10. Let $a, b \in \mathbf{R}, a<b$, the function $g(x)=\max _{a \leq t \leq b}|x+t|(x \in \mathbf{R})$ (where $\max _{a \leq t \leq b}$ denotes the maximum value of the expression $|x+t|$ for $x \in \mathbf{R}$ when $t \in[a, b]$), then the minimum value of $g(x)$ is
Answer $\frac{b-a}{2}$.
Analysis For every $x \in \mathbf{R}$, the function $f(t)=x+t$ is a linear function, so on any finite closed interval, the maximum and minimum values of the function $|f(t)|$ are reached at the endpoints of the interval, thus we have
$$
g(x)=\max _{a \leq t \leq b}|x+t|=\max _{a, b}\{|x+a|,|x+b|\}
$$

Since the intersection point's x-coordinate $c$ of the graphs of the functions $y=|x+a|, y=|x+b|$ satisfies
$$
-(c+b)=c+a \Rightarrow c=-\frac{a+b}{2}
$$

We get $g(x)=\left\{\begin{array}{ll}|x+a| & , x \leq c \\ |x+b| & , x>c\end{array}\right.$, its graph consists of two line segments, and $\min _{x} g(x)=g(c)=\frac{b-a}{2}$.