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If f(x) is continuous on an open interval, it is di erentiable there.

Guest Sep 3, 2014

Best Answer 

 #5
avatar+92788 
+5

As one of my calculus teachers would have said.....differentiability implies continuity, but continuity does not imply differentiability.....

To see why, look at the absolute value function at x = 0. Note that, it is continuous at that point but the derivative is undefined. (What is the "slope" of the function at that point .....1  or -1 ??)

We can see this in one more way......the absolute value of a number "x" is defined as √(x2).....and taking the derivative of this gives us  .......   (1/2) (x2)-1/2 *(2x)  =  x / √(x2) =  x / lxl     ........and note that, this derivative is undefined at x = 0  !!!!

 

CPhill  Sep 3, 2014
 #1
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Sorry it was a true or false question.

Guest Sep 3, 2014
 #2
avatar+4472 
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AzizHusain  Sep 3, 2014
 #3
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Thank you.

Guest Sep 3, 2014
 #4
avatar+94118 
+5

I'm just thinking.

If f(x) is continuous in an open region does that mean it could be a relation that is not  a function.

Like a circle or an s shape?

Relations that are not functions are not differentiable either.  

Melody  Sep 3, 2014
 #5
avatar+92788 
+5
Best Answer

As one of my calculus teachers would have said.....differentiability implies continuity, but continuity does not imply differentiability.....

To see why, look at the absolute value function at x = 0. Note that, it is continuous at that point but the derivative is undefined. (What is the "slope" of the function at that point .....1  or -1 ??)

We can see this in one more way......the absolute value of a number "x" is defined as √(x2).....and taking the derivative of this gives us  .......   (1/2) (x2)-1/2 *(2x)  =  x / √(x2) =  x / lxl     ........and note that, this derivative is undefined at x = 0  !!!!

 

CPhill  Sep 3, 2014
 #6
avatar+94118 
0

The absolute value of x is the one Aziz gave us as a counter example. (from the site he sent us to)

My query/consideration was a tad different.

Melody  Sep 3, 2014

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