Continuous on [a,b] and differentiable on (a,b) - Why the different brackets?

acb

Can someone please explain why we always say a function is continuous on a closed [a,b] and differentiable on an open interval (a,b)??


Can we not have continuous functions on open intervals and differentiable on closed intervals?

Thanks for your help in advance!

sponsoredwalk

What are the conditions for a limit to exist?

What is a derivative?

acb

limit exists if sequences from the left and right equal the same thing and its not undefined

A derivative tells you how fast something is changing

sponsoredwalk

lol, yeah it does Well the definition of a derivative is:

[latex] \lim_{\Delta x \to 0} \ \frac{f(x \ + \ \Delta x) \ - \ f(x)}{\Delta x}[/latex]

& you see that a limit is involved in this description. Because you know that a
limit is only defined if both the left & right limits exist & are equal, at an
endpoint you'll only have one side of a limit defined on the interval if it's
closed. There are one-sided limits but I don't know how defining a one
sided limit in specific theorems @ each endpoint will affect the theorem
tbh, I'm sure it's either because the theorem fails @ the endpoint or it's
just laziness on the part of the authors :P

LeixlipRed

Understanding the formal epsilon-delta definition of a limit will make this more clear. Try here.

acb

Thank you sponsoredwalk - though I didnt really follow what you said

Looked at those videos you suggested LeixlipRed- found delta-E much easier to understand as this is what we'd be using in college.
It was good revision but i still dont get why its closed brackets for continous and open brackets for differentiable

MathsManiac

For the function to be differentiable at a point (x, f(x)), this limit must exist:

[latex] \lim_{\Delta x \to 0} \ \frac{f(x \ + \ \Delta x) \ - \ f(x)}{\Delta x}[/latex]

The limit only exists if the limit from the left and the limit from the right both exist and are equal.

At the endpoint of the interval, the limit from one side doesn't exist, since the function is not defined beyond the point concerned. (At the left endpoint, the limit from the left doesn't exist, because the function is undefined for x<a, which means that you cannot meaningfully refer to [latex] f(x \ + \ \Delta x)[/latex] when x= a and [latex] \Delta x [/latex] is negative.)

Since we cannot say that the left-limit and right-limit exist and are equal at x=a, we therefore cannot say that the limit exists at x=a.

sponsoredwalk

If you try to take the derivative of a function on the domain of [3,6) at
the point x = 3 you can't find the left hand limit, er... whatever limit
approaches from 0 up to 1 up to 2 up to 2.5 up to 2.99 up to 2.99999....
etc... as it's all outside the domain! Therefore you have failed to meet the
criteria for a limit to exist at this point because nothing exists past 3 so
there simply isn't a derivative here. Now, you can make an addendum that
a right-hand limit exists, i.e. the limit as x goes from 4 down to 3.5 down to
3.0001 down to 3.0000000000001 down to.... that you're approaching a
limit of 3 so we have a limit of 3 here but the derivative just isn't
applicable.

Now, if you want to take the derivative at 6 you can't either, it's just not
in the domain of the function. However, and an ε - δ limit would make it
clearer, you can get as close as you want to 6 without actually touching 6
to within some specified distance so you can always get a limit &
derivative no matter how close you are to 6. The fact that the domain is
open around 6 & the basic infinity of the number line you'll always
get a derivative or limit. No matter how close you want to get to 6 I can
get closer :cool: Pick a point & I'll show you one closer to 6 with a limit on
both sides that exists, ergo a derivative. At x = 3 you can't get any further
from 3 than 3 to the left so eventually there is a point where the limit is
one-sided so there is no derivative.

equivariant

acb wrote: »

Can someone please explain why we always say a function is continuous on a closed [a,b] and differentiable on an open interval (a,b)??

We don't - more on this below.

My guess is that you have come accross this when reading about the mean value theorem. In that case, the hypothesis insist that the function is continuous on [a,b] and differentiable on (a,b) becasue these are the weakest hypothesis that are necessary to ensure the conclusion of the theorem.

Can we not have continuous functions on open intervals and differentiable on closed intervals?

Thanks for your help in advance!

It is perfectly sensible to talk about a function that is differentiable on a closed interval [a,b]. However, the definition of differentiability at the endpoints of the interval (a and b) uses a one sided limit (left or right limit as appropriate). For example, it makes perfect sense to say that the function [latex] f(x) = x^{\frac{3}{2}}[/latex] is differentiable on the closed interval [latex] [0,\infty) [/latex] as long as we realise that the derivative at 0 is a one sided derivative.

acb

Thanks EVeryone for posting- its starting to become clearer!!:D
Just need to go off now and try to take all that in!

acb

Now, if you want to take the derivative at 6 you can't either, it's just not
in the domain of the function. However, and an ε - δ limit would make it
clearer, you can get as close as you want to 6 without actually touching 6
to within some specified distance so you can always get a limit &
derivative no matter how close you are to 6. The fact that the domain is
open around 6 & the basic infinity of the number line you'll always
get a derivative or limit. No matter how close you want to get to 6 I can
get closer :cool: Pick a point & I'll show you one closer to 6 with a limit on
both sides that exists, ergo a derivative. At x = 3 you can't get any further
from 3 than 3 to the left so eventually there is a point where the limit is
one-sided so there is no derivative.[/QUOTE]

Are we saying then that we cant take the derivative at 3 or 6 because its an open interval and we cant include them, we can only take the limit going to 3 from the right hand side, or 6 but only from the left hand side.

I know when we check limits we are supposed to check from the right and the left and they should equal. So are we saying that because we cant check 3 from the left (as outside our interval) thats the derv doesnt exist?:o.

sponsoredwalk

Yeah exactly! Well, for 6 it's more complicated, any point you pick below
6 will have a derivative as no matter what point (a/b) you pick I'll always
be able to find an [(a ± 1)/b] which will allow me to form a secant line
& therefore a limiting process to compute a derivative. However, in
[3,6) you have to remember that 6 is not in the domain! That's what that
notation means, you have every point up to 6 but not 6 in it.