Example 2 As shown in Figure 9-1, in the pyramid $P-ABC$, if the edge $AP=x$, and the lengths of the other edges are all 1, let the volume of the pyramid $P-ABC$ be $V(x)$. Find:
(1) The expression for $V(x)$ and determine its domain;
(2) For what value of $x$ does $V(x)$ achieve its maximum value? What is the maximum value?
Analysis: The second question (2) is based on the first question (1), so the key is to find the expression for $V(x)$. In geometry, when dealing with isosceles triangles with a common base, it is generally useful to take the midpoint of the common base and draw an auxiliary line through the vertex and the midpoint.

Solution (1) Take the midpoint of $B C$ as $D$, and connect $A D$ and $P D$. Then $A D \perp B C$ and $P D \perp B C$, so $B C \perp$ plane $P A D$, and $P D = A D = \sqrt{1 - \left(\frac{1}{2}\right)^{2}} = \frac{\sqrt{3}}{2}$. Take the midpoint of $P A$ as $E$, and connect $D E$. Then $D E \perp P A$, and $D E = \sqrt{P D^{2} - P E^{2}} = \sqrt{\frac{3}{4} - \left(\frac{x}{2}\right)^{2}} = \frac{1}{2} \sqrt{3 - x^{2}}$. Therefore, $S_{\triangle P A D} = \frac{1}{2} \cdot P A \cdot D E = \frac{1}{2} \cdot x \cdot \sqrt{3 - x^{2}} \cdot \frac{1}{2} = \frac{1}{4} x \sqrt{3 - x^{2}} (0 < x < \sqrt{3})$.
And $V(x) = V_{C P A D} + V_{B-P A D} = 2 V_{C-P A D} = 2 \cdot \frac{1}{3} \cdot \frac{x}{4} \sqrt{3 - x^{2}} \cdot \frac{1}{2}$
$$
= \frac{x}{12} \sqrt{3 - x^{2}} (0 < x < \sqrt{3}).
$$
(2) $V(x) = \frac{1}{12} x \sqrt{3 - x^{2}} = \frac{1}{12} \sqrt{x^{2} (3 - x^{2})} = \frac{1}{12} \sqrt{-(x^{2} - \frac{3}{2})^{2} + \frac{9}{4}}$, when $x^{2} = \frac{3}{2}$, i.e., $x = \frac{\sqrt{6}}{2} \in (0, \sqrt{3})$, $V(x)$ max $= \frac{1}{12} \cdot \frac{3}{2} = \frac{1}{8}$.

Alternatively, for $x^{2} (3 - x^{2}) \leqslant \left[\frac{x^{2} + (3 - x^{2})}{2}\right]^{2} = \left(\frac{3}{2}\right)^{2}$, equality holds when and only when $x^{2} = 3 - x^{2}$, i.e., $x^{2} = \frac{3}{2}$, yielding the maximum value of $V(x)$.