7. Let $f(x)=m x^{2}+(2 n+1) x-m-2(m, n \in \mathbf{R}, m \neq 0)$ have at least one zero in $[3,4]$, then the minimum value of $m^{2}+n^{2}$ is $\qquad$ .
Solve:
Rearrange $f(x)$ as the linear equation in terms of $m, n$: $\left(x^{2}-1\right) m+2 x \cdot n+x-2=0$, then $\sqrt{m^{2}+n^{2}}_{\text {min }}=\frac{|x-2|}{\sqrt{\left(x^{2}-1\right)^{2}+(2 x)^{2}}}=\frac{|x-2|}{x^{2}+1}$. When $x \in[3,4]$, $\sqrt{m^{2}+n^{2}}_{\min } \xlongequal{t=x-2 \in[1,2]} \frac{t}{t^{2}+4 t+5}=\frac{1}{t+\frac{5}{t}+4} \in\left[\frac{1}{10}, \frac{2}{17}\right]$. Therefore, when $x=3$, the minimum value of $m^{2}+n^{2}$ is $\frac{1}{100}$.