The points on the graph of the function $x \mapsto 2 x^{3}-2 x$ defined on the interval $[-1,1]$ that are farthest apart are which two points?
Let's delve into the arithmetic and geometric means inequality written for the numbers $x^{2}, 1-x^{2}$ (where $x^{2} \geq 0$, $1-x^{2} \geq 0$):

$$
\frac{1}{2}=\frac{x^{2}+\left(1-x^{2}\right)}{2} \geq \sqrt{x^{2}\left(1-x^{2}\right)}=|x| \sqrt{1-x^{2}}
$$

Therefore,

$$
\sqrt{1-x^{2}} \geq 2|x|\left(1-x^{2}\right)=2\left|x\left(1-x^{2}\right)\right|=|f(x)|
$$

(1) precisely means that the curve of $f(x)$ on the interval $[-1,1]$ is contained within the unit circle centered at the origin. Since the distance between any two points on a closed disk is at most the diameter of the circle, the distance between any two points on the curve of $f$ is at most 2. This can be achieved, for example, with the points $A=(-1 ; 0)$ and $D=(1 ; 0)$.

To actually achieve this bound, both points must lie on the circumference of the circle, and they must be opposite each other. A different solution can be obtained if $x \neq \pm 1$ (and $\sqrt{1-x^{2}} \neq 0$).

In (1), equality holds precisely when equality holds in the arithmetic-geometric mean inequality, i.e., $x^{2}=1-x^{2}$, so $x= \pm \frac{1}{\sqrt{2}}$. This gives two points: $B=\left(-\frac{1}{\sqrt{2}} ; \frac{1}{\sqrt{2}}\right)$ and $C=\left(\frac{1}{\sqrt{2}} ;-\frac{1}{\sqrt{2}}\right)$. These points indeed lie on the circle opposite each other and are also on the curve of $f$. Thus, we have two solutions: the pairs of points $A, D$ and $B$, $C$.

Based on the work of Péter Szabó (Fazekas M. Főv. Gyak. Gimn., 12th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_0046d50b53c5d91e04b8g-1.jpg?height=510&width=616&top_left_y=946&top_left_x=749)