Problem 1 If $f(x)=|x-a|+|x-b|(a<b)$, find the minimum value of this function.
Solution 1 (Geometric Method) By the geometric meaning of absolute value, the value of $|x-a|+|x-b|$ can be seen as the sum of distances from point $x$ to points $a$ and $b$ on the number line. Therefore, its minimum value is achieved when $x \in[a, b]$, and is $|a-b|$, which is $b-a$.

Solution 2 (Algebraic Method) Remove the absolute value and convert to a piecewise function to find the minimum value. It is easy to get
$$
f(x)=\left\{\begin{array}{ll}
-2 x+a+b, & x \leqslant a ; \\
b-a, & ab .
\end{array}\right.
$$

By solving each segment, we get $f(x)_{\min }=b-a$, which is achieved at any point in the interval $[a, b]$.