Consider the function f(x) ={ 1, if x<= 0 2, if x>0}
Show that the differential equation x dot = f(x) has no solution on any open interval about t knot =0. Could someone help me out? I am not sure how to do this. I tried integrating for f(x)=1 and f(x)=2 to find x(t), but I don't know where to go from there.

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Is there a way to tell without knowing what the graph looks like though?

the graph is not needed. just check the value of the function at x=0.

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