Question

Can anyone please explain to me in as simple as terms as possible what this theorem means?
(calculus)
If f(x) is differentiable at x=a then f(x) is continuous at x=a. I understand the basic concept (I think) but I don't understand when it comes to the composition?...

Answer 1

Could you give maybe an example of a composition you don't agree with or something? I think we're at a good starting place.

Answer 2

Well it's not that I don't agree with it.. i just don't fully understand what it means?

Answer 3

\[ \lim_{x \rightarrow a} f(x)-f(a)/x-a = f'(a)\]

Answer 4

The theorem is basically saying that if the limit as you approach that point is the same as the value at that point, then we're just saying that's what it means to be continuous.

You could have a removable discontinuity there, so although the limit approaches that point from both sides, the actual point doesn't exist there, so it's not continuous. Or maybe the function is evaluated at that point but the limit from both sides doesn't approach that point, like a jump discontinuity. I don't know if that answers your question or not since there's sort of a deep end we can fall into here that I'm trying to avoid haha.

Answer 5

I do understand what your saying. completely. Which is kind of where my confusion lies in not so much what you just said, but what i don't really comprehend is the proof when you have to write it out?... I guess im looking for like a breakdown of each step in the proof?... to truly understand the equation form of it. (i.e. as x approaches a)

Answer 6

in my math book it has a proof and breaks it down, but it goes through these weird steps and I don't really understand what the steps mean? if that makes sense?

Answer 7

Personally I am not the person to answer this question for you because I don't believe that there is any value in the proof and set theory imitation version of calculus, it's just a bunch of ad hoc arguments that are meaningless formalism... Sorry.

Answer 8

well, i do appreciate it. Could i ask you one other question. Besides x= \[\left| x \right|\] is there another example of a function that is continuous on its domain but not differentiable?

Answer 9

at a point?

Answer 10

Although kind of the same sort of thing, this might be what you want: https://www.desmos.com/calculator/6dy2bksacn

Answer 11

y = x^(2/3)

Answer 12

ok. Sure. So really unless the function has a corner there's not typically any other variation of a function that is going to be undifferentiable unless it's piecewise or sqrt x.?

Answer 13

https://www.desmos.com/calculator/rlchoeozow

Answer 14

so if I put down what the proof from my book said to prove my question earlier... would you be able to elaborate on some of the steps for me?

Answer 15

To prove that f is continuous at a, we have to show that limit x>a f(x)=f(a). We do this by showing that the differences f(x)-f(a) approaches 0 as x approaches a. The given information is that f is differentiable at a, that is,
f ' (a) =limx>a f(x)-f(a)/x-1 exists.
to connect the given and the unknown, we divide and multiply f(x) - f(a) by x-a (which we can do when x doesn't equal a):
f(x)-f(a)= f(x)-f(a)/x-a * (x-a)
thus using the product law we can write
lim x>a [f(x)-f(a)] = lim x>a f(x)-f(a)/x-a* (x-a)
= lim x>a f(x)-f(a)/x-a * lim x>a (x-a)
= f ' (a) * 0
= 0

thats what the books shows.... but what confuses me is why you multiply by x-a and why it becomes zero?....

Answer 16

zero is the slope?

Answer 17

can you write out the proof step by step with spaces

Answer 18

its difficult to read that

Answer 19

or take a screen shot of the proof

Answer 20

zero is the slope?