Some problems in "differential mean value theorem and derivative application" of higher numbers 1. Let f (x) be continuous on [0,1], derivable in (0,1), and f (0) = f (1) = 0, f (1 / 2) = 1 / 2. It is proved that for any C ∈ (0,1), there exists ξ∈ (0,1) such that f '(ξ) = C 2. It is known that f (x) is differentiable in R, and (x →∞) Lim f '(x) = E, (x→∞)lim [(x+c)/(x-c)]=(x→∞)lim [f(x)-f(x-1)] Finding the value of C

1. Let f (x) = f (x) - CX, it is easy to know that f (x) is continuous on [0,1] and differentiable in (0,1)
And f (0) = f (1) = 0, f (1 / 2) = 1 / 2, C ∈ (0,1)
Then f (1) = f (1) - C = - C < 0
F(1/2)= f(1/2) - 1/2 c = 1/2 (1-c)＞ 0
From the zero value theorem, we can see that there exists a η∈ (1 / 2, 1) such that f (η) = 0
And f (0) = f (0) - 0 = 0
For f (x) on [0, η], by using Rolle's theorem, there exists a ξ ∈ (0, η) contained in (0,1) such that f '(ξ) = 0, that is, f' (ξ) = C
2. Let x ∈ R, then f (x) is differentiable in the interval [X-1, x] and continuous in the interval (x-1, x)
According to Lagrange mean value theorem, there is a point ξ ∈ (x-1, x),
Let f '(ξ) = [f (x) - f (x-1)] / [x - (x-1)]
That is, (x →∞) Lim f '(x) = (x →∞) Lim [f (x) - f (x-1)] = E
So (x →∞) Lim [(x + C) / (x-C)] ^ x = e
The solution is C = 1 / 2
(Note: [(x + C) / (x-C)] should omit an X power, otherwise it can't be solved. You can refer to the title again.)

Prove the boundedness theorem of continuous virtual function with interval nest theorem: if f (x) is continuous on [a, b], then f (x) is bounded on [a, b]

If f (x) is unbounded on [a, b], let [a, b] = [A1, B1], half of which, at least one of the two closed intervals makes f (x) unbounded, let it be
[A2, B2]. And then divide it into [A3, B3]. And so on. Get a closed interval nest
[A1, B1] > (borrowing, meaning to include) [A2, B2] > .|[an,bn]|＝（b-a）/2^n→0.
F (x) is unbounded on every [an, BN]
From the interval nest theorem, there exists ξ ∈ every [an, BN]. Of course, ξ ∈ [a, b]. Let f (ξ) = C. ∵ f (x) be continuous on [a, b], and there exists δ > 0
When x ∈ (ξ - δ, ξ + δ), f (x) ∈ (C-1, C + 1),
Note that | (ξ - δ, ξ + δ) | = 2 δ. · let (B-A) / 2 ^ N0 ＜ δ. Then [an0, bn0] ＜ (included in) (ξ - δ, ξ + δ)
When x ∈ [an0, bn0], f (x) ∈ (C-1, C + 1), which contradicts "f (x) is unbounded on every [an, BN]"
Let f (x) be bounded on [a, b]

Can we prove that decreasing function increasing function = decreasing function by theorem?

Let f (x) be a decreasing function in the domain and G (x) be an increasing function in the domain. That is to say, for any X1 f (x2) in the domain and G (x1) < g (x2), make a new function H (x) = f (x) - G (x) because H (x1)

A is a rational number that is not 1. Let's make 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-A = - 1. For example, the difference reciprocal of - 1 is 1 / 1 - (- 1) = 1 / 2. It is known that A1 = - 1 / 3, A2 is the difference reciprocal of A1, A3 is the difference reciprocal of A2 By analogy, find A2009. A2 = 1 / (1-a1) = 1 / (1 + 1 / 3) = 3 / 4, A3 = 1 / (1-a2) = 1 / (1-3 / 4) = 4, A4 = 1 / (1-a3) = 1 / (1-4) = - 1 / 3 = A1, so A5 = A2, A6 = A3, a7 = A4 = A1, so this takes three as a cycle, 2009 / 3 remainder is 2, so A2009 = 3 / 4
1 / (1-3 / 4) = 4, how does 3 / 4 come from / what does it mean

Because A2 = 1 / (1-a1) = 1 / (1 + 1 / 3) = 3 / 4, so A3 = 1 / (1-a2) = 4, 3 / 4 is A2, of course, obtained through A1. As long as we calculate the first four, we will find that their values appear every three times. According to the law, we can know the remaining 2 of 2009 / 3, so A2009 = A2 = 3 / 4

Ask for advice
When the aircraft accelerates on the runway, the maximum acceleration can be 5 meters per second, and the take-off speed is 50 meters per second. If the aircraft taxis for 100 meters and takes off, what is the initial speed of the aircraft? Suppose that an aircraft carrier is not equipped with an ejection system, but the aircraft can take off normally on it, what is the minimum length of the aircraft? (root can be reserved)

(1)vt^2-v0^2=2ax
50^2-v0^2=2*5*100
v0=10√15
(2) If there is no ejection system, VT ^ 2 = 2aX
x=50*50/2*5=250m
The length of the ship should be at least 250m

Five out of seven x minus seven out of nine equals eleven out of nine

Sliding friction and maximum static friction
What is the sliding friction f = μ n
Or the maximum static friction f = μ n

Sliding friction f = μ n (μ is dynamic friction coefficient)
Maximum static friction f = μ on (μ o is static friction coefficient)

Univariate linear equation: 5 (3-2x) - 12 (5-2x) = - 17

5(3-2x)-12(5-2x)=-17
15-10x-60+24x=-17
-10x+24x=-17-15+60
14x=28
x=2

A and B run 40 km / h and B 45 km / h respectively from ab. the two vehicles meet at 35 km away from the midpoint and AB2 is far away from each other

Let t meet in time
Then 45t-40t = 35 * 2
t＝14
AB＝（45+40）*14＝1050

2 (a-b) {3 power of 2 / 1 (B-A)} 3 power

2(a-b)[2/(1(b-a)^3 ) ]^3=2(a-b)[2^3/1(b-a)^9]=-[2^4/1(b-a)^8]=-16/(b-a)^8