= Solution
Let $\omega=\sqrt{k/m}$, $b=\sqrt{\hbar/(m\omega)}$ and $\xi=x/b$. With $\psi_0=\pi^{-1/4}b^{-1/2}e^{-\xi^2/2}$, the next two normalized harmonic-oscillator <wavefunctions> are $\psi_1=\sqrt2\xi\psi_0$ and $\psi_2=(2\xi^2-1)\psi_0/\sqrt2$. Their energies are $E_n=(n+1/2)\hbar\omega$, with respectively zero, one and two <wavefunction nodes>, alternating even and odd <parity>. The <Schrödinger equation> gives
$$
\psi_n''(x)=b^{-2}[\xi^2-(2n+1)]\psi_n(x).
$$
For <harmonic-oscillator wavefunction inflections>, \b[the inflections include both <classical turning points> and simple <wavefunction> <wavefunction nodes>]. In dimensionless coordinates the complete sets are
$$
\boxed{n=0:\ \xi=\pm1;\qquad
n=1:\ \xi=0,\pm\sqrt3;\qquad
n=2:\ \xi=\pm1/\sqrt2,\pm\sqrt5}.
$$
The curvature changes sign at every listed point; none of these states has a <wavefunction node> coinciding with its turning points.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2012/paper-1-oscillator-inflections.png]
{title=Original lowest harmonic-oscillator wavefunctions with turning-point and node inflections distinguished}
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