3. Let planar vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ satisfy
$$
1 \leqslant|\boldsymbol{\alpha}|, |\boldsymbol{\beta}|, |\boldsymbol{\alpha}+\boldsymbol{\beta}| \leqslant 3 \text {. }
$$

Then the range of $\boldsymbol{\alpha} \boldsymbol{\beta}$ is $\qquad$
3. $\left[-\frac{17}{2}, \frac{9}{4}\right]$.

Notice,
$$
\begin{array}{l}
\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=\frac{1}{2}\left(|\boldsymbol{\alpha}+\boldsymbol{\beta}|^{2}-|\boldsymbol{\alpha}|^{2}-|\boldsymbol{\beta}|^{2}\right) \\
\geqslant \frac{1-9-9}{2}=-\frac{17}{2}, \\
\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=\frac{1}{4}\left(|\boldsymbol{\alpha}+\boldsymbol{\beta}|^{2}-|\boldsymbol{\alpha}-\boldsymbol{\beta}|^{2}\right) \leqslant \frac{9}{4} .
\end{array}
$$

For example: Let $|\boldsymbol{\alpha}|=|\boldsymbol{\beta}|=t,|\boldsymbol{\alpha}+\boldsymbol{\beta}|=4-t$, $\frac{4}{1+\sqrt{2}} \leqslant t \leqslant 3$, then the range of $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}$ covers $\left[-\frac{17}{2}, 0\right]$; Let $|\boldsymbol{\alpha}|=|\boldsymbol{\beta}|=t,|\boldsymbol{\alpha}+\boldsymbol{\beta}|=3, \frac{3}{2} \leqslant t \leqslant \frac{3}{\sqrt{2}}$, then the range of $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}$ covers $\left[0, \frac{9}{4}\right]$.
Therefore, the range of $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}$ is $\left[-\frac{17}{2}, \frac{9}{4}\right]$.