11.3. Find the maximum length of a horizontal segment with endpoints on the graph of the function $y=x^{3}-x$
Answer: 2.

Solution 1. A horizontal segment of length $a>0$ with endpoints on the graph of the function $y=x^{3}-x$ exists if and only if the equation $(x+a)^{3}-(x+a)=x^{3}-x$ has at least one solution for the given value of the parameter $a$. Expanding the brackets, combining like terms, and dividing by $a>0$, we obtain the quadratic equation $3 x^{2}+3 a x+a^{2}-1=0$, which is solvable when $D=12-3 a^{2} \geq 0$, from which $0<a \leq 2$. Therefore, the length of the desired segment does not exceed 2. When $a=2$, the solution to the equation is $x=-1$, from which it follows that the length 2 is achieved for the segment with endpoints $(-1,0)$ and $(1,0)$ on the graph of the function $y=x^{3}-x$.

## Solution 2.

As in Solution 1, we obtain the equation $3 x^{2}+3 a x+a^{2}-1=0$, which we consider as a quadratic equation in $a$ with parameter $x: a^{2}+3 x a+3 x^{2}-1=0$. We find its roots $a_{1,2}=\frac{-3 x \pm \sqrt{4-3 x^{2}}}{2}$, and since $a$ is positive, we consider only the one with the plus sign: $a=\frac{\sqrt{4-3 x^{2}}-3 x}{2}$. This function of $x$ is defined for $|x| \leq \frac{2}{\sqrt{3}}$ and is positive for $-\frac{2}{\sqrt{3}} \leq x \leq \frac{1}{\sqrt{3}}$. Its derivative, equal to
$a^{\prime}(x)=-\frac{3 x+3 \sqrt{4-3 x^{2}}}{2 \sqrt{4-3 x^{2}}}$, is zero at $x=-1$, is greater than zero to the left, and less than zero to the right. Therefore, its value is maximal at $x=-1$ and equals $a_{\max }=2$. Indeed, in this case, the segment of length 2 connects on the x-axis the two roots $x_{1}=-1$ and $x_{2}=1$ of the equation $x^{3}-x=0$.

Grading criteria. The answer 2 and an example of a segment of such length are provided: 1 point. Lack of an explicit example in the solution: minus 2 points.