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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.

Differential Equations
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continuity on an closed interval is the first requirement for differentiability in the open interval. The function is discontinuous on any closed interval containing 0, so it is not differentiable there (at zero). take a look at the graph of \(f(x)\) |dw:1359845550730:dw|
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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for example, if f(x) = { 2x+5, x<0, x^2+3 for x>=0 at 2(0)+5=5, (0)^2+3=3 since the two values are different, the function is discontinuous at x=0, so it's not differentiable there.
is it like saying the limit going to 0 is different, so the limit does not exist?
yes, that's right. the correct argument is to use limits.
okay, that makes sense. thanks!

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