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Mathematics 14 Online
OpenStudy (anonymous):

how can you tell if a function is continuous and differentiable ? Looking for quick answer

OpenStudy (anonymous):

There is no quick way to know, other than from experience. Basically, you have to know the definitions, and analyze the function in question.

OpenStudy (anonymous):

the only quick way, I suppose, is knowing from experience.

OpenStudy (anonymous):

for example x / (2-x) would that be continuous or differentiable ?

OpenStudy (anonymous):

can you tell just by looking at it ?

OpenStudy (anonymous):

We can see it is continuous since lim x->a is equal to f(a). Differentiability at a point requires f'(a) exist. Does it for this function ?

OpenStudy (anonymous):

a is 2 right ?

OpenStudy (anonymous):

thats what you are refering a to ?

OpenStudy (anonymous):

'a' can be any number in the domain.

OpenStudy (anonymous):

yes so you took the 2 from the function right ?

OpenStudy (anonymous):

i mean 2 from the function would equal a ?

OpenStudy (anonymous):

2 from the function? you mean the 2 in: x/(2-x) ?

OpenStudy (anonymous):

yes it would be plugged in for a ?

OpenStudy (anonymous):

No. The number 'a' is a value of x. So if f(x) = x/(2-x), then f(a) = a/(2-a). Note, if a =2, then a would not be in the domain since f is not defined at x = 2.

OpenStudy (anonymous):

hey are you there ? i just graphed x/(2-x) and it wasn't continuous

OpenStudy (anonymous):

It is continuous over its domain (no holes, and no truncated points)

OpenStudy (anonymous):

no according to the graph it isn't continous it starts at another direction that isn't continuity right ?

OpenStudy (anonymous):

look http://www.math24.net/discontinuous-functions.html

OpenStudy (anonymous):

I see. Yes, there is an asymptote at x = 2, which means the function is not continuous at that point, but it is also not defined there. At all other points, f is continuous and differentiable.

OpenStudy (anonymous):

Okay thanks for clarifying!

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