# Then is $f_a$ continious?

Excuse me for the bad title, here's the question

Given a differentiable function defined on R. For a given number $a$, $\forall x\in \mathbb R, x\neq a$, by mean value theorem, there exists a $\xi$ between $a$ and $x$ such that $f'(\xi)=\frac {f(x)-f(a)}{x-a}$. Suppose it's unique for all $x\neq a$. Therefore $\forall x\neq a$ we define $f_a$ as $f_a(x)=\xi$. Then is $f_a$ continious?

• How can there be a unique number between two unequal real numbers? Nov 21 '14 at 2:49
• Maybe what you mean is $\forall x \in \mathbb R, x\neq a\,\exists\,\xi$ between $a$ and $x$ such that $f_a(x)=\xi$? Nov 21 '14 at 2:51
• I'm supposing so Nov 21 '14 at 2:53
• Is $f_a$ defined at the point $a$? In particular does one need to show that $f_a$ is continous on all of $\mathbb{R}$ or on $\mathbb{R}\backslash\{a\}$? Nov 21 '14 at 2:53
• Any suggestions? Nov 21 '14 at 2:57

The mean value theorem says that for any $x\ne a$, there exists $\xi \in [a,x]$ such that $$\frac{f(x)-f(a)}{x-a} = f'(\xi).$$ I believe the question is: if the choice of $\xi$ is unique, then if we define $f_a(x)$ to be that $\xi$ for a given $x$, then is $f_a$ continuous? I also assume that "differentiable" means "the derivative exists and is continouus".
To me it seems that the assumption that the $\xi$ is always unique implies that $f'$ is monotonic. If that's true, then $(f')^{-1}$ exists and is continuous, and we have $$f_a(x) = (f')^{-1} \bigg( \frac{f(x)-f(a)}{x-a} \bigg).$$ As a composition of continuous functions, this is clearly continuous everywhere except $x=a$. If we further define $f_a(a)=a$, then it matches its limit there and hence is continuous as well.
Can anyone (dis)prove the claim that $f'$ is monotonic under the uniqueness assumption?
• I agree that you've shown that no ray emanating from $(a,f(a))$ can intersect the graph in two other points. What's the proof that this property implies that $f$ is concave (or convex) everywhere? Nov 21 '14 at 5:00