Mean value theorem $e^{f(t)}-e^{g(t)}=e^{h(t)}(f(t)-g(t)).$ Consider $f,g:\mathbb{R}\to \mathbb{R}$ two continuous functions.
For each $t\in \mathbb{R}$, using the mean value theorem, one can choose a number $h(t)$ such that
$$e^{f(t)}-e^{g(t)}=e^{h(t)}(f(t)-g(t)).$$
Can we choose the numbers $h(t)$ so that $h:\mathbb{R}\to \mathbb{R}$ is a continuous function?
 A: If $f(t) \ne g(t)$ then
$$
\begin{align}
e^{f(t)} - e^{g(t)} &= e^{g(t)} \frac{e^{f(t)-g(t)}-1}{f(t)-g(t)} \cdot (f(t) - g(t)) \\
&= e^{g(t)} u(f(t) - g(t)) \cdot (f(t) - g(t))
\end{align}
$$
with $u(x) = \frac{e^x-1}{x}$ for $x \ne 0$. We can now define $u(0) = 1$,
then $u$ is continuous and positive on all of $\Bbb R$, and
$$
e^{f(t)} - e^{g(t)} = e^{g(t)} u(f(t) - g(t)) \cdot (f(t) - g(t))
$$ 
holds for all $t$. Therefore
$$
h(t) = g(t) + \ln (u(f(t) - g(t))) =
\begin{cases}
\ln\left(\frac{e^{f(t)} - e^{g(t)}}{f(t)-g(t)}\right)& \text { if } f(t) \ne g(t) \\
g(t) & \text{ if }f(t) = g(t)
\end{cases}
$$
is a continuous solution. This can also be written as
$$
h(t) = \ln M_{lm}(e^{f(t)}, e^{g(t)})
$$
where $M_{lm}$ is the logarithmic mean.

More generally, this works for all strictly convex differentiable functions $F: \Bbb R \to \Bbb R $, and continuous functions $f, g$.
For $x \ne y$ let $c(x, y)$ be the unique solution of
$$
 F(x) - F(y) = F'(c(x, y)) (x-y) \, .
$$
$c(x, y)$ is uniquely determined because $F'$ is strictly increasing. We also know that $c(x,y)$ lies between $x$ and $y$, i.e.
$$
\min(x, y) \le c(x, y) \le \max(x, y) \, .
$$
It follows that
$$
 h(t) = \begin{cases}
c(f(t), g(t) & \text{ if } f(t) \ne g(t) \\
g(t) & \text{ if } f(t) = g(t)
\end{cases}
$$
is continuous and satisfies
$$
F(f(t)) - F(g(t)) = F'(h(t)) \cdot (f(t) -g(t))
$$
for all $t$.
