## mathlilly 2 years ago 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.

1. sirm3d

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|

2. mathlilly

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

3. sirm3d

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

4. sirm3d

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.

5. mathlilly

is it like saying the limit going to 0 is different, so the limit does not exist?

6. sirm3d

yes, that's right. the correct argument is to use limits.

7. mathlilly

okay, that makes sense. thanks!