The minimum value of the function defined by $f\left(x\right)=$ maximum $\{x.x+1,2-x\}$ is

$f\left(x\right)=\max \{x,x+1,2-x\}$
we plot $f\left(x\right)$ value in the graph below.

minimum value of $f\left(x\right)$ is find at the point of intersection between line $y=2-x$ and $y=x+1$
$$\begin{gathered} \therefore 2-x=x+1 \\ 2x=1 \\ x=\frac{1}{2} \end{gathered}$$
$$\begin{gathered} \therefore y=2-x \\ =2-1/2 \\ =\frac{4-1}{2} \\ =\frac{3}{2} \end{gathered}$$