Example 10 (1994 Shanghai Competition Question) The range of the function $y=\sqrt{1994-x}+\sqrt{x-1993}$ is . $\qquad$
Solve: Fill in $[1, \sqrt{2}]$. Reason: First find the domain: $1993 \leqslant x \leqslant 1994$.
$$
\text { Let } \begin{aligned}
y^{\prime} & =\frac{1}{2 \sqrt{1994-x}} \cdot(1994-x)^{\prime}+\frac{1}{2 \sqrt{x-1993}} \cdot(x-1993)^{\prime} \\
& =-\frac{1}{2 \sqrt{1994-x}}+\frac{1}{2 \sqrt{x-1993}}=0,
\end{aligned}
$$

Thus, we get the critical point $x=\frac{1994+1993}{2}$.
Calculate the function values at the critical point and the endpoints of the interval, we get $f\left(\frac{1994+1993}{2}\right)=\sqrt{2}, f(1993)=1, f(1994)=$ 1 , so $y_{\operatorname{max}}=\sqrt{2}, y_{\min }=1$, hence the range is $1 \leqslant y \leqslant \sqrt{2}$.