3. Given that for any $m \in\left[\frac{1}{2}, 3\right]$, we have $x^{2}+m x+4>2 m+4 x$.
Then the range of values for $x$ is $\qquad$ .
3. $x>2$ or $x2 m+4 x \\
\Leftrightarrow m(x-2)+(x-2)^{2}>0 .
\end{array}
$$

Let $g(m)=m(x-2)+(x-2)^{2}$, where,
$$
m \in\left[\frac{1}{2}, 3\right] \text {. }
$$

The problem is converted to $g(m)$ being always greater than 0 for $m \in\left[\frac{1}{2}, 3\right]$. Thus, $\left\{\begin{array}{l}g\left(\frac{1}{2}\right)>0, \\ g(3)>0 .\end{array}\right.$ Solving this yields $x>2$ or $x<-1$.