## Task 3B - 331233B

For each integer $n$ with $n \geq 0$, let $f_{n}$ be the function defined by

$$
f_{n}(x)=x^{3}+(n+3) \cdot x^{2}+2 n \cdot x-\frac{n}{n+1}
$$

for all real $x$.

Determine all integers $n$ with $n \geq 0$ for which the following holds:

All roots of $f_{n}$ lie in an interval of length 3.
The function $f(x)$ behaves in principle as follows:

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2339.jpg?height=486&width=472&top_left_y=742&top_left_x=792)

This means that at the "outer" zeros $x_{\min}$ and $x_{\max}$, the slope is positive. For $n=0$, it is immediately clear that the zeros are $x_{\min}=-3$ and $x_{\max}=0$ (in this case, $x=0$ is a double zero). The interval between the smallest and largest zero (with "small" and "large" referring to the $x$-value) is thus exactly 3 in length.

For very large $n$, the function approximates to

$$
g(x)=x^{3}+(n+3) x^{2}+2 n x
$$

One solution is again $x=0$. The others are:

$$
\begin{gathered}
x^{2}+(n+3) x+2 n=0 \\
x=-\frac{1}{2}(n+3) \pm \frac{1}{2} \sqrt{(n+3)^{2}-8 n}=-\frac{1}{2}(n+3) \pm \frac{1}{2} \sqrt{n^{2}-2 n+9}
\end{gathered}
$$

For large $n$, one can approximate:

$$
\begin{gathered}
x_{1} \approx-\frac{1}{2}(n+3)-\frac{1}{2}(n-1)=-n-1 \\
x_{2} \approx-\frac{1}{2}(n+3)+\frac{1}{2}(n-1)=-2
\end{gathered}
$$

It is clear that the approximate solution $x \approx -2$ lies between the other two zeros, because the largest zero must be $x_{\max} > 0$ since $f(0) = -\frac{n}{n+1} < 0$.

From this, it follows that

$$
x_{\min} < -2 < x_{\max}
$$

Since $x_{\max} > 0$, the interval width $d$ is at least

$$
d = x_{\max} - x_{\min} > n + 1
$$

Thus, for $n \geq 2$, the allowed interval width is exceeded. It remains to check whether the interval width is maintained for $n=1$. We do this by simply substituting $n=1$ and $x=-3$. It turns out that $f_{1}(-3) = \frac{5}{2} > 0$. This means that for $n=1$, the smallest zero $x_{\min} < -3$, so the interval width of 3 is not maintained for $n=1$ either. Therefore, $n=0$ is the only solution.