differentiability of piece-wise defined functions

[eng_boody]

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

 

[albbg]

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+

 

[eng_boody]

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+

View attachment 109868

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

 

[albbg]

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