Let $f: [a, b]\to\mathbb{R}$ be continuous on the interval $[a,b]$. Assume that $f$ is differentiable on the open interval $(a,b)$. Assume that there is a number $c \in\mathbb{R}$ so that $$\lim_{x\to b^-}f'(x)=c$$ Show that $f$ is differentiable in the endpoint $b$ with $f '(b)=c$.
Hint: use the mean value theorem. I hope you can help me, thanks :-)
Let's apply the mean value theorem on every segment $[x,b]$, $x \in [a,b)$ : you know, because $f$ is differentiable of $(x,b)$ and continuous in $x$ and $b$ that there exists $c_x \in (x,b)$ such that $$\frac{f(x)-f(b)}{x-b} = f'(c_x)$$
Now let $x$ tend to $b$. Because $c_x \in (x,b)$, you have that $c_x$ also tends to $b$, so $f'(c_x)$ tends to $c$. So the quantity $$\frac{f(x)-f(b)}{x-b}$$ has a finite limit when $x$ tends to $b$, which means that $f$ is differentiable in $b$, and $f'(b)$ is the value of this limit, i.e. $f'(b) =c$.
