Application of lagrange mean value theorem Let $f$ be continuous on $[a,b]$ , differentiable on ($a,b$), and its given $f(a)=a, f(b)=b$. We need to prove the following:


*

*$f'(c1)+f'(c2)=2,$ , for some  $c1,c2 \in (a,b)$, and $c1\neq c2$.

*$\frac{1}{f'(c1)}+ \frac{1}{f'(c2)}=2,$ , for some  $c1,c2 \in (a,b)$, and $c1\neq c2$.

*$ f'(c)=\frac{1}{b-c} + \frac{1}{a-c}$ , for some  $c\in (a,b).$
I got the insight for the 1st one: since we need to apply the theorem twice, it is reasonable to split the intervals into two sub intervals$(a,\frac{a+b}{2})$ and $(\frac{a+b}{2},b)$. After applying the Lagrange mean value theorem on each of these intervals and adding, we easily prove 1.
I thought of a similar argument for 2, but the reciprocals make things messy.
I am absolutely clueless about 3.
Edit: option 3 seems similar to cauchy mean value theorem, but I am still confused on how to proceed.

For $2$, since $f(a) = a$ and $f(b) = b$, with $a \lt b$, since $f$ is continuous, note the intermediate value theorem says there's a point, call it $a \lt d \lt b$, where
$$f(d) = \frac{a + b}{2} \tag{1}\label{eq1A}$$
Now, use the Lagrange mean value theorem on the sub-intervals $(a, d)$ and $(d, b)$ to get for that some $c_1 \in (a, d)$ you have
$$f'(c_1) = \frac{f(d) - f(a)}{d - a} \implies \frac{1}{f'(c_1)} = \frac{d - a}{f(d) - f(a)} = \frac{d - a}{\frac{b + a}{2} - a} = \frac{2(d - a)}{b - a} \tag{2}\label{eq2A}$$
and for some $c_2 \in (d, b)$ you have
$$f'(c_2) = \frac{f(b) - f(d)}{b - d} \implies \frac{1}{f'(c_2)} = \frac{b - d}{f(b) - f(d)} = \frac{b - d}{b - \frac{b + a}{2}} = \frac{2(b - d)}{b - a} \tag{3}\label{eq3A}$$
Thus, you then get
$$\begin{equation}\begin{aligned}
\frac{1}{f'(c_1)}+ \frac{1}{f'(c_2)} & = \frac{2(d - a)}{b - a} + \frac{2(b - d)}{b - a} \\
& = \frac{2d - 2a + 2b - 2d}{b - a} \\
& = \frac{2(b - a)}{b - a} \\
& = 2
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Of course, you also have that $c_1 \neq c_2$.
For $3$, consider
$$g(x) = \ln(b - x) + \ln(x - a) + f(x), \; x \in (a,b) \tag{5}\label{eq5A}$$
which is continuous and differentiable on $(a,b)$. You also have
$$\begin{equation}\begin{aligned}
g'(x) & = -\frac{1}{b - x} + \frac{1}{x - a} + f'(x) \\
& = -\frac{1}{b - x} - \frac{1}{a - x} + f'(x)
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Note since $f(x)$ is continuous on a closed interval, it's bounded. This means that as $x \to a^{+}$, you have $g(x) \to -\infty$ and as $x \to b^{-}$, you have $g(x) \to -\infty$. From \eqref{eq5A}, you have
$$\begin{equation}\begin{aligned}
e_m & = g\left(\frac{b+a}{2}\right) \\
& = \ln\left(b - \frac{b+a}{2}\right) + \ln\left(\frac{b+a}{2} - a\right) + f\left(\frac{b+a}{2}\right) \\
& = 2\ln\left(\frac{b-a}{2}\right) + f\left(\frac{b+a}{2}\right)
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Consider any $e \lt e_m$. Thus, since $g(x) \to -\infty$ as $x \to a^{+}$, by the intermediate value theorem, there's a point $d_1 \in \left(a, \frac{b+a}{2}\right)$ where $f(d_1) = e$, and likewise (since $g(x) \to -\infty$ as $x \to b^{-}$) there's a point $d_2 \in \left(\frac{b+a}{2},b\right)$ where $f(d_2) = e$. Thus, by Lagrange's mean value theorem, there's a $c \in (d_1,d_2)$ such that
$$g'(c) = \frac{f(d_2) - f(d_1)}{d_2 - d_1} = \frac{e - e}{d_2 - d_1} = 0 \tag{8}\label{eq8A}$$
Thus, from \eqref{eq6A}, you get
$$g'(c) = 0 = -\frac{1}{b - c} - \frac{1}{a - c} + f'(c) \implies f'(c) = \frac{1}{b - c} + \frac{1}{a - c} \tag{9}\label{eq9A}$$
