Thread: I need help with piecewise smooth functions

Piecewise smooth functions..

Originally Posted by sarideli18

Indicate which of the following functions are piecewise smooth on [-1,1] and why:

i.) h(x)=x^2ln|x|

ii.) q(x)=xln|x|

iii.) r(x)= 2/(1-x) if x≠1 and 0 if x=0

iv.) s(x)= 1/(2-x) if x≠1 and 0 if x=0

The definition of peicewise smooth is that a function is differentable on the domain except at a finite number of points. Also at these points both the left and right derviatve must exist.

So for i)

h(x) is continous and differentable everywhere except at x=0. For simplicity note that h(x) is an even function so we only have to test the derviative on one side.



This limit can be resolved using L'hosipitals
by L.H



So now this is the form infinity divided by infinity so we take the derivative to get



So the limit is 0 and both the left and right derviative exist at that point. Also the function can be made continous if we define h(0)=0.

Now try the 2nd one.

Well, I just found that the first and fourth ones are piecewise smooth, and the second one is not piecewise smooth. But I am stuck on the third one. Since the function itself (2/(1-x)) and 0) and its derivative (2/(x-1)^2 and 0) are continuous I think it is piecewise smooth. But there is a jump at x=1. That confuses me.

I mean, the limit from left is infinite and the limit from right is 0. It is the same case as the derivative of the function. Therefore, I think it is not piecewise smooth, but I am not sure.

Originally Posted by sarideli18

I mean, the limit from left is infinite and the limit from right is 0. It is the same case as the derivative of the function. Therefore, I think it is not piecewise smooth, but I am not sure.

You are correct the derivative cannot be infinty at a point so it does not exist.

I just found that the first and fourth ones are piecewise smooth, and the second one is not piecewise smooth.

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