Is there such a function?

Is there a function satisfying these 2 conditions?
1- the function is differentiable everywhere in (a, b).
2- the derivative of the function is discontinous somewhere in (a, b).

(a, b) is an open region: a < x < b.
prove or make a counterexample.

Hi,
Its not possible to have such a function.

~Kalyan.

Sure, there is. Just take the triangle function on the interval [-1, 1]:

\[\Lambda(x) = \left\{ \begin{array}{l l} 1+x & \quad \mbox{if }x \leq 0\\ 1-x & \quad \mbox{if }x >0\\ \end{array} \right.\]

The derivative to this function is not continuous:
\[\frac{d\Lambda}{dx}(x) = \left\{ \begin{array}{l l} 1 & \quad \mbox{if }x \leq 0\\ -1 & \quad \mbox{if }x >0\\ \end{array} \right.\]

The derivative exists everywhere in the interval but is discontinuous at x=0.

jayc said:

Sure, there is. Just take the triangle function on the interval [-1, 1]:

\[\Lambda(x) = \left\{ \begin{array}{l l} x & \quad \mbox{if }x \leq 0\\ 1-x & \quad \mbox{if }x >0\\ \end{array} \right.\]

The derivative to this function is not continuous:
\[\frac{d\Lambda}{dx}(x) = \left\{ \begin{array}{l l} 1 & \quad \mbox{if }x \leq 0\\ -1 & \quad \mbox{if }x >0\\ \end{array} \right.\]

The derivative exists everywhere in the interval but is discontinuous at x=0.

Click to expand...

Hi

Are you sure about that your, function is discontinuos there shouldn't be any value for derivate in point 0.

Regards

The function is continous and the derivative exists everywhere in the interval. Since we difined the function to be \[x\] on the interval \[x\leq 0\], the corresponding derivative also exists on the same interval. Therefore, the derivative at x=0 is 1.

jayc

jayc,

What's your interval? Have you drawn that function? It's discontinuous in any open interval that includes 0.

According to Calculus, a function has derivative at a point iff it is differentiable. Therefore, the question becomes, find a function that has derivative everywhere in (a,b) but the derivative is not continuous at some point in (a,b). Here are two examples (both for interval (-1,1)):

1. f(x)=x^2 * sin(1/x) when x>0 or <0, f(0)=0.

This is a continuous function and it can be easily proved that the function has derivative everywhere. Actually, when x>0 or < 0,

f'(x)=2x * sin(1/x) -cos(1/x)

while when x=0,

f'(0)=0.

Notice that the derivative at x=0 should be obtained separately from the definition of the derivative, which is
lim(Δx->0) (f(Δx)-f(0))/Δx=0

Obviously, f'(x) exists everywhere in (-1,1) but is not continuous at 0 (become oscilatory when x->0).

2. f(x)=x^2 * sin(1/(x^2)) when x>0 or <0, f(0)=0.

This function is similar to the previous one, and you can prove (similarly) that the derivative exists everywhere in (-1,1). However, not only is the derivative discontinuous, but it is even also unbounded around 0.

main_road said:

jayc,

What's your interval? Have you drawn that function? It's discontinuous in any open interval that includes 0.

Click to expand...

Wow, I'm sorry. Now I realize my mistake. I meant for it to be a triangle function and I missed the 1+x for x<=0. When I drew it in my head, it looked like a triangle =) The function I meant to write was this

\[\Lambda(x) = \left\{ \begin{array}{l l} 1+x & \quad \mbox{if }-1 < x \leq 0\\ 1-x & \quad \mbox{if }0<x<1\\ 0 & \quad \mbox{otherwise} \end{array} \right.\]

The derivative to this function is not continuous:
\[\frac{d\Lambda}{dx}(x) = \left\{ \begin{array}{l l} 1 & \quad \mbox{if }x \leq 0\\ -1 & \quad \mbox{if }x >0\\ \end{array} \right.\]

The derivative exists everywhere in the interval but is discontinuous at x=0.

jayc said:

The derivative to this function is not continuous:
\[\frac{d\Lambda}{dx}(x) = \left\{ \begin{array}{l l} 1 & \quad \mbox{if }x \leq 0\\ -1 & \quad \mbox{if }x >0\\ \end{array} \right.\]

The derivative exists everywhere in the interval but is discontinuous at x=0.

