Differential mean value theorem problem It is known that f (x) is second-order differentiable on [a, b], a (a, f (a)), B (B, f (b)). Line segment AB intersects y = f (x) curve at another point C. verification: existence μ ∈ (a, b), so that the second derivative f '' (x) = 0, there is no reward,

Differential mean value theorem problem It is known that f (x) is second-order differentiable on [a, b], a (a, f (a)), B (B, f (b)). Line segment AB intersects y = f (x) curve at another point C. verification: existence μ ∈ (a, b), so that the second derivative f '' (x) = 0, there is no reward,

Let the coordinates of point C be (C, f (c)), it is easy to know that a is known by Lagrange mean value theorem and exists ξ 1 ∈ (a, c), so that f '（ ξ 1)=[f(c)-f(a)]/(c-a)
existence ξ 2 ∈ (C, b), so that f '（ ξ 2)=[f(b)-f(c)]/(b-c)
ξ 1< ξ two
Since point C ∈ line segment AB, [f (c) - f (a)] / (C-A) = [f (b) - f (c)] / (B-C) = slope of line segment ab
So f '（ ξ 1)=f'( ξ 2)
By Rolle's theorem, existence μ ∈( ξ 1, ξ 2) So that f ''（ μ)= [f'( ξ 2)-f'( ξ 1)]/( ξ 2- ξ 1)=0
Namely existence μ ∈ (a, b), so that f '' (x) = 0

(1) A > b > 0,0 > C > d launch a / C > b / D (2) A > b > 0 roll out 1 / a > 1 / b (3) A > b > 0 deduces the nth power of a > the nth power of B Please help judge right and wrong and analyze it casually!

1 -d>-c
-ad>-bc
ad1/a
3 multiply countless a > b to get

Let x > 0, Y > 0, and X + y = 1, verify (1 + 1 x）（1+1 y）≥9．

Proof: it is necessary to prove that (1 + 1x) (1 + 1y) ≥ 9 is true, -------- (1 point) because x > 0, Y > 0, and X + y = 1, y = 1-x > 0. ----------------- (1 point) only need to prove that (1 + 1x) (1 + 11 − x) ≥ 9, -------- (1 point) is the proof that (1 + x) (2-x) ≥ 9x (1-x), --

1. Known function f (x) = (x) ²- 4X + 4) / x, X belongs to [1, + 00), find the minimum value of F (x) and the corresponding value of X 2. If a + B = 3, the minimum value of a power of 2 + B power of 2 is________ 3. Known f (x) = 3 + lgx + 4 / lgx, (0

1、f(x)=(x ²- 4x+4)/x＝x+4/x-4>=2√4-4=0
When x = 4 / x, take the equal sign; when x = 2, the minimum value is 0
2. 2 ^ A + 2 ^ b > = 2 √ 2 ^ A * 2 ^ B = 2 √ 2 ^ (a + b) = 2 √ 2 ^ 3 = 4 √ 2 (minimum)
3、f(x)=3+lgx+4/lgx
0=2√4=4
F (x) = 3 - (- lgx-4 / lgx) < = 3-4 = - 1 (max.)

Let a, B and C be positive real numbers, and a + B + C = 1, then a ^ 2 + B ^ 2 + C ^ 2 + X √ ABC

a ²＋ b ²＋ c ²＋ x√abc≤1=(a+b+c) ² ∴x√abc≤2ab+2ac+2bc≤2(a ²＋ b ²＋ c ²)
∴x≤2(a ²＋ b ²＋ c ²)/ √ ABC equals sign holds if and only if a = b = C = 1 / 3
∴x≤2√3

What are the conditions for the existence of limits? When does the limit not exist? When does the function limit not exist?

sequence limit
Definition: let | xn | be a sequence of numbers, if there is a constant a, for any given positive number ε (no matter how small it is), there is always a positive integer n, so that when n > N, |xn - A|