Intuitive approach towards the auxiliary function of Lagrange's mean value theorem proof

Question

I have often seen many people using another auxiliary function $g(x)$ such that $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}\cdot x$ where $f(x)$ is our original function continuous in $[a,b]$, differentiable in $(a,b)$ and $g(x)$ satisfies all the condition of Rolle's theorem, to prove Lagrange's mean value theorem. What is the intuition or methodology to arrive at that function $g(x)$?

Answer 1

Usually the auxiliary function is $$g(x)=f(x)-\left(\frac{f(b)-f(a)}{b-a}\cdot (x-a)+f(a)\right)$$ which is the subtraction of the original function $f(x)$ and the function related to the secant line through the points $(a,f(a))$ and $(b,f(b))$. In this way, $g(a)=g(b)=0$ and we may apply Rolle's theorem.

The same works if we replace $g(x)$ with $h(x)=g(x)+c$ for any constant $c\in \mathbb{R}$. Then $h(a)=h(b)=c$ and again we may apply Rolle's Theorem. Note that $g'(x)=h'(x)$ and the rest of the proof of Lagrange's theorem remains the same.

The function $$h(x)=f(x)-\frac{f(b)-f(a)}{b-a}\cdot x$$ has just that form with $$c=-\frac{f(b)-f(a)}{b-a} a+f(a)=\frac{bf(a)-af(b)}{b-a}.$$ In another way, this $h(x)$ is the function $g(x)$ where we removed all the additive constants.