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The fundamental theorem of calculus gives $y^{\prime }(x)=e^{-x^2}>0$, so the graph is strictly increasing. Its second derivative is $y^{\prime \prime }(x)=-2x{e}^{-x^2}$: it curves upward for $x<0$ and downward for $x>0$, with an inflection point at zero.

By the Gaussian integral and the evenness of its integrand,Thus the only axis intersection is $(0,\sqrt{\pi }/2)$; there is no finite $x$-axis intersection because the integral is strictly positive for every finite $x$. The error function gives $y(x)=\frac{\sqrt{\pi }}{2}(1+\mathrm{erf}x)$.

Differentiate the flux:On $-1<z_0<1$, $\Phi$ is strictly increasing. Its limiting endpoint values are $\Phi (-1)=0$ and $\Phi (1)=4\pi$, and $\Phi (0)=2\pi$. It has horizontal limiting tangents at the endpoints, curves upward for $z_0<0$, curves downward for $z_0>0$, and has an inflection point at zero.

The derivative graph is a downward-opening parabola, approaching zero at both endpoints and attaining its maximum $3\pi$ at $z_0=0$.

In a symmetry-plane cross-section, write $\mathbf{p}=r(\theta )\mathbf{n}(\theta )$, $v=1/r$, and $\mathbf{t}=\mathbf{n}\prime$. The elastic wave surface is $\mathbf{g}=v\mathbf{n}+v\prime \mathbf{t}$. Its derivative vanishes when $v+v\prime \prime =(r^2+2r{\prime }^2-rr\prime \prime )/r^3=0$, exactly an inflection point of the regular slowness surface. At a simple zero the second and third derivatives of $\mathbf{g}$ are linearly independent, giving an ordinary cusp. Smooth nonconvex slowness surface sheets can therefore generate cusped wavefronts.

The liquid-gas critical point of the Van der Waals equation occurs where its two subcritical stationary points merge into a horizontal inflection point. The conditions are $(\partial p/\partial v{)}_T=({\partial }^{2}p/\partial v^2{)}_T=0$. The short calculation isDividing these relations gives $2/(v_c-b)=3/v_c$, henceSubstitution gives the Van der Waals law of corresponding states and the parameter-independent ratio $p_c{v}_c/(k_B{T}_c)=3/8$. That ratio belongs to this equation of state; reducing a real fluid's variables does not enforce it.^[ref]

An exact rearrangement of the reduced critical Van der Waals isotherm makes its flatness visible:The first nonzero departure near $x=1$ is $-3(x-1{)}^3/2$. On the critical particle volume approached from $t>1$, $(\partial q/\partial x{)}_{x=1}=-6(t-1)$ andThe divergent isothermal compressibility and cubic flattening explain why the critical region is mechanically soft. These are mean-field theory predictions; thermal fluctuations limit their use for real fluids very close to a thermodynamic critical point.