Several concepts:

1) continuous: for a point x in the domain? What if? ε& gt； 0,? δ& gt； 0, making:? X, satisfy |x-x? |<δ, both have |f(x)-f(x? |<ε, it is said that f(x) is in x? Continuous.

2) Lower supremum: sets E and B satisfy:? X∈E, x≥b, and? ε& gt； 0,? X'∈E, so x'

Preparation theorem:

Existence theorem of supremum on 1): If E is a set that is not empty and has a lower bound, then E has a lower supremum.

This theorem is very important, but it is only slightly involved in this proof, and it has nothing to do with what you want to master. If you are interested, you can refer to the first volume of Mathematical Analysis written by Wu Shengjian of Peking University Press, which contains detailed introduction and proof.

2) Zero theorem: the continuous function f(x) is continuous on [a, b], and f(a)? f(b )& lt； 0, and then what? C∈(a, b) makes: f(c)=0.

This theorem can be proved by using the existence theorem of supremum and the finite covering theorem.

Lagrange's mean value theorem: If f(x) is continuous on [a, b] and takes derivatives everywhere on (a, b), then? C∈(a, b), therefore:

f'(c)=(f(a)-f(b))/(a-b)。

This theorem is divided into two propositions.

(1) Rolle theorem: f(x) is continuous on [a, b] and derivable on (a, b), and f (a) = f (b) = 0, then? C∈(a, b) makes: f'(c)=0.

Prove: suppose:? c∈(a，b)，f'(c)？ 0.

What if? C∈(a, b), f ′ (c) >; 0, then F increases monotonically at (a, b), so F is discontinuous and contradictory at point B; Similarly, it can't happen? c∈(a，b)，f '(c)& lt； 0.

So:? a & ltx？ & ltx？ & ltb, so that: f(x? )f(x？ )& lt0.

According to the zero theorem. c∈(x？ ，x？ ), f'(c)=0, contradictory.

Therefore, the above proposition holds. #

(2) The proof of the mean value theorem.

Proof: take F(x)=(f(b)-f(a))/(b-a)? (x-a)+f (a)-f (x)。 -This kind of function construction is very useful.

Verifiable: F(a)=F(b)=0.

So, according to Rolle's theorem, C∈(a, b), like this:

F'(c)= 0 =(f(b)-f(a))/(b-a) - f'(c)。

Then: f'(c)=(f(b)-f(a))/(b-a).

Theorem proof. #