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Simple Calculus Stuff


Amaton
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Before I start, just want to say that this is not homework. I'm studying single-variable calculus and I sometimes have trouble getting certain things. (btw, might have more questions after this one)

 

So. Let's say I have a function [math]f(x)[/math]. I'm aware that one can show it is continuous at a certain point, but how do I prove that it's continuous on an interval, say [math][a, b][/math]?

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Before I start, just want to say that this is not homework. I'm studying single-variable calculus and I sometimes have trouble getting certain things. (btw, might have more questions after this one)

 

So. Let's say I have a function [math]f(x)[/math]. I'm aware that one can show it is continuous at a certain point, but how do I prove that it's continuous on an interval, say [math][a, b][/math]?

You just have to prove that it is continuous for every [math]x \in [a,b][/math]. In the proof just pick an arbitrary number [math]x \in [a,b][/math] and show that the funktion [math]f[/math] is continuous at that point.

 

Also if you know that the funktion [math]f: \mathbb{R} \rightarrow X[/math] is continuous, then the restriction [math]f_{\vert [a,b]}[/math] is automatically also continuous.

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You just have to prove that it is continuous for every [math]x \in [a,b][/math]. In the proof just pick an arbitrary number [math]x \in [a,b][/math] and show that the funktion [math]f[/math] is continuous at that point.

 

Okay, that makes sense. But I'm a little confused with this part: "pick an arbitrary number [math]x \in [a,b][/math] and show that the function [math]f[/math] is continuous at that point". Is this suggesting that showing a single point being continuous shows something about the entire interval?

 

Could I also just exclusively look for discontinuities in the interval, where I can come to a conclusion based on a presence or lack thereof?

 

Also if you know that the funktion 0e9f4d2133005fb7bbaf8cdf703f0c23-1.png is continuous, then the restriction f3999124b99ff9f53400c5e79aba3cc4-1.png is automatically also continuous.

 

I thought as much. Thanks.

Edited by Amaton
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Okay, that makes sense. But I'm a little confused with this part: "pick an arbitrary number [math]x \in [a,b][/math] and show that the function [math]f[/math] is continuous at that point". Is this suggesting that showing a single point being continuous shows something about the entire interval?

 

Could I also just exclusively look for discontinuities in the interval, where I can come to a conclusion based on a presence or lack thereof?

Ah, but by picking an arbitrary number you are not showing that the function is continuous at a single point, you are showing that the function is continuous at every point in the set you picked your point from (think about it!). This, however, is just the definition of the function being continuous on the whole interval, because continuity is a property that is defined pointwise.

 

If you want to show that the function is not continuous, you can of course just pick a specific point, and show that it is not continuous at that point. Observe that you (usually) can't pick an arbitrary point when doing this. You have to look at the function and try to figure out where is behaves "badly". Pick that point, and show that it is not continuous. Observe that then you have shown only that the function is not continuous at that specific point. We still don't know anything about the continuity of the function at any other point. This is still enough to show that the function is not continuous on any set containing that specific point, because (again) the continuity was defined pointwise.

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Ah, but by picking an arbitrary number you are not showing that the function is continuous at a single point, you are showing that the function is continuous at every point in the set you picked your point from (think about it!).

 

Okay, I think there's a misunderstanding here on my part. You say to pick an arbitrary point and evaluate its continuity. I first took this to mean taking some random, specific point and evaluating the continuity for just that one, specific point in the interval. But I don't see how that works for this... So I'm guessing you're talking about an arbitrary point for the interval in general?

Edited by Amaton
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Okay, I think there's a misunderstanding here on my part. You say to pick an arbitrary point and evaluate its continuity. I first took this to mean taking some random, specific point and evaluating the continuity for just that one, specific point in the interval. But I don't see how that works for this... So I'm guessing you're talking about an arbitrary point for the interval in general?

Yes. To pick an arbitrary point is not the same as to pick any specific point. You have to show continuity at a point, and the only thing you can assume about the point is that it is in the set (eg. [0,1]). The point of this is that when you have shown how the argument is done, then you can use the same argument for any specific point. In other words you have shown that the arguments works for all points in the set, thus the continuity of the function is proved.

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Yes. To pick an arbitrary point is not the same as to pick any specific point. You have to show continuity at a point, and the only thing you can assume about the point is that it is in the set (eg. [0,1]). The point of this is that when you have shown how the argument is done, then you can use the same argument for any specific point. In other words you have shown that the arguments works for all points in the set, thus the continuity of the function is proved.

 

Okay. That makes sense. I was just confused earlier from the misunderstanding. So to sum it up, I should be able to prove this using an epsilon-delta approach for an arbitrary [math]x[/math], which represents all [math]x[/math] in the interval [math]I[/math] (right?).

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Okay. That makes sense. I was just confused earlier from the misunderstanding. So to sum it up, I should be able to prove this using an epsilon-delta approach for an arbitrary [math]x[/math], which represents all [math]x[/math] in the interval [math]I[/math] (right?).

Correct.

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