We must compare the two functions
$$y_1=|x^2-2x|$$
and
$$y_2=\frac{x}{2}$$
to find where the maximum function changes branches, since such points can create corners and hence non-differentiability.
On the interval
$$[0,2]$$
we have
$$x^2-2x\le0$$
Therefore,
$$y_1=2x-x^2$$
Setting
$$y_1=y_2$$
gives
$$2x-x^2=\frac{x}{2}$$
$$4x-2x^2=x$$
$$3x-2x^2=0$$
$$x(3-2x)=0$$
Hence,
$$x=0,\frac32$$
On the interval
$$[2,3]$$
we have
$$x^2-2x\ge0$$
Therefore,
$$y_1=x^2-2x$$
Setting
$$y_1=y_2$$
gives
$$x^2-2x=\frac{x}{2}$$
$$2x^2-5x=0$$
$$x(2x-5)=0$$
Hence,
$$x=\frac52$$
Now analyze the maximum function.
For
$$x\in[0,\frac32]$$
we have
$$f(x)=2x-x^2$$
For
$$x\in[\frac32,\frac52]$$
we have
$$f(x)=\frac{x}{2}$$
Notice that at
$$x=2$$
the maximum continues along
$$\frac{x}{2}$$
since
$$|x^2-2x|=0$$
and
$$\frac{x}{2}=1$$
Thus no corner occurs at
$$x=2$$
For
$$x\in[\frac52,3]$$
we have
$$f(x)=x^2-2x$$
Therefore the branch changes occur at
$$x=\frac32$$
and
$$x=\frac52$$
Both points are continuous but not differentiable.
Hence the required sum is
$$\frac32+\frac52$$
$$=\frac82$$
$$=4$$
Final Answer :
$$4$$
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