On the method of constructing auxiliary function for a class of middle value theorem proof problem

Let's start with the proof of the $lagrange$ theorem.

Almost all mathematical textbooks (such as advanced mathematics, Mathematical analysis), in proving this theorem, use geometric meanings to construct functions $\varphi (x) $$=$ $f (x)-F (a)-\frac{f (b)-F (A)}{b-a} (x-a) $. Then the $rolle$ theorem is used to prove it.

Naturally, in the face of a problem of proof, especially the proof of the median theorem, few people will think of geometrical meaning. In a sense, this method is not worthy of promotion, do you think of every one of these questions? In fact, many proofs can not find the geometrical meaning to illustrate.

Now we are trying to construct the appropriate auxiliary function using theoretical analysis, so we will first restate the $lagrange$ median theorem:

Set the function $f (x) $ in $[a,b]$ continuous, in $ (a, b) $, then $\exists \xi$$\in$$ (A, b) $, making $f ' (\xi) =\frac{f (b)-F (a)}{b-a}$.

To prove this equation, we might consider $f ' (x) =\frac{f (b)-F (a)}{b-a}$ as a differential equation, easy to calculate $f (x) =\frac{f (b)-F (a)}{b-a}x+c$

Solve the $c=f (x)-\frac{f (b)-F (a)}{b-a}x$. So we take the auxiliary function for $f (x) =f (x)-\frac{f (b)-F (a)}{b-a}x$ (why?). Because the constant derivation is zero! That's why we're going to get $c$ out of the equation and make it a function! ）

Readers may wish to try to calculate $f (a) $ with $f (b) $, they are obviously equal, thus the proposition is certified!

From the above process, we can see that the process of constructing the function is completely using the differential equation, which is the method discussed in this paper. Most tutoring book constructors do not explain why, so the reader will be mistaken for "brainwave" to come out, is a "magical" things, in fact, otherwise. To get a better understanding of this approach, let's look at one more example.

Set function $f (x) $ in $[a,b]$ continuous, in $ (a, b) $, wherein $a>0$ and $f (a) =0$, test proves: $\exists \xi$$\in$$ (A, b) $, makes $f (\xi) =\frac{b-\xi}{a}f ' (\ XI) $

Proof: We still look at it as a differential equation: $f (x) =\frac{b-x}{a}f ' (x) $, easy-to-calculate $f (x) =c (b-x) ^{-a}$. To solve the $c$ and make it $f (x) $, we found the auxiliary function: $F (x) =f (x) (b-x) ^{a}$!

Easy to verify, $F (a) =f (b) =0$, then the $rolle$ theorem is used to obtain the certificate.

In fact, this kind of problem can be treated by differential equation, so the key is to solve the differential equation

On the method of constructing auxiliary function for a class of middle value theorem proof problem