Example 4. Let $x$ be a real number, $y=|x-1|+|x-1|$. Among the following four conclusions:
(1) $y$ has no minimum value;
(2) there is only one $x$ that makes $y$ reach its minimum value;
(3) there are a finite number of $x$ (more than one) that make $y$ reach its minimum value;
(4) there are infinitely many $x$ that make $y$ reach its minimum value. The correct one is ( ).
(A) (1)
(B) (2)
(C) (3)
(D) (4).
(1993, National Junior High School Competition)
Solution 1
$$
y=|x-1|+|x+1|=\left\{\begin{array}{l}
-2 x, x \leqslant-1, \\
2,-1 \leqslant x \leqslant 1, \\
2 x, x \geqslant 1 .
\end{array}\right.
$$

Its graph is shown in the figure.
From the graph, it can be seen that when $x$ takes any value in the range $-1 \leqslant x \leqslant 1$, $y$ achieves its minimum value of 2. Therefore, the correct choice is D.

Solution 2 By the geometric meaning of absolute value, $y=|x-1|+|x+1|$ represents the sum of the distances from point $x$ on the number line to -1 and 1.
Obviously, when $x \in[-1, 1]$, this sum of distances is minimized and is 2. Therefore, the correct choice is D.
Regarding the maximum and minimum values of functions involving absolute values of linear expressions, there is the following important conclusion:
Let $a_{1}<a_{2}<\cdots<a_{n}$, then for the function
$$
y=\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right| \text {, }
$$
(1) If $n$ is even, then when $x$ takes values in the range $a_{\frac{n}{2}} \leqslant x \leqslant a_{\frac{n}{2}+1}$, we have
$$
\begin{aligned}
y_{\text {min}}= & \left(a_{\frac{n}{2}+1}+a_{\frac{n}{2}+2}+\cdots+a_{n}\right) \\
& -\left(a_{\frac{n}{2}}+\cdots+a_{2}+a_{1}\right) .
\end{aligned}
$$
(2) If $n$ is odd, then when $x=a_{\frac{n+1}{2}}$, we have
$$
\begin{aligned}
y_{\text {min}}= & \left(a_{\frac{n+3}{2}}+a_{\frac{n+5}{2}}+\cdots+a_{n}\right) \\
& -\left(a_{\frac{n+1}{2}}+\cdots+a_{2}+a_{1}\right) .
\end{aligned}
$$