Proof of differential mean value theorem in Advanced Mathematics Let a > b > 0, prove that A-B / a < ln a / b < A-B / b

Proof of differential mean value theorem in Advanced Mathematics Let a > b > 0, prove that A-B / a < ln a / b < A-B / b

Let f (x) = LNX, then f '(x) = 1 / x,
Applying Lagrange mean value theorem to f (x) on interval [b, a], LNA LNB = (a-b) / C, where a > C > b > 0,
So (a-b) / A

Application of differential mean value theorem in Advanced Mathematics It is proved that the equation x ^ 5 + X-1 = 0 has only one root

1. Root:
Let f (x) = x ^ 5 + X-1, then f (x) is continuous on [0,1], f (0) < 0, f (1) > 0, so from the zero point theorem, f (x) has a zero point in (0,1) ξ, That is, the equation x ^ 5 + X-1 = 0 has roots ξ
2. Root unique
Let the equation have another root η,η ≠ ξ, It may be assumed that η＞ξ, Then in[ ξ,η] Using Rolle's theorem, existence ζ ∈( ξ,η), Make f '（ ζ)＝ 0. And f '(x) = 5x ^ 4 + 1 > 0. Contradiction
So there's only one equation

Multivariate function differentiation Let x + Z = YF (x ^ 2-z ^ 2), seek

Partial derivation of X at both ends: 1 + z'x = YF '(x ^ 2-z ^ 2) (2x-2zz'x) [1 + 2yzf' (x ^ 2-z ^ 2)] z'x = 2xyf '(x ^ 2-z ^ 2) - 1z'x = [2xyf' (x ^ 2-z ^ 2) - 1] / [1 + 2yzf '(x ^ 2-z ^ 2)] partial derivation of Y at both ends: z'y = f (x ^ 2-z ^ 2) + YF' (x ^ 2-z ^ 2) (- 2zz'y) [1 + 2yzf '(x ^ 2-z ^ 2)] z'y = f (x ^ 2 -

Known quadratic function y = x2 + 4x + k-1 (1) If the parabola has two different intersections with the x-axis, find the value range of K; (2) If the vertex of the parabola is on the X axis, find the value of K

(1) ∵ the image of quadratic function y = x2 + 4x + k-1 has two intersections with the X axis
∴b2-4ac=42-4 × one × （k-1）=20-4k＞0
∴k＜5，
Then the value range of K is k < 5;
(2) According to the meaning of the question:
  4ac−b2
4a=4(k−1)−16
four × 1=0，
The solution is k = 5

Given the quadratic function y = X2 - [M2 + 8] x + 2 [M2 + 6], let the vertex of the parabola be a and intersect with the X axis at two points B and C, ask whether there is a real number m so that the triangle ABC is an isosceles right angle? If it exists, find the value of M; If not, please explain the reason

Y = x ^ - [M ^ + 8] x + 2 [M ^ + 6] (^ represents Square)
=(x-(m^+8)/2)^-(m^+8)^/4+2(m^+6)
=(x-(m^+8)/2)^-(m^+4)^/4
Then the coordinates of point a are ((m ^ + 8) / 2, - (m ^ + 4) ^ / 4)
Let X1 and X2 be the X coordinates of BC two points respectively
Then X1 + x2 = m ^ + 8, X1 * x2 = 2 (m ^ + 6)
Because AB = AC, if ABC is an isosceles right triangle, BC must be an oblique edge. At this time, the vertical line segment from point a to X axis is also the center line of BC edge, that is, the absolute value of y value of point a is 1 / 2 of BC line segment
That is, 2 | - (m ^ + 4) ^ / 4 | = | x1-x2 | (1)
(x1-x2)^=(x1+x2)^-4x1*x2=(m^+8)^-4*2(m^+6)=(m^+4)^
Then | x1-x2 | = m ^ + 4 (because m ^ + 4 > 0)
From (1), (m ^ + 4) ^ / 2 = m ^ + 4, (m ^ + 4) ^ = 2 (m ^ + 4), m ^ + 4 = 2, m ^ = - 2
Since m is a real number and m ^ > = 0, m ^ = - 2 does not exist
That is, there is no real number m, so that triangle ABC is an isosceles right triangle

It is known that the parabola y = (M2-4) x2 + (M + 2) x + 3, when what is the value of M, is this function a quadratic function?

X is a quadratic function ² Coefficient not equal to 0
m ²- 4≠0
(m+2)(m-2)≠0
M ≠ - 2 and m ≠ 2