Example 6 (1998 National High School Mathematics League Question) If $f(x) (x \in \mathbf{R})$ is an even function with a period of 2. And when $x \in [0,1]$, $f(x) = x^{\frac{1}{110}}$, then the arrangement from smallest to largest of $f\left(\frac{98}{19}\right) \cdot f\left(\frac{101}{17}\right) \cdot f\left(\frac{104}{15}\right)$ is $\qquad$ .
Given that for $x \in \mathbf{R}, k \in \mathbf{Z}$, we have $f(2 k+x)=f(x)=f(-x)$,
thus $f\left(\frac{98}{19}\right)=f\left(6-\frac{16}{19}\right)=f\left(-\frac{16}{19}\right)=f\left(\frac{16}{19}\right), f\left(\frac{101}{17}\right)=f\left(6-\frac{1}{17}\right)=f\left(-\frac{1}{17}\right)=$ $f\left(\frac{1}{17}\right), f\left(\frac{104}{15}\right)=f\left(6+\frac{14}{15}\right)=f\left(\frac{14}{15}\right)$. Since $x \in[0,1]$, $f(x)=x$ is an increasing function.
Therefore, $f\left(\frac{1}{17}\right)<f\left(\frac{16}{19}\right)<f\left(\frac{14}{15}\right)$. That is, $f\left(\frac{101}{17}\right)<f\left(\frac{98}{19}\right)<f\left(\frac{104}{15}\right)$.
Note: The key to this problem is to use the periodicity of the function to transform the three function values to the same monotonic interval $[0,1]$ for comparison, which is a common method for comparing sizes using the monotonicity of functions.

This problem can also be solved using the graphical method as follows: By the property of even functions, the graph on $[-1,0]$ is symmetric to the graph on $[0,1]$ about the $y$-axis. Since the function has a period of 2, the graph on $[-1,0]$ is the same as the graph on $[1,2]$. Hence, we get Figure 1-2-4. Let the three known function values be denoted as $A$, $B$, and $C$. Then these three points are distributed on either side of the value 6. It is easy to see that (point is the lowest, $A$ point is the second, and $B$ point is the highest.