Click to expand...

jayc,

The derivative does not exist at x=0. That the derivative exists means that both the right derivative and the left derivative exist and are equal. In your case, both the right and the left derivatives exist at x=0 but they are not equal! So, the derivative does not exist.

mainroad is right...

derivative does not exist.... :|

Added after 7 minutes:

This is not possible:

Proof: By contradiction

Suppose such a function exists, call it f(x)

and suppose f'(x) is discontinous at A ;


=> lim f'(x) as x --> A- from left side
is NOT EQUAL TO lim f'(x) as x---> A+ from right side

the above statement is true since f'(x) is discontinous at A .

but wait! , the statement also implies that f'(x) is not differentiable at A.
why?
because the definition for differentiability of f(x) imples that
lim f'(x) is equal when you approach from both sides for all x in the interval.
see definition give above in previous posts.

so CONTRADICTION! .

there you go... its not possible.

Eecs4ever,

WHen you start the proof by contradiction, you assume that "f'(x) is discontinous at A", which is fine. However, you are running some logic issues later. Here is the faulty part of your proof:

eecs4ever said:

...

but wait! , the statement also implies that f'(x) is not differentiable at A.
why?
because the definition for differentiability of f(x) imples that
lim f'(x) is equal when you approach from both sides for all x in the interval.
...

Click to expand...

The differentiability of f(x) at A does NOT imply "lim f'(x) is equal when you approach from both sides for all x in the interval ...", which is implied by the continuity of the derivative f'(x) at A. Instead, the differentiability of f(x) at A IMPLIES

lim(Δx->0+) (f(A+Δx)-f(A))/Δx=lim(Δx->0-) (f(A+Δx)-f(A))/Δx

which is different from "lim f'(x) is equal when you approach from both sides". The difference is that one limit process is taken for f(x) while the other is for f'(x).


Besides, your following argument is the same as above which is incorrect:

eecs4ever said:

...

=> lim f'(x) as x --> A- from left side
is NOT EQUAL TO lim f'(x) as x---> A+ from right side
...

Click to expand...

To the best of my knowledge, i seem to remember what mainroad said is right.

"That the derivative exists means that both the right derivative and the left derivative exist and are equal."

the limit from the right side is:
"lim x->A+ f'(x) " = = lim(Δx->0+) (f(A+Δx)-f(A))/Δx

There is nothing wrong with what mainroad said. I guess here we are running into a concept issue. Here are the important concepts:

1. the derivative exists at A:

left derivative = lim(Δx->0-) (f(Δx+A)-f(A))/Δx = lim(Δx->0+) (f(Δx+A)-f(A))/Δx = right derivative

(notice that the formula says nothing abou the continuity)

2. the derivative is continuous at A:

left limit = lim (Δx->0-) f'(Δx+A) = lim (Δx->0+) f'(Δx+A) = right limit

(notice that, when you use f(x) or f(Δx+A), you have implicitly assumed that the derivative exists).


We all know that #2 is stronger than #1, therefore, #2 => #1. alitavakol's question is that, can we possibly go from #1 to #2 or #1 => #2? My two examples say that we cannot.

As I mentioned in the previous post, when you talked about the right and the left derivatives, you actually used #2, instead of #1. Check again what you said, "...lim f'(x) is equal when you approach from both sides for all x in the interval", which is exactly what #2 tells. #2 is not about right or left derivatives, but it is about continuity of derivative. You should use #1.

Only Steve10 said the correct answer. thanks to him.
for others, I advise them to read the question carefully.

There is no such function!
Please , this is a math, not word games:?:

I have found this somewere
"Thus there is a link between continuity and differentiability: If a function is differentiable at a point, it is also continuous there. Consequently, there is no need to investigate for differentiability at a point, if the function fails to be continuous at that point."


Also
https://en.wikipedia.org/wiki/Derivative


I may be wrong . Did not read the question carefully.Sorry.

I think you are right by the following:

zox11 said:

I may be wrong . Did not read the question carefully.Sorry.

Click to expand...

But now you may want to read it more carefully and post a right one.

I am not sure anymore. I do not want to write something wrong ,again.
But, for me, I know the answer, or think I know.

Thanks.