2. Let the function $f(x)=\frac{x^{2}+x+16}{x}(2 \leqslant x \leqslant a)$, where the real number $a>2$, if the range of $f(x)$ is $[9,11]$, then the range of values for $a$ is $\qquad$ .
Solve:
$f(x)=\frac{x^{2}+x+16}{x}=x+\frac{16}{x}+1 \geqslant 9$, equality holds when $x=4$, thus $a \geqslant 4$.
Also, $\frac{x^{2}+x+16}{x}=11 \Rightarrow x^{2}-10 x+16=0 \Rightarrow x_{1}=2, x_{2}=8$,
Therefore, the range of $a$ is $[4,8]$.