Mean value theorem and the continuity of step length Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}, f\in C^1$. Then for two points $a,b \in \mathbb{R}$, we know by mean value theorem we know that there exists $t \in (0, 1)$ such that
$$
f^\prime(a+t(b-a)) = \frac{f(b)-f(a)}{b-a}
$$
We can view $t$ as a parameter collectively decided by $a,b$ and $f(\cdot)$. Now, suppose that we hold $a$ to be fixed and move $b$, what can be said about the changes in $t$ corresponding to the changes in $b$ (e.g. can we argue if $t(a,b)$ is continuous)? 
Thanks in advance for your hints/suggestions!
 A: After thinking about this a bit, I realized it's a simple application of the implicit function theorem.  Let
$$
    \newcommand{\R}{\mathbb{R}}
    F(a,b,t) = f(b) - f(a) - f'(a+t(b-a))(b-a)
$$
Let $U = \{(a,b): b > a\} \subset \R^2$ and $V = (0,1) \subset \R$.  The mean value theorem says that for all $(a_0,b_0) \in U$ there exists $t_0 \in V$ such that $F(a_0,b_0,t_0) = 0$.  
The derivatives of $F$ are:
\begin{align*}
    \newcommand{\pderiv}[2]{\frac{\partial #1}{\partial #2}}
    \pderiv{F}{a} &= - f'(a) + f'(a+t(b-a)) - f''(a+t(b-a))(1-t)(b-a) \\
    \pderiv{F}{b} &=  f'(b) - f'(a+t(b-a)) - f''(a+t(b-a))t(b-a) \\
    \pderiv{F}{t} &= -f''(a+t(b-a))(b-a)^2
\end{align*}
If $f$ is $C^2$, then $F$ is $C^1$, and that is smooth enough to apply IFT.  The invertibility criterion for the IFT is that $\pderiv{F}{t}(a_0,b_0,t_0) \neq 0$.  Since $b_0 \neq a_0$, this is guaranteed if only $f''(a_0+t(b_0-a_0)) \neq 0$.
To summarize, if $f$ is $C^2$, $\frac{f(b_0)-f(a_0)}{b_0-a_0} = f'(a+t(b-a))$ and $f''(a_0 + t_0(b_0-a_0) \neq 0$, then $t$ is a $C^1$ function of $(a,b)$ in a neighborhood of $(a_0,b_0)\cap U$.  Moreover,
\begin{align*}
    \pderiv{t}{a} &= -\frac{\pderiv{F}{a}}{\pderiv{F}{t}}
        = \frac{f'(a+t(b-a))-f'(a)}{f''(a+t(b-a))(b-a)^2} - \frac{1-t}{b-a} \\
    \pderiv{t}{b} &= -\frac{\pderiv{F}{b}}{\pderiv{F}{t}}
        = \frac{f'(b) - f'(a+t(b-a))}{f''(a+t(b-a))(b-a)^2} - \frac{t}{b-a} \\
\end{align*}
