IS IT CONTINUOUS?

[FlorianK]

NEW QUESTION! Is a function still considered continuous if at a certain point, the line is physically broken/discontinuous, but the x values are continuous.
For example a hybrid functoin where:
f(x) {
x^2 [-∞, 5)
-2x [5, ∞)
Or does it the line have to be completely unbroken to be a continuous function?
 

As long as there are y-values for every x-value over the interval you are looking at then the function will be continous over this interval. So in your case this funtion WILL be continous. However the derivative won't be in your case because at the point x=5 there won't be a gradient, because the gradient does not exist at sharp-bends, breaks or end points.

[diligent18]

It is essential for a function to be smooth at any given point in order to be continuous. You take limit from both sides and see if they are the same. For the example above, if you take limit at x=5, you will see they are not the same. So this function is not differentiable at x=5

NEW QUESTION! Is a function still considered continuous if at a certain point, the line is physically broken/discontinuous, but the x values are continuous.
For example a hybrid functoin where:
f(x) {
x^2 [-∞, 5)
-2x [5, ∞)
Or does it the line have to be completely unbroken to be a continuous function?
 

As long as there are y-values for every x-value over the interval you are looking at then the function will be continous over this interval. So in your case this funtion WILL be continous. However the derivative won't be in your case because at the point x=5 there won't be a gradient, because the gradient does not exist at sharp-bends, breaks or end points.

These were the two contradicting ideas I had.
Both of you reinforced one of them, so I'm still stuck.
I need a third opinion.

[availn]

NEW QUESTION! Is a function still considered continuous if at a certain point, the line is physically broken/discontinuous, but the x values are continuous.
For example a hybrid functoin where:
f(x) {
x^2 [-∞, 5)
-2x [5, ∞)
Or does it the line have to be completely unbroken to be a continuous function?
 

As long as there are y-values for every x-value over the interval you are looking at then the function will be continous over this interval. So in your case this funtion WILL be continous. However the derivative won't be in your case because at the point x=5 there won't be a gradient, because the gradient does not exist at sharp-bends, breaks or end points.

I don't think so, doesn't continuous mean it has to be "smooth"? http://en.wikipedia.org/wiki/Continuous_function

[polar]

if the function was continuous, then its left-hand limit and its right hand limit are equal - in this case they are not


obviously, the function isn't smooth either but if the function was smooth, then the left-hand limit and the right hand limit of the gradient function are also equal - in this case they are not


therefore, the function isn't continuous at x=5 and isn't smooth either.

[BigAl]

if the function was continuous, then its left-hand limit and its right hand limit are equal - in this case they are not


obviously, the function isn't smooth either but if the function was smooth, then the left-hand limit and the right hand limit of the gradient function are also equal - in this case they are not


therefore, the function isn't continuous at x=5 and isn't smooth either.

This was what I tried to explain..but I don't know how to type mathematical expressions here

[diligent18]

if the function was continuous, then its left-hand limit and its right hand limit are equal - in this case they are not


obviously, the function isn't smooth either but if the function was smooth, then the left-hand limit and the right hand limit of the gradient function are also equal - in this case they are not


therefore, the function isn't continuous at x=5 and isn't smooth either.

I don't understand..
What's the difference between smooth and continuous?
Please elaborate further, in simple terms preferably, my inferior brain can't seem to grasp it.

[kamil9876]

I'm guessing that by "smooth" we mean differentiable (some people prefer it to mean infinitely differentiable, but i doubt that is the point here).

Continous and differentiable are not the same, for example is a continous function of everywhere (yes, even at ) but it is not differentiable at . So we can have continous functions that aren't differentiable (hence not smooth). But every differentiable function will be continous (go back to the limit definition of the derivative if you want to see why).

Aside: What's frightening is that you can have functions which are continous everywhere but differentiable nowhere (not like the absolute value, which is continous everywhere and differentiable everywhere except just one point).

[daniel034]

If the y value at the open circle is equal to the y value at the closed circle the function is continuous. If the gradient of the two points is not equal then the derivative function is discontinuous. For example y=abs(x) is continuous but the derivative is discontinuous because the gradients aren't equal at 0.

[kamil9876]

Quiz: is the function given by for and for differentiable/smooth at ?

[charmanderp]

Quiz: is the function given by for and for differentiable/smooth at ?

It's no differentiable. It has an endpoint at x=0 so I wouldn't say it's smooth either.

[BubbleWrapMan]

A function is continuous at a point c if and the limit exists

Recall that exists if and both exist and are equal.

A visual representation of this is literally that the line depicting the graph is continuous, i.e. it has no breaks at that point.

A function is differentiable at a point c if exists and the function is continuous at c.

[kamil9876]

Quiz: is the function given by for and for differentiable/smooth at ?

It's no differentiable. It has an endpoint at x=0 so I wouldn't say it's smooth either.

Correct, I was just waiting for someone to fall into the trap "oh but the left side derivative=right side derivative therefore differentiable".

[TrueTears]

Aside: What's frightening is that you can have functions which are continous everywhere but differentiable nowhere (not like the absolute value, which is continous everywhere and differentiable everywhere except just one point).

one of the most famous examples: http://en.wikipedia.org/wiki/Weierstrass_function

[kamil9876]

A function is differentiable at a point c if exists and the function is continuous at c.

You can have a differentiable function where that limit doesn't exist (you can have functions with derivatives which aren't even continous). Hopefully this example works:

for and otherwise.

Clearly is differentiable at , so just need to manually check the derivative at . So need to compute the limit:



  as

So is indeed differentiable at

Now let's compute for x>0 the derivative



which goes to infinity as goes to . So the limit doesn't exist (one side is infinity the other is 0)

Aside: What's frightening is that you can have functions which are continous everywhere but differentiable nowhere (not like the absolute value, which is continous everywhere and differentiable everywhere except just one point).

one of the most famous examples: http://en.wikipedia.org/wiki/Weierstrass_function

In one assignment we showed that every continous function can be approximated by a continous but nowhere differentiable function.

[BubbleWrapMan]

Okay I need to study calculus more haha