Continuity, Differentiable at a point and limits

[Dark Horse]

I was wondering, is the function abs(x-2) continuous or discontinuous? Does the limit exist at this point, and if so, what is the significance of this in terms of being differentiable at a point?

Thanks

[luken93]

Anywhere that there is a cusp (sharp point that changes direction), a function is not continuous, and cannot be differentiated...

There is probably someone who will have a better understanding than I, but thats the basics of it

[98.40_for_sure]

The graph of |x-2| is continuous at any x value. However the point (2,0) is not differentiable. Continuous meaning the limit approaching from the left and the right is equal to the y value at that point. Differentiable meaning it has a definable gradient, which a "cusp" does not have.

[TrueTears]

consider a particle, travelling to x = 2, if we can parametrize the path with some variable t, namely, r(t) = <x(t), y(t)> then r'(t) = 0 when x = 2, the result follows

[98.40_for_sure]

consider a particle, travelling to x = 2, if we can parametrize the path with some variable t, namely, r(t) = <x(t), y(t)> then r'(t) = 0 when x = 2, the result follows

Bro... we don't understand chinese

[TrueTears]

i assume you would know since u are a fan of FTC + you do spesh and i assume you have covered vector functions, if not then you better revise!

[98.40_for_sure]

care to explain the last bit of what you said?

"then r'(t) = 0 when x = 2, the result follows"

[TrueTears]

its the same as the interpretation as the vector functions you've learnt in spesh, haven't you learnt it?

the result is what you have stated earlier in this thread.

r(t) is the position vector

r'(t) is the the derived vector.

the rest is trivialities from spesh

[98.40_for_sure]

its the same as the interpretation as the vector functions you've learnt in spesh, haven't you learnt it?

the result is what you have stated earlier in this thread.

r(t) is the position vector

r'(t) is the the derived vector.

the rest is trivialities from spesh

Lol wut? what does that have to do with continuity, differentiability and limits

[TrueTears]

http://en.wikipedia.org/wiki/Parametric_continuity#Parametric_continuity

functions don't always have to be expressed in terms of x and y, by parametrization and seeing things from a different perspective yields alot more results (as you have shown yourself in your sigs, so im suprised you didnt know )


let me put it in english for you (aka non rigorous maths language) consider a particle as a car, if the path it travels on is smooth then it shouldn't need to "stop" when it turns around a corner, but here the cusp at x = 2 makes the particle stop and turn thus not smooth and thus non differentiable.

[98.40_for_sure]

Yeah, that's just too much... i think my explanation suffices for VCE level

[TrueTears]

nah u clearly surpass VCE level with your superior knowledge of

[98.40_for_sure]

I dream about FTA and FTC in my sleep. Soon i will invent my own theorem. Just you watch, i will be the next big thing. Bigger than pytharogas.

[TrueTears]

known as the "do arts" theorem

[sajib_mostofa]

The most simple definition I have of a continuous function is that you don't take your pen off the paper when you draw the graph