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

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 $f$ is differentiable on $(a,b)$ except possibly at $x_o \in (a,b)$ and is continuous on $[a,b]$; assume $lim(x \rightarrow x_o) f'(x)$ exists. Prove that $f$ is differentiable at $x_o$ and $f'$ is continuous at $x_o$.
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? $f'$ continuous at $x_0$ means that $\lim_{x \to x_0}f'(x) = f'(x_0)$. the right side of this equation gives us problems when we are in the case where $f$ is not differentiable at $x_0$

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

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

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

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

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

It is an oddly phrased question though

4. Originally Posted by Showcase_22
For the first part, since $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then, by MVT, $\exists \ x_0 \in (a,b) \ s.t \ f'(x_0)=\frac{f(b)-f(a)}{b-a}$.
How do we know that the $x_o$ that you used in the MVT is the same $x_o$ 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 $\exists c \in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$, but if $x_o \in c$ defines a single point, then we can clearly have that $c \neq x_o$.

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

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

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

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

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

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

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

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