7. Find all positive real numbers $c$ such that the graph of $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=$ $x^{3}-c x$ has the property that the circle of curvature at any local extremum is centered at a point on the $x$-axis.
Answer: $\frac{\sqrt{3}}{2}$
Solution: The equation $0=f^{\prime}(x)=3 x^{2}-c$ has two real roots: $\pm \sqrt{c / 3}$. Let $a:=\sqrt{c / 3}$. As $f^{\prime \prime}(-a)=-6 \sqrt{c / 3}<0, f$ has a unique local maximum at $x=-a$.

Because $f$ has half-turn symmetry about the origin, it suffices to consider this local extremum. The radius of curvature at any local extremum is
$$
r(x)=\frac{1}{\left|f^{\prime \prime}(x)\right|}=\frac{1}{6|x|},
$$
so the condition in the problem is equivalent to
$$
\begin{array}{c}
r(-a)=f(-a) \\
\frac{1}{6 a}=-a\left(a^{2}-c\right) \\
1=6 a^{2}\left(c-a^{2}\right)=2 c(2 c / 3) \\
c=\sqrt{3} / 2 .
\end{array}
$$