If the function f(x) satisfies the following conditions in the interval [a, b]:

(1) continuous in [a, b]

(2) can be derived in (a, b)

Then there is at least one point f' (c) = [f (b)-f (a)]/(b-a) a.

Prove that the original function f(x)={[f(b)-f(a)]

Let the auxiliary function g (x) = f (x)-{[f (b)-f (a)]/(b-a)} x.

It is easy to prove that this function satisfies the condition in this interval:

1.G(a)= G(b)；

2.G(x) is continuous in [a, b];

3.G(x) is differentiable in (a, b).

This is the condition of Rolle's theorem, which is proved by Rolle's theorem.

Extended data:

Theorem expression

If the function f(x) satisfies:

(1) is continuous on the closed interval [a, b];

(2) Derivable in the open interval (a, b);

Then at least one point in the open interval (a, b) satisfies the equation.

In other forms, there is a formula called finite increment formula.

We know that the differential of a function is an approximate expression of the increment δy of the function. Generally, only when |δx | is small, the similarity between dy and δy will be improved. However, the finite increment formula gives an accurate expression of the function increment Δ y when the independent variable x takes the finite increment Δ x (| Δ x | is not necessarily small), which is the value of this formula.

Auxiliary function method:

Knowing the continuously differentiable in the open interval, an auxiliary function is constructed.

It is available because it is continuous in the world and differentiable in the open interval. According to Rolle's theorem, there must be something to make it. Is this available? Deformity? Prove theorem.

References:

Baidu encyclopedia-Lagrange mean value theorem