Why are functions not differentiable at holes?

[CuriousBanker]

I understand at cusps, corners, etc, because the negative and positive directions do not agree with each other.

But what about at jump discontinuity on a graph? Why wouldn't a function be differentiable there? I understand that from the definition of differentiable that it just isn't, but I don't get WHY. Using the epsilon-delta definition, you still can get within delta for any given delta, it's just that it doesn't happen to work at that exact point...but why does it have to? Derivatives are as delta x approaches zero...as long as it is approaching zero, and not actually zero, the derivative still holds up...so f(x)+dx does not equal f(x) even if it is very, very close.

Not sure if I worded this properly but maybe somebody will get what I mean?

 

[gopher_p]

By definition, $f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$. In order for the term inside the limit to make sense, $f$ must be defined at $a$. A necessary condition for the limit to exist is that $\lim_{x\rightarrow a}(f(x)-f(a))=0$. So $f$ has to be continuous at $a$ if it's going to stand a chance at being differentiable at $a$.

 

[Mark44]

I understand at cusps, corners, etc, because the negative and positive directions do not agree with each other.

But what about at jump discontinuity on a graph? Why wouldn't a function be differentiable there? I understand that from the definition of differentiable that it just isn't, but I don't get WHY. Using the epsilon-delta definition, you still can get within delta for any given delta, it's just that it doesn't happen to work at that exact point...but why does it have to? Derivatives are as delta x approaches zero...as long as it is approaching zero, and not actually zero, the derivative still holds up...so f(x)+dx does not equal f(x) even if it is very, very close.

Not sure if I worded this properly but maybe somebody will get what I mean?

Consider this simple function with a jump discontinuity at 0:

f(x) = 0 for x ≤ 0 and f(x) = 1 for x > 0

Obviously the function is differentiable everywhere except x = 0.

$$f'(0) = \lim_{h \to 0}\frac{f(0 + h) - f(0)} h$$

For negative values of h, the limit above exists and is equal to 0. However, if h is positive, what happens with the limit for small but positive values of h? You should be able to convince yourself that the limit doesn't exist, hence the derivative doesn't exist. For it to exist, both one-sided limits have to exist and be equal.

 

[micromass]

A geometric answer: the derivative is just the slope of the tangent line. But given a function with a jump discontinuity, for example

What would you consider the tangent line at the discontinuity? There is no reasonable definition here.

 

[CuriousBanker]

Yeah I get those examples. I meant something more like f(x)=2x when x does not equal 5, and f(x) = 5 when x=5. Then in that case, using epsilon delta, both negative and positive directions approach f(x)=10 as x gets closer and closer to 5...the only value is when f = exactly 5 that it does not work. But with epsilon delta, if you get within delta of 5 you can get within epsilon/2 of 10...so it still holds up, so why can't it work then?

 

[hilbert2]

^ If we'd redefine derivative as $f'(x)=lim_{h \rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}$ then the function would be differentiable in such a point. However, I've never seen such a definition being used.

 

[johnqwertyful]

Yeah I get those examples. I meant something more like f(x)=2x when x does not equal 5, and f(x) = 5 when x=5. Then in that case, using epsilon delta, both negative and positive directions approach f(x)=10 as x gets closer and closer to 5...the only value is when f = exactly 5 that it does not work. But with epsilon delta, if you get within delta of 5 you can get within epsilon/2 of 10...so it still holds up, so why can't it work then?

No, they do NOT in fact approach 10. From one side (the right side), the slope approaches +∞, the other side it approaches -∞.

Look at it graphically, it might make sense.

 

[Boorglar]

Well I think that your definition (or intuitive notion) of the derivative is not exactly the same as the standard one. Most mathematicians will define the derivative of a function f at a point x as:

$f'(x) = \lim_{h \rightarrow 0 } \frac{ f(x+h) - f(x) }{ h }$.
Using that definition, your function with "holes" won't be differentiable because f(5) = 5 and for h ≠ 0, $\frac{ f(5 + h ) - f(5) } { h } = \frac{ 2( 5 + h ) - 5 }{ h } = \frac{ 5 + 2h }{ h } = \frac{ 5 }{ h } + 2$ which obviously diverges. This is because your secant lines have one endpoint "stuck inside the hole" and thus they will become more and more "vertical" as the other endpoint approaches 5.

Although I agree, there is some sense in which this function still looks like it could have a slope at x = 5. It doesn't, by the derivative definition. But our eyes are accustomed to continuity, and since the slope is 2 everywhere except at 5, we like to "imagine" that the function still has a tangent with slope 2 at x = 5.

This notion can be stated more rigorously by saying that, although $f'(5)$ does not exist, $\lim_{ x \rightarrow 5 } f'(x)$ does exist. This is as nice as it gets for non-differentiable functions. Most non-differentiable functions will look less "smooth" because their slopes don't converge to a limit. Say, for the absolute value function, the corner at x = 0 has -1 and 1 and the two possible slopes, but the limit of the derivatives as x approaches 0 from both sides does not exist.

P.S. This is not a jump discontinuity. This type of discontinuity is called a "removable discontinuity", because we could modify the value of f at 5 and make it continuous there.

 

[CuriousBanker]

Ahh, I get it now...since we are approaching 5, and there is a hole, the line keeps slanting more and more vertical...I see now, thanks everybody

 

[Mireia]

I understand at cusps, corners, etc, because the negative and positive directions do not agree with each other.

But what about at jump discontinuity on a graph? Why wouldn't a function be differentiable there? I understand that from the definition of differentiable that it just isn't, but I don't get WHY. Using the epsilon-delta definition, you still can get within delta for any given delta, it's just that it doesn't happen to work at that exact point...but why does it have to? Derivatives are as delta x approaches zero...as long as it is approaching zero, and not actually zero, the derivative still holds up...so f(x)+dx does not equal f(x) even if it is very, very close.

Not sure if I worded this properly but maybe somebody will get what I mean?

If a function is differentiable at point x, then it is continuous at that point. By counter-reciprocal, if a function is not continuous at point x, then it is not differentiable either.

 

[Stephen Tashi]

But with epsilon delta, if you get within delta of 5 you can get within epsilon/2 of 10...so it still holds up, so why can't it work then?

It's not clear what you mean by "with epsilon delta". Yes, the definition of derivative does traditionally mention the variables epsilon and delta, but it doesn't define any process called "using epsilon delta". I think you are using that phrase to stand for some intuitive way of visualizing limits. The definition of derivative doesn't answer an intuitive question. The definition means what it says. Mathematics is legalistic. The mathematical answer about why your example isn't differentiable at x = 5 is that the definition of derivative implies that your function isn't differentiable at x = 5. Case closed! We can consider the question of "Why is the definition of derivative chosen to be the way it is?" That is a sociological question because it concerns choices made by human beings. Some of the answers you are getting explain that the standard definition of derivative is convenient for mathematicians.