It is proved that the function f (x) = LNX / X is an increasing function on (0, e)

It is proved that the function f (x) = LNX / X is an increasing function on (0, e)

F`(x)=(1/x*x-lnx)/x^2=(1-lnx)/x^2
(0,e),lnx＜1
1-lnx ＞ 0 x ^ 2 ＞ 0, that is, f ` (x) ＞ 0
The function f (x) = LNX / X is an increasing function on (0, e)

Function f (x) = (a + 1) LNX + ax * x + 1, let a be less than or equal to - 2, prove that any x1, X2 is greater than 0, | f (x1) - f (x2) | is greater than or equal to 4| x1-x2|

f（x）=（a+1）lnx+ax^2+1
Then f '(x) = (a + 1) / x + 2aX
From a < = - 2, x > 0
（a+1）/x<0,2ax<0
Then | f '(x) | = | (a + 1) / x + 2ax| = | (a + 1) / x| + | 2ax| > = 2 root signs [(a + 1) / X * 2aX] = 2 root signs [2 (a + 1) a]
Y = (a + 1) a is a subtractive function when a < = 2, Ymin = (- 2 + 1) (- 2) = 2
Then | f '(x) | = 2 root numbers [2 (a + 1) a] > = 2 root numbers [2 * 2] = 4
Then | f (x1) - f (x2) | > = 4 | x1-x2 |

The function f (x) = LNX ax, a is a constant. If the function f (x) has two zeros x1, X2, try to prove that x1x2 > e ^ 2

First find the derivative y '= 1 / x-a, let y' = 0, x = 1 / A, and the maximum value of the function at 1 / A is - LNA + 1 > 0, and get 00 to get x2 > 2 / a-x1
Let g (x) = ln (2 / A-X) - A (2 / A-X) - (LNX ax),
G '(x) = 1 / (X-2 / a) + 2a-1 / x = 2A (x-1 / a) ^ 2 / [x (X-2 / a)], G' (x) 0 on (0,2 / a),
So ln (2 / a-x1) - A (2 / a-x1) > 0, it is proved

Given that a is a constant, the function f (x) = x (LNX ax) has two extreme points x1, X2 (X10), f (x2) > - 1 / 2 B, f (x1)

Given that a is a constant, the function f (x) = x (LNX ax) has two extreme points x1, X2 (X10), f (x2) > - 1 / 2 B, f (x1) f (1 / E) = - 1 / E. when a ≠ 0, f (x) = xlnx ax ^ 2 = = > F '(x) = lnx-2ax + 1 = 0 = = > A = (LNX + 1) / (2x) let a (x) = (LNX + 1) / (2x) make a' (x) = - 2lnx / (4x ^ 2) = 0 = = > x = 1 when 01

Given the function f (x) = LNX ax + (1-A) / X-1, let g (x) = x ^ 2-2bx + 4, when a = 1 / 4, if for any 0 < X1 < 2, there is 1 ≤ x2 ≤ 2 to make f (x1) Make f (x1) ≥ g (x2), and find the value range of B

The problem only needs to be transformed into minf (x1) > = maxg (x2). When a = 1 / 4, f (x) = lnx-x / 4 + 3 / (4x) - 1. Find the derivative f '(x) = - (x-3) (x-1) / (4x ^ 2). Let f' (x) = 0, get the unique stagnation point x = 1, (note 0

Hurry! Given the function f (x) = A / X-1 + LNX, ∃ x0 > 0, so that f (x0) ≤ 0 is true, find the value range of real number a Use the function method, that is, find the derivative of F (x), not the separation constant method, thank you For detailed process, thank you!

Given the function f (x) = A / X-1 + LNX, ∃ x0 > 0, so that f (x0) ≤ 0 is true, find the value range of real number a
Analysis: ∵ function f (x) = A / X-1 + LNX, and its definition domain is x > 0
Let f '(x) = - A / x ^ 2 + 1 / x = (x-a) / x ^ 2 = 0 = = > x = a
When A0, f (x) increases monotonically in the definition domain;
When a > 0
f’’(x)=2a/x^3-1/x^2==>f”(a)>0
‡ f (x) takes the minimum value at x = a
∃ x0 ＞ 0, so that f (x0) ≤ 0 is established
F(a)=a/a-1+lna=lnaa