Piecewise smooth functions..
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.
$\displaystyle \lim_{h \to 0}\frac{(0+h)^2\ln(0+h)}{h} = \lim_{h \to 0}h\ln(h)$
This limit can be resolved using L'hosipitals
by L.H
$\displaystyle \frac{\ln(h)}{\frac{1}{h}}$
So now this is the form infinity divided by infinity so we take the derivative to get
$\displaystyle \frac{\frac{1}{h}}{\frac{-1}{h^2}}=-h$
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 just found that the first and fourth ones are piecewise smooth, and the second one is not piecewise smooth.
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