1. Let the polynomial $f(x)$ satisfy: for any $x \in \mathbf{R}$, we have
$$
f(x+1)+f(x-1)=2 x^{2}-4 x .
$$

Then the minimum value of $f(x)$ is $\qquad$
- 1. - -2 .

Since $f(x)$ is a polynomial, we know that $f(x+1)$ and $f(x-1)$ have the same degree as $f(x)$. Therefore, $f(x)$ is a quadratic polynomial (let's assume $f(x) = ax^2 + bx + c$). Then,
$$
\begin{array}{l}
f(x+1) + f(x-1) \\
= 2ax^2 + 2bx + 2(a + c) = 2x^2 - 4x.
\end{array}
$$

Thus, $a = 1, b = -2, c = -1$.
Hence, $f(x) = x^2 - 2x - 1 = (x-1)^2 - 2 \geqslant -2$.