(a + B-5) (a-b + 5) with the formula of square difference

(a + B-5) (a-b + 5) with the formula of square difference

To solve the problem of proving the mean value theorem of calculus of higher numbers,
Let f (x) be second-order differentiable on [0,1], and f (0) = f (1). It is proved that there exists at least one ξ ∈ (0,1) such that f '' (ξ) = 2F '(ξ) / (1 - ξ)

F (0) = f (1), so there exists a such that f '(a) = 0
Let f (x) = f '(x) (1-x) ^ 2,
F (a) = f (1), so there exists ξ, a

Calculus, mean value theorem
Proof question:
When x > 0, X / (1 + x)

Look to the right first
Divide by two, take e as the base index at the same time, and then expand e ^ X by McLaughlin (in fact, it is proved that the growth rate of e ^ x is greater than 1 + x)
ln(1+x)/x=(1+x)/e^x=(1+x)/(1+x+x^2/2+x^3/6+.)

Lagrange theorem of calculus
Use Lagrange theorem to prove the following inequality: under double root sign, 3 is greater than 3 minus 1 divided by X, where x is greater than 0 and X is not equal to 1

Let f (x) = 2 √ 3-3 + 1 / X
f’(x)=-1/x^2
By means of Lagrange mean value theorem, the existence of
ξ∈(x,1),x1
f(x) =f(1) +f'(ξ)*(x-1)
=2√3-3+1- (x-1)/ξ^2
=2√3-2-(x-1)/ξ^2
When x1, 12 √ 3-2-1 / 4
>0
Sure enough, a lot of waste! But I learned how to use Lagrange mean value theorem