3. If $f(x)$ is an odd function on the interval $\left[t, t^{2}-2 t-2\right]$, then the value of $t$ is $\qquad$
3. -1 .

The domain of an odd function is symmetric about the origin, and the right endpoint of the interval is not less than the left endpoint. Therefore,
$$
\left\{\begin{array} { l } 
{ - t = t ^ { 2 } - 2 t - 2 > 0 , } \\
{ t \leqslant t ^ { 2 } - 2 t - 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t^{2}-t-2=0, \\
t<0, \\
t^{2}-3 t-2 \geqslant 0 .
\end{array}\right.\right.
$$

Solving this, we get $t \doteq-1$.