Intro to Analysis (Differentiation)

[bloynoys]

Homework Statement

Prove or disprove:
Suppose f:[a,b]->R is continuous. If f is diff on interval (a,b) and f'(x) has a limit at b, then f is diff at b.

Homework Equations

We say that f is differentiable at x0 to mean that there exists a number A such that:
f(x)=f(x0)+A(x-x0)+REM
where,
lim(x->x0) REM(x)/(x-x0) = 0

The Attempt at a Solution

We will prove f is diff at b by showing that that there exists a number A so that f(x)=f(b)+A(x-b)+REM
so,
lim(x->x0) REM(x)/(x-b) = 0

I have gotten good at normal proofs in this course and am very confused on how to build proofs with this diff theorem. I know that this is true, but am confused how to fashion the plan and how to start the proof. How do I establish the existence of an A that satisfies this?

Thanks!

 

[Dick]

Write a difference quotient for f'(b) and use the Mean Value Theorem. That works, doesn't it?

 

[bloynoys]

Hmm. How would I go about doing that? I know by the mean value theorem there is a c in (a,b) that is satisfied by f'(c)=f(b)-f(a)/b-a but how can I show that since f'(c) exists f'(b) exists since it is not for every c in the interval?

 

[Dick]

Hmm. How would I go about doing that? I know by the mean value theorem there is a c in (a,b) that is satisfied by f'(c)=f(b)-f(a)/b-a but how can I show that since f'(c) exists f'(b) exists since it is not for every c in the interval?

f'(b)=limit x->b (f(b)-f(x))/(b-x). There is a c such that f'(c) equals that quotient in (x,b).

 

[bloynoys]

So I can say that because f'(x) has a limit at b the function limit x->b (f(b)-f(x))/(b-x)=f'(b)?

So a proof would go something like:

We will prove f is diff at b by showing that f'(b)=f'(b)=limit x->b (f(b)-f(x))/(b-x).

Since f'(x) has a limit at b, we know limit x->b (f(b)-f(x))/(b-x)=f'(b).

Thus, we know that the function is diff. at b.

 

[Dick]

So I can say that because f'(x) has a limit at b the function limit x->b (f(b)-f(x))/(b-x)=f'(b)?

So a proof would go something like:

We will prove f is diff at b by showing that f'(b)=f'(b)=limit x->b (f(b)-f(x))/(b-x).

Since f'(x) has a limit at b, we know limit x->b (f(b)-f(x))/(b-x)=f'(b).

Thus, we know that the function is diff. at b.

You need to mention the mean value theorem and show where you used it.

 

[bloynoys]

Alright I have been working on this proof all dayish trying to get everything work out right.

I have for my plan:

Suppose x<c<b

We will prove f'(c) has a limit m at b by showing that for every ε>0 there exists an δ>0 so that for every x in (a,b) 0<abs(x-b)<δ where abs(f'(c)-m)<ε.

Consider ε>0 arb.
Since by the mean value theorem and the def of derivative we know that limit x->b (f(b)-f(x))/(b-x)=lim c->b (f'(c)) and (f(b)-f(a))/(b-a)=f'(c) as a<x<c<b. We know that there is a δ that satisfies this. Choose such a δ.

Consider x in (a,b) arb.

for such an x,

abs(f'(c)-m))...

Then I don't know how to build that final inequality to make it work.


Any suggestions?

 

[Dick]

Alright I have been working on this proof all dayish trying to get everything work out right.

I have for my plan:

Suppose x<c<b

We will prove f'(c) has a limit m at b by showing that for every ε>0 there exists an δ>0 so that for every x in (a,b) 0<abs(x-b)<δ where abs(f'(c)-m)<ε.

Consider ε>0 arb.
Since by the mean value theorem and the def of derivative we know that limit x->b (f(b)-f(x))/(b-x)=lim c->b (f'(c)) and (f(b)-f(a))/(b-a)=f'(c) as a<x<c<b. We know that there is a δ that satisfies this. Choose such a δ.

Consider x in (a,b) arb.

for such an x,

abs(f'(c)-m))...

Then I don't know how to build that final inequality to make it work.


Any suggestions?

You don't need epsilons or deltas. You know f'(b)=limit x->b (f(b)-f(x))/(b-x). The MVT tells you that there is a c(x) in (x,b) such that f'(c(x))=(f(b)-f(x))/(b-x). As x->b it must be that c(x)->b. Since you know f'(c(x)) approaches a limit as c(x)->b doesn't that tell you the difference quotient must have a limit?

 

[bloynoys]

Since you know f'(c(x)) approaches a limit as c(x)->b doesn't that tell you the difference quotient must have a limit?

May be a dumb question but we know there is a limit at b, but trying to say if it is differentiable. How can we make the leap that there is a limit at b to that it is differentiable at b?

I am following all of your logic (hopefully!) but that is the part I am hung up about. Thanks!

 

[Dick]

May be a dumb question but we know there is a limit at b, but trying to say if it is differentiable. How can we make the leap that there is a limit at b to that it is differentiable at b?

I am following all of your logic (hopefully!) but that is the part I am hung up about. Thanks!

If the limit of the difference quotient exists then f(x) is differentiable at b. At least, it has a one sided derivative. There's not much you can say about the other side because f is only defined on [a,b].

 

[Whovian]

I feel like disproving it. Take [itex]\displaystyle f\left(x\right)=\sqrt[3]{x}[/itex]. Take a = -1, b = 0. Is it differentiable on (-1,0)? Yes. Note that it's open. Is it continuous? Yes. f'(a) as a->0 does have a limit, -∞. (Yes, this is considered a limit.) But ... it's not differentiable at 0.