(1) (2006 - Zhejiang) For $a, b \in \mathbf{R}$, denote $\max \{a, b\}=\left\{\begin{array}{ll}a, & a \geqslant b, \\ b, & a<b,\end{array}\right.$ The minimum value of the function $f(x)=$ $\max \{|x+1|,|x-2|\}(x \in \mathbf{R})$ is $\qquad$
Let $|x+1| \geqslant|x-2|$, then we have $x^{2}+2 x+1 \geqslant x^{2}-4 x+4$, solving this yields $x \geqslant \frac{1}{2}$. Therefore, $f(x)=\left\{\begin{array}{ll}x+1, & x \geqslant \frac{1}{2}, \\ 2-x, & x<\frac{1}{2},\end{array}\right.$ When $x=\frac{1}{2}$, $f(x)$ achieves its minimum value $\frac{3}{2}$.