Let $n$ be a given positive integer. Determine the minimum of the polynomial $f(x)=x^{2 n}+2 x^{2 n-1}+3 x^{2 n-2}+\cdots+(2 n+1-k) x^{k}+\cdots+2 n x+(2 n+1)$ defined on the real numbers.
I. solution. The polynomial can be transformed as follows:

$$
\begin{gathered}
\left(x^{n}+x^{n-1}\right)^{2}+2\left(x^{n-1}+x^{n-2}\right)^{2}+\cdots+(n-k)\left(x^{k+1}+x^{k}\right)^{2}+\cdots+n(x+1)^{2}+n+1= \\
\quad=(x+1)^{2}\left(x^{2 n-2}+2 x^{2 n-4}+\cdots+(n-k) x^{2 k}+\cdots+(n-1) x^{2}+n\right)+n+1
\end{gathered}
$$

In the first form, $x^{2 k+1}$ only arises from the written term, with a coefficient of $2(n-k) ; x^{2 k}$ arises from the written term and the following one if $k \geq 1$, so its coefficient is $n-k+n+1-k=2 n+1-2 k$. Finally, the constant term is $n+n+1=2 n+1$. Thus, we have indeed rewritten the polynomial $f$ in another form. The correctness of the second transformation is evident.

From the obtained form, it is clear that $f(-1)=n+1$, and elsewhere the value of $f$ is greater than this. Therefore, the sought minimum is $n+1$, which the polynomial attains at $-1$.

II. solution. We prove by induction that the value of the polynomial at $-1$ is $n+1$, and elsewhere it is greater. Denote the polynomial by $f_{n}$.

$$
f_{1}(x)=x^{2}+2 x+3=(x+1)^{2}+2 \geq 2
$$

From this, it is clear that the statement is true for $n=1$.

Assume that the statement is true for some value $k-1$ of $n$. From the obvious

$$
f_{k}(x)=f_{k-1}(x) x^{2}+2 k x+2 k+1=\left(f_{k-1}(x)-k\right) x^{2}+k(x+1)^{2}+k+1
$$

we get that $f_{k}(-1)=k+1$, and at any other point, the second term on the right side, as well as the first term by the assumption, is positive.

Thus, the correctness of the statement is inherited from $k-1$ to $k$. Therefore, the statement is true for all $n$.

Remark. We can also perform the inductive proof based on the transformation

$$
\begin{gathered}
f_{n}(x)=f_{n-1}(x)+x^{2 n}+2 x^{2 n-1}+2\left(x^{2 n-2}+x^{2 n-3}+\cdots+x+1\right)= \\
=f_{n-1}(x)+1+x^{2 n}-1+2\left(\left(x^{2 n-1}+x^{2 n-2}\right)+\left(x^{2 n-3}+x^{2 n-4}\right)+\cdots+(x+1)\right)= \\
=f_{n-1}(x)+1+(x+1)((x-1)+2)\left(x^{2 n-2}+x^{2 n-4}+\cdots+x^{2}+1\right)= \\
f_{n-1}(x)+1+(x+1)^{2}\left(x^{2 n-2}+x^{2 n-4}+\cdots+x^{2}+1\right)
\end{gathered}
$$

III. solution. We can solve the problem using the derivative as well. It is clear that for non-negative $x$ values, the polynomial is positive, and it increases with increasing $x$. For negative $x$ values, we write the polynomial in another form.

$$
x f(x)-f(x)=x^{2 n+1}+x^{2 n}+x^{2 n-1}+\cdots+x-2 n-1=\frac{x^{2 n+2}-1}{x-1}-(2 n+2)
$$

From this, we get

$$
f(x)=\frac{x^{2 n+2}-(2 n+2) x+2 n+1}{(x-1)^{2}}
$$

We find the derivative as the product of the numerator and the $\frac{1}{(x-1)^{2}}$ function:

$$
\begin{gathered}
f^{\prime}(x)=\frac{(2 n+2) x^{2 n+1}-(2 n+2)}{(x-1)^{2}}-\frac{2 x^{2 n+2}-(2 n+2) x+2 n+1}{(x-1)^{3}}= \\
=\frac{2 n\left(x^{2 n+2}-1\right)-(2 n+2)\left(x^{2 n+1}-x\right)}{(x-1)^{3}}= \\
=2(x+1) \frac{n\left(x^{2 n}+x^{2 n-2}+\cdots+x^{2}+1\right)-(n+1) x\left(x^{2 n-2}+x^{2 n-4}+\cdots+x^{2}+1\right)}{(x-1)^{2}}
\end{gathered}
$$

Here, for negative $x$, both terms of the numerator and the denominator are positive, and the factor $x+1$ changes from negative to positive at $x=-1$. Therefore, the function attains its minimum at $x=-1$. The value of the minimum (e.g., from the last form of $f(x)$) is $n+1$.