differentiability (1 Viewer)

Aesytic

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Hey guys,

Draw a sketch of a possible curve that is differentiable in the domain -a<=x<=a

How would you sketch this, if the endpoints of a curve aren't differentiable? Would you have a continuous curve and then just mark the points x=a and x=-a, and make the part in between those values differentiable?
 
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tohriffic

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Hey guys,

Draw a sketch of a possible curve that is differentiable in the domain -a<=x<=a

How would you sketch this, if the endpoints of a curve aren't differentiable? Would you have a continuous curve and then just mark the points x=a and x=-a, and make the part in between those values differentiable?
Does that mean the graph can be differentiated outside that given domain? Which is probably the question you are asking.. and to be honest, I think that is the only way you can do it because you're absolutely correct, end points cannot be differentiated. :)
 

Aesytic

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Yeah, I think that's the only way, but the marking guidelines [it's from my school's past paper] say that the answer is "Any continuous smooth curve with endpoints at x=a and x=-a", and 1 mark is awarded for "clearly indicated endpoints". That kind of threw me off, is it saying that the curve shouldn't be continuous for all real x, and if it is, then wouldn't it be not differentiable throughout because of the endpoints not being differentiable?
 

tohriffic

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Yeah, I think that's the only way, but the marking guidelines [it's from my school's past paper] say that the answer is "Any continuous smooth curve with endpoints at x=a and x=-a", and 1 mark is awarded for "clearly indicated endpoints". That kind of threw me off, is it saying that the curve shouldn't be continuous for all real x, and if it is, then wouldn't it be not differentiable throughout because of the endpoints not being differentiable?
Could you email your teacher and enquire about that question? Answers even school test paper answers can be wrong.. or.. maybe we're missing something. D: I kinda wanna know how that works. :L
 

Aesytic

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Yeah, I'm probably gonna ask him tomorrow, just before my 2 unit assessment lol
 

Aesytic

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Simple, open circles at x=+/-a.
but the question asked for the domain -a<=x<=a, so they should be filled-in circles. sorry, don't know how to use latex :/
anyway, my teacher was away today and I got a substitute, but a lot of my friends who are smarter than me suggested drawing a curve that was continuous for all real x and to just have the part between x=a and x=-a differentiable, with the points on the curve where x=+/- a filled in, so I'm guessing that would be the best way to go about that answering that question
 

tohriffic

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That sounds reasonable enough but I still find that a very iffy question. > ^ <

Thanks for that Aesytic! At least I can sleep easy lol.
 

cutemouse

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If I remember my first/second year university maths correctly, differentiability is very hard to prove (continuity, however is not).

But, it should be any "smooth" curve with the endpoints defined...
 

D94

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but the question asked for the domain -a<=x<=a, so they should be filled-in circles. sorry, don't know how to use latex :/
anyway, my teacher was away today and I got a substitute, but a lot of my friends who are smarter than me suggested drawing a curve that was continuous for all real x and to just have the part between x=a and x=-a differentiable, with the points on the curve where x=+/- a filled in, so I'm guessing that would be the best way to go about that answering that question
It doesn't matter if the domain is -a<=x<=a. The end points can still be open circles because the drawn curve is within the domain.
 

Sy123

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You cannot differentiate any points on asymptotes because there are no points on an asymptote, and cant be differentiated
Here is a poor MS Paint job by me to give an example of such a graph
Where x=a and x=-a are vertical asymptotes of the graph
 

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