Rolle's theorem, Lagrange's mean value theorem and Cauchy's mean value theorem are generally applied to what questions?

Rolle's theorem, Lagrange's mean value theorem and Cauchy's mean value theorem are generally applied to what questions?

Cauchy mean value theorem actually includes Rolle's theorem and Lagrange's mean value theorem. The key is to use them flexibly according to the needs of the topic. The topic of proving the existence of zero derivative may be rolle, and proving that the property of the derivative function of a function may be Lagrange. If a complex relationship or the relationship between the derivative functions of two functions is involved, Cauchy mean value theorem is needed

Proof of limit of higher number
1. Prove that limxn = 0 is Lim | xn | = 0
2. Let the sequence {xn} be bounded, limyn = 0, and prove limxnyn = 0 with the definition of sequence limit

It's very simple
1. Evidence: sufficiency
Because Lim | xn | = 0, for any t > 0, there is a positive integer n, and for all n > n with - TN, there will always be │ yn │ n when there is n

The proof limit of higher numbers
It is proved that LIM [(4-3x) / (6x-5)] = 1
x→1
Note: it needs proof!

lim [(4-3x)/(6x-5)]=1
=（4-3）/（6-5）
=1

What is the monotone increasing interval of the function f (x) = sinxcosx?

f(x)=sinxcosx=1/2*sin2x
If sin 2x increases, then 2K π - π / 2

How many feet is a square meter?

1 sq m = 10.7639104167 sq ft

Given the square of a + the square of B - A + 4B + 4 and 1 / 4 = 0, find the value of a and B

Let f (x), G (x) be continuous on [a, b], have second derivative in (a, b) and have equal maximum, f (a) = g (a), f (b) = g (b). It is proved that there exists ξ ∈ (a, B) such that f ″ (ξ) = g ″ (ξ)

Let f (x) = f (x) - G (x), then f (x) is continuous on the surface, has second derivative in (a, b), and f (a) = f (b) = 0. (1) if f (x), G (x) get the maximum at the same point C in (a, b), then f (c) = g (c) ﹥ f (c) = 0, then from Rolle's theorem, we can get that there exists ξ 1 ∈ (a, c), ξ 2 ∈ (C, b), such that f '(ξ 1) = f' (ξ 2) = 0 By using Rolle's theorem, we can get that there exists ξ ∈ (ξ 1, ξ 2) such that f ″ (ξ) = 0, that is, f ″ (ξ) = g ″ (ξ). (2) if f (x), G (x) have the maximum values at different points C1 and C2 in (a, b), then f (C1) = g (C2) = m, then f (C1) = f (C1) - G (C1) > 0, f (C2) = f (C2) - G (C2) < 0, then we can get that there exists C3 ∈ (C1, C2) ）From Rolle's theorem, we can get that there exists ξ 1 ∈ (a, C3), ξ 2 ∈ (C3, b) such that f ′ (ξ 1) = f ′ (ξ 2) = 0. By using Rolle's theorem, we can get that there exists ξ ∈ (ξ 1, ξ 2) such that f ″ (ξ) = 0, that is, f ″ (ξ) = g ″ (ξ)

Simple operation of 4 and 17:11 plus 8 and 7:4 plus 5 and 17:6 plus 7 and 7:3

4 and 17:11 + 8 and 7:4 + 5 and 17:6 + 7 and 7:3
=4+8+5+7+11/17+4/7+6/17+3/7
=24+2
=26

Seven fifths is a few more than the reciprocal of three-thirds

Seven fifths is a few more than the reciprocal of three-thirds
（7/5-2/3）÷2/3
=11/15÷2/3
=11/10
Seven fifths is 10 / 11 more than the reciprocal of three fifths

80+x=(56-x)*3 x=x

80+x=(56-x)*3
80+x=168-3x
4x=88
x=22