Should we use the Lagrangian median theorem to prove the problem?

The Lagrangian median theorem is a differentially related theorem, and in this problem there is no differentiation, only continuous, and it is not clear whether it is derivable.

Therefore the Lagrangian median value theorem cannot be used.

Seeing continuity, the intermediate value theorem (or its special case: the zero value theorem) is generally considered

Proof: Let g(x) = c1·f(x1)+c2·f(x2)-(c1+c2)·f(x).

Then g(x) is also continuous in the closed interval [a,b].

g(x1)=c1·f(x1)+c2·f(x2)-(c1+c2)·f(x1)=c2·[f(x2)-f(x1)]

g(x2)=c1·f(x1)+c2·f(x2)-(c1+c2)·f(x2)=c1·[f(x1)-f(x2)]

C1 and C2 are any normal numbers, i.e., C1>0, C2>0

g(x1) and g(x2) aliases.

From the zero value theorem, we know that there is at least a little d in [a,b], so that g(d)=0

That is, c1·f(x1)+c2·f(x2)-(c1+c2)·f(d)=0

C1·f(x1)+c2·f(x2)=(c1+c2)·f(d) is completed.

Take advantage of the properties of continuous functions on closed intervals.

Because the function is continuous in a closed interval, there must be a maximum and a minimum value, which is set to m, mc1+c2)m<=c1·f(x1)+c2·f(x2)<=(c1+c2)·m, note: c1, c2 are normal numbers, so the magnitude can be guaranteed to be in the sign direction.

From the intermediate value theorem, there is at least a little d in [a,b], such that c1·f(x1)+c2·f(x2)=(c1+c2)·f(d)

The Lagrangian median theorem is generally used to find "there is a point such that f'(ξc"of the formula.

For this kind of unequal sign, we generally use monotonicity + maximum value to solve it!

The process and results are as follows.

This type of question is not understood in Lagrange. It is a monotonionic solution for derivation.

Remembering f(x) =sinx - x + x 2 2, then f(0) =0,f'(x) =cosx - 1 + x, f''(x) =sinx + 1 ≥ 0, f'(x) monotonically increases, then when x > 0, f'(x) >f'(0) =0, f(x) increases monotonically, giving f(x) =sinx - x + x 2 2 > 0, i.e. sinx > x - x 2 2

<> below you can do it on your own.

Probably not, the Lagrangian median theorem cannot find the limit.

<> can't be calculated using the Lagrangian median theorem, so it is not used.

Lagrange seems to be generally used for proof questions.

First of all, we must know the meaning of this median, which is represented by k, and according to the Lagrangian median value theorem, the median value is satisfied.

f'(k) =arctan(b)/b

Because f'(x)=1/(1+x^2)

f'(k)=1/(1+k^2) =arctanb/bk^2 = b/arctanb -1 = b-arctanb)/blim k^2/b^2 = lim k^2/b^2 = lim (b-arctanb)/b^3

lim (1 - 1/(1+b^2))/3b^2lim b^2/3b^2 =1/3

<> testimony to talk about the process of the Ming Dynasty, I saw the attack on the waiter and missed the picture above.

<> such as the loss of the source of the loss, oh calendar hahaha.

i.e. [(x+1)lnx-(1+1)ln1)] (x-1)>2Let f(x)=(x+1)lnx, then f'=(x+1) x+lnx,f(x) satisfies the lagrange theorem at (1,x), i.e., [(x+1)lnx-(1+1)ln1)] (x-1)=f'(z)=(z+1) z+lnz, 12, i.e. g(z)=1 z+lnz>1 g(1)=1, g'=(z-1) z 2>0, so g(z) increases, i.e., g(z)>1 holds, so it can be proved.

Let f(x)=lnx-2+4 (x+1).

f'(x)=1/x-4/(x+1)^2=((x+1)^2-4x)/x(x+1)^2=(x-1)^2/x(x+1)^2>0

By the Lagrangian median value theorem:

f(x)-f(1)=f'(c)(x-1)>0, c between 1 and x, f(x)> f(1)=0

lnx-2+4/(x+1)

lnx>2-4/(x+1) =2(x-1)/(x+1) （x>1）

First of all, since the line equation for the points (a, f(a)) and the points (b, f(b)) is like this: y=[ (f(b)-f(a)) (b-a)]x-a)+f(a).

So the constructor is the relationship between the distance d and x of two curves: h(x)=f(x)-y (curve minus straight line).

Since the beginning and end points of the two lines coincide, the condition h(a)=h(b) must be met by Rohr's theorem, and then it can be proved immediately by Rohr's theorem.

Ideas: 1. The Lagrangian median theorem is actually a generalization of Roll's theorem (or the general situation), and Cauchy's median theorem is the generalization of the Lagrangian median value theorem (or special cases).

2. The condition of Rawl's theorem f(a)=f(b) means that the line of the point (a, f(a)) and the point (b, f(b)) is parallel to the coordinate axis, and then the extreme point of the function f(x) is found (equivalent to finding f'(k)=0) is a special case.

Whereas the case of the Lagrangian median theorem is the general case of Rolle's theorem. The line of ( a, f(a)) and the point ( b, f(b) ) is already at an angle to the x-axis, so it is necessary to transform its coordinate axis when constructing the function. Then, like Rohr's theorem, just find the extreme point of the function h(x).

One idea (without using a constructor, this idea should be more straightforward): let k0=(f(b)-f(a)) (b-a), which is a critical slope, if the slope of all points on this curve is less than k0, then it is obvious that the y-axis increment from a to b is <(b-a)*k0 then the function value of point b should be less than f(b)! , so this is contradictory, and in the same way it can be shown that the slope of all points on the curve cannot be greater than k0, which is good, then there are two points c and d satisfying:

The slope of c is greater than k0 and the slope of d is less than k0, i.e., f'(c)=k1>k0, f'(d) k2Actually, I think that no matter what the way of proof is, the ultimate idea is the continuity of the derivative.

Let f(x)=sinx,g(x)=x;

f(x)-f(y)] g(x)-g(y)]=f( ) Derivative of g( ).

That is: 丨sinx-siny丨 丨x-y丨=丨cos 丨 1 i.e.: 丨do the key sinx-siny丨 repentance丨x-y丨, x and pure no coincidence y y are arbitrary real numbers.