2m(r+bm^2+cm^4)=0.
$$
For $b>0$, $r\geq0$ makes every term of $f$ nonnegative, so $m=0$ is the unique <global minimum>. For $r<0$, its curvature $f''(0)=2r$ is negative. Setting $x=m^2$ gives $cx^2+bx+r=0$, whose unique positive root is
$$
x_+=\frac{-b+\sqrt{b^2-4cr}}{2c}.
$$
At a nonzero stationary point,
$$
f''(m)=4x(b+2cx),\qquad f(m)=-\frac b2x^2-\frac{2c}{3}x^3.
$$
Thus both $m=\pm\sqrt{x_+}$ are stable and have lower <free energy> than $m=0$. The <equilibrium magnetization> is
$$
\boxed{m_{\rm eq}=\begin{cases}
0,&T\geq T_c,\\
\displaystyle\pm\left[\frac{-b+\sqrt{b^2+4ac(T_c-T)}}{2c}\right]^{1/2},&T<T_c.
\end{cases}}
$$
For fixed $b>0$ and $T\to T_c^-$, expand the square root:
$$
m_{\rm eq}^2=\frac{a(T_c-T)}b+O((T_c-T)^2).
$$
It tends continuously to zero. Hence the <continuous phase transition> has the <order-parameter critical exponent>
$$
\boxed{|m_{\rm eq}|\sim\sqrt{\frac ab}(T_c-T)^{1/2},\qquad\beta=\frac12.}
$$
This is the <mean-field critical exponent>; fluctuation corrections beyond <Landau theory> are not included.
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