differentiability of piece-wise defined functions

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eng_boody

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pls i'd like to ask how to test for differentiability of a function, what is that related to one sided derivative in case of piece wise defined functions and in lim h->0+ or lim h->0- what does the positive and negative sign beside the zero indicate GRAPHICALLY .. thanks
 

First of all you have to check the continuity of the function at each joint. f.i, if the function f(x) is defined as:

2*x-1 for x<=3
x^2-1 for x>3

first check the continuity in 3 that means the limit ot the first function when x-->3 has the same value of the
limit (when x-->3) of the second function. In our example if x=3 we have

5
8

the the function is not continuos hence not differentiable.

Let's try with:

2*x-1 for x<=3
x^2-4 for x>3

now it's continuos. We have now to check the differentiability. Calculate the first derivative:

2 for x<=3
2*x for x>3

again check both the subfunctions have the same limit for x-->3. We have:

2
6

the this function is continuous but not differentiable.
If, instead we have

6*x-13 for x<=3
x^2-4 for x>3

the limit for x-->3 is 5 in both cases. The first derivative is:

6
2*x

then the limit of the derivative for x-->3 is 6 in both cases. The function is differentiable.

When I applied the limit to the first of the two subfunction on each f(x) I was coming from 0 towards 3, because that subfunction is defined for x<=3. So I'm moving from left to right that means I'm using all values a little bit less than 3 then I'll write lim x-->3-. When instead I applied the same limit to the second subfunction, I was moving from right to left, that means I was using all values a little bit greater than 3 then I'll write lim x-->3+

 
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must the function be continuous before differentiate both branches ?? , I mean if we apply that to the function f(x)=(x-2)+4 x<2
(x-2)+5 x>=2
we will find the same derivatives though the function is not differentiable at x=2 , is that true ??

thank
 

Yes, a function to be differentiable in a point xo must be continuous in xo
 

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