Example 1 (2004 National College Entrance Examination Hunan Paper) Let $f(x)$ and $g(x)$ be odd and even functions defined on $\mathbf{R}$, respectively. When $x>0$, and $g(-3)=0$, the solution set of the inequality $f(x) \cdot g(x)<0$ is (.
A. $(-3,0) \cup(3,+\infty)$
B. $(-3,0) \cup(0,3)$
C. $(-\infty,-3) \cup(3,+\infty)$
D. $(-\infty,-3) \cup(0,3)$
Solution: Choose D. Reason: Since $f(x)$ and $g(x)$ are odd and even functions defined on $\mathbf{R}$, respectively, $f(x) \cdot g(x)$ is an odd function on $\mathbf{R}$. Also, $x0$, i.e., $[f(x) \cdot g(x)]^{\prime}>0$, so $f(x) \cdot g(x)$ is an increasing function on $(-\infty, 0)$, and thus $f(x) \cdot g(x)$ is also an increasing function on $(0, +\infty)$.

From $g(-3)=0$, we get $f(-3) \cdot g(-3)=0$. Therefore, the solution set for $f(x) \cdot g(x)<0$ is $(-\infty, -3) \cup (0,3)$.