Example 5 Let the function $f_{0}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed figure formed by the graph of $y$ $=f_{2}(x)$ and the $x$-axis is $\qquad$
(1989, National High School Mathematics Competition)
Analysis: If it is not easy to directly draw the graph of $y=f_{2}(x)$, but we know the relationship between the graphs of $y=f(x)$ and $y=|f(x)|$. If we follow the sequence
$$
\begin{array}{l}
f_{0}(x)=|x| \rightarrow y=f_{0}(x)-1 \\
\rightarrow f_{1}(x)=\left|f_{0}(x)-1\right| \rightarrow y=f_{1}(x)-2 \\
\rightarrow f_{2}(x)=\left|f_{1}(x)-2\right|
\end{array}
$$

to perform graphical transformations, it is easier to draw the graph of $y=f_{2}(x)$.
It is easy to get the answer as 7.