7. The minimum value of the function $f(x)=\frac{x^{2}}{8}+x \cos x+\cos (2 x)(x \in$ $\mathbf{R}$ ) is $\qquad$
7. -1 .

Solution: Since $f(x)=\frac{x^{2}}{8}+x \cos x+\cos (2 x)=\frac{1}{8}\left(x^{2}\right.$
$$
\begin{array}{l}
\left.+8 x \cos x+16 \cos ^{2} x\right)-1 \\
\quad=\frac{1}{8}(x+4 \cos x)^{2}-1 \geqslant-1,
\end{array}
$$

and the line $y=x$ intersects with $y=-4 \cos x$,
$\therefore$ the minimum value of the function is -1.