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 !!!!
False. This answers your question:
http://oregonstate.edu/instruct/mth251/cq/Stage5/Lesson/diffVsCont.html
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.
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 !!!!