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Give an example of a function which is continuous on the whole real line but fails to be differentiable at exactly 2 points.
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This isn't too bad of a question. Just remember one thing: You can define a function piecewise:
f(x) =
x^2 if x > 0
-x if x <= 0
This function is continuous. It is definitely differentiable at everywhere but 0. Is it differentiable at 0?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Another example:
This one is continuous everywhere and not differentiable at exactly two points (which is what is required).
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Another example:
This one is continuous everywhere and not differentiable at exactly two points (which is what is required).
Oh!! Thanks a lot.:lol:
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Youre welcome. The important question though is: do you know why it is not differentiable at those two points where its not differentiable?
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Youre welcome. The important question though is: do you know why it is not differentiable at those two points where its not differentiable?
I think the graph has holes on (-1,1).
Am I right?
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JaneFairfax wrote:Youre welcome. The important question though is: do you know why it is not differentiable at those two points where its not differentiable?
I think the graph has holes on (-1,1).
Am I right?
No, you are not. The graph having "holes" would mean the function is not continuous. That is not the case here. Have you drawn the graph? I recommend you do that.
Here's another one with absolute value. Enjoy!
y = | 6 - x^2 |
First figure it out without the absolute value signs, then anything below the
x-axis, bounce it up to the positive area above the x-axis.
igloo myrtilles fourmis
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