1. ## Rolle/MVT Problem

I've been stating at this problem for a while and have no idea where to even start. Not looking for an answer, but really just a nudge in the right direction.

The homework is in a section covering Rolle's Theorem and the Mean Value Theorem.

Suppose $\displaystyle f$ is differentiable on $\displaystyle (a,b)$ except possibly at $\displaystyle x_o \in (a,b)$ and is continuous on $\displaystyle [a,b]$; assume $\displaystyle lim(x \rightarrow x_o) f'(x)$ exists. Prove that $\displaystyle f$ is differentiable at $\displaystyle x_o$ and $\displaystyle f'$ is continuous at $\displaystyle x_o$.

2. Originally Posted by Junesong
I've been stating at this problem for a while and have no idea where to even start. Not looking for an answer, but really just a nudge in the right direction.

The homework is in a section covering Rolle's Theorem and the Mean Value Theorem.

Suppose $\displaystyle f$ is differentiable on $\displaystyle (a,b)$ except possibly at $\displaystyle x_o \in (a,b)$ and is continuous on $\displaystyle [a,b]$; assume $\displaystyle lim(x \rightarrow x_o) f'(x)$ exists. Prove that $\displaystyle f$ is differentiable at $\displaystyle x_o$ and $\displaystyle f'$ is continuous at $\displaystyle x_o$.
maybe it is me, but there seems to be something off about this question. first it says we are not (or might not be) differentiable at a point, and then it asks us to prove that we are? $\displaystyle f'$ continuous at $\displaystyle x_0$ means that $\displaystyle \lim_{x \to x_0}f'(x) = f'(x_0)$. the right side of this equation gives us problems when we are in the case where $\displaystyle f$ is not differentiable at $\displaystyle x_0$

3. Since $\displaystyle \lim_{x \rightarrow x_0}f'(x)$ exists the second part of the question follows immediately. This shows $\displaystyle f'(x)$ is sequentially continuous and is therefore continuous.

For the first part, since $\displaystyle f(x)$ is continuous on $\displaystyle [a,b]$ and differentiable on $\displaystyle (a,b)$ then, by MVT, $\displaystyle \exists \ x_0 \in (a,b) \ s.t \ f'(x_0)=\frac{f(b)-f(a)}{b-a}$.

We already know that $\displaystyle f'(x_0)=\lim_{x \rightarrow x_0} f'(x)$ from the question. Therefore we have:

$\displaystyle f'(x_0)=\lim_{x \rightarrow x_0}f'(x)=\frac{f(b)-f(a)}{b-a}$.

From the question $\displaystyle f(b),f(a)$ and $\displaystyle b-a$ are defined ($\displaystyle f(b)$ and $\displaystyle f(a)$ are defined since $\displaystyle f(x)$ is continuous and $\displaystyle x_0 \neq a,b$ from how we defined $\displaystyle x_0$. Also, $\displaystyle b-a \neq 0$ from how we defined a and b.) This gives the conclusion that $\displaystyle f'(x_0)$ exists so $\displaystyle f(x)$ is differentiable at $\displaystyle x_0$ as required.

It is an oddly phrased question though

4. Originally Posted by Showcase_22
For the first part, since $\displaystyle f(x)$ is continuous on $\displaystyle [a,b]$ and differentiable on $\displaystyle (a,b)$ then, by MVT, $\displaystyle \exists \ x_0 \in (a,b) \ s.t \ f'(x_0)=\frac{f(b)-f(a)}{b-a}$.
How do we know that the $\displaystyle x_o$ that you used in the MVT is the same $\displaystyle x_o$ mentioned in the problem? As Jhevon stated, the problem is worded very oddly and that may be a lot of the problem in me grasping it.

MVT states that $\displaystyle \exists c \in (a,b)$ such that $\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$, but if $\displaystyle x_o \in c$ defines a single point, then we can clearly have that $\displaystyle c \neq x_o$.

I was thinking, and I don't know if this also holds, but we know by MVT that $\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$. For any other point $\displaystyle x \in (a,b)$ such that $\displaystyle x \neq c$, is there guaranteed a subinterval $\displaystyle x \in (m,n) \subset (a,b)$ such that $\displaystyle f'(x)=\frac{f(n)-f(m)}{n-m}$?

Sorry if this seems jumbled a bit, but I greatly appreciate all the help.

5. oh, I thought $\displaystyle x_0$ was a generic point on $\displaystyle (a,b)$.

What I wrote still holds, turn the $\displaystyle x_0$ in my proof to a $\displaystyle c$ and show that it's continuous and differentiable for $\displaystyle c$. Therefore $\displaystyle x_0$ can equal $\displaystyle c$ so $\displaystyle f(x)$ must be differentiable at $\displaystyle x_0$.

(If every $\displaystyle f(t)$ is continuous and differentiable where $\displaystyle t \in (a,b)$ then if $\displaystyle x_0 \in (a,b)$ then $\displaystyle f(x_0)$ must be must continuous and f must be differentiable at $\displaystyle x_0$).

$\displaystyle x_0 \in c$
This is incorrect. $\displaystyle \in$ can only be used for sets but c is a point.

6. Originally Posted by Showcase_22
This is incorrect. $\displaystyle \in$ can only be used for sets but c is a point.
I definitely meant $\displaystyle x_o \in (a,b)$, not $\displaystyle x_o \in c$. That's what I get for working at 7am...

Thanks for all the help, it's much appreciated.