Function f (x) = LG [(kx-1) / (x-1)], k > 0. If f (x) monotonically increases on [10, + ∞], find the value range of K

Function f (x) = LG [(kx-1) / (x-1)], k > 0. If f (x) monotonically increases on [10, + ∞], find the value range of K

LGU is added when u > 0,
When x ∈ [10, + ∞), G (x) = (kx-1) / (x-1) > 0
(kx-1) / (x-1) = K + (k-1) / (x-1) single increase = > (k-1) < 0 = > 0 (kx-1) > 0 = > k > 1 / 10
Thus, 1 / 10 < K < 1

Known function f (x) = x * e ^ (KX), (K ≠ 0) (1) Find the tangent equation of curve y = f (x) at point (0, f (0)) (2) Find f (x) monotone interval (3) If f (x) monotonically increases in (- 1,1), find the range of K ---------------------------------------- Please use the relevant knowledge of higher derivative to solve the problem. There should be a concise process, Write the final answer

(1) Because f (0) = 0, the point (0, f (0)) is (0,0), take the derivative of F: F '(x) = e ^ (KX) + kxe ^ (KX) = (1 + KX) e ^ (KX) f' (0) = 1 + 0 = 1, then get the tangent equation y-0 = 1 * (x-0) y = x (2) 2-1, because no matter how KX is taken, e ^ (KX) > 0, because K ≠ 0 makes f '= 0 get x = - 1 / k2-2, k > 0, when x > - 1 / K, f

Known function f (x) = e ^ x-kx ², X ∈ R, if k = 1 / 2, prove that when x ∈ (0, + ∞), f (x) > 1

Derivative of F (x) = e ^ x-x
Derivative of F (x) = e ^ X-1
When x ∈ (0, + ∞), the derivative of the derivative of F (x) is greater than 0, that is, the derivative of F (x) is an increasing function, so the derivative of F (x) > F '(0) = 0, so f (x) is an increasing function, that is, f (x) > F (0) = 1

Known function f (x) = e ^ x-kx ^ 2 (x ∈ R) (1) If k = 1 / 2, verify that when x ∈ (0, + ∞), f (x) > 1 (2) If f (x) monotonically increases in the interval (0, + ∞), try to find the value range of K (3) Verification: [(2 / 1 ^ 4) + 1] [(2 / 2 ^ 4) + 1] [(2 / 3 ^ 4) + 1]... [(2 / N ^ 4) + 1] < e ^ x (n ∈ Z *) The first two questions don't need to be written to me. Just take them out to show you whether they have any effect on the third question. You can directly give me the third question process (3) Verification: [(2 / 1 ^ 4) + 1] [(2 / 2 ^ 4) + 1] [(2 / 3 ^ 4) + 1]... [(2 / N ^ 4) + 1] < e ^ 4 (n ∈ Z *). The third question should be this! Thank you for reminding me

It can be proved by mathematical induction as follows:
The original inequality is
[(2/1^4)+1][(2/2^4)+1][(2/3^4)+1]···[(2/n^4)+1]＜e^4 (n∈Z*)
Multiply both sides by (1.2.3 ··· n) ^ 4 at the same time
(2+1^4)(2+2^4)(2+3^4)···(2+n^4)＜(e·1·2·3···n)^4 (1)
In the following, it is proved by mathematical induction that inequality (1) holds for any n ∈ Z *
1 ° when n = 1
2+1^4=3＜16=2^4＜e^4
That is, 2 + 1 ^ 4 < e ^ 4 is established
2 ° let inequality (1) hold when n = K (K ≥ 1, K ∈ Z *)
(2+1^4)(2+2^4)(2+3^4)···(2+k^4)＜(e·1·2·3···k)^4
Established, then
When n = K + 1
(2+1^4)(2+2^4)(2+3^4)···(2+k^4)[2+(k+1)^4]＜[(e·1·2·3···k)^4]·[2+(k+1)^4]＜[(e·1·2·3···k)^4]·[(k+1)^4]=[e·1·2·3···k·(k+1)]^4
That is, inequality (1) holds when n = K + 1
It can be obtained by combining 1 ° and 2 °
For any n ∈ Z *, inequality (1) holds
Therefore, the original inequality [(2 / 1 ^ 4) + 1] [(2 / 2 ^ 4) + 1] [(2 / 3 ^ 4) + 1] · [(2 / N ^ 4) + 1] < e ^ 4 is constant for any n ∈ Z *

Function f (x) = lgkx − 1 X − 1 (K ∈ R, and k > 0) (1) Find the definition domain of the function (2) If the function f (x) monotonically increases on [10, + ∞), find the value range of K

(1) K > 0 and KX − 1
x−1＞0．
When 0 < K < 1, the definition field is {x | x < 1 or x > 1
k}； When k = 1, the definition field is {x|x ≠ 1}; When k > 1, the definition field is {x | x > 1 or x < 1
k}；
(2) ∵ function f (x) monotonically increases on [10, + ∞),
∴y=kx−1
x−1=k+k−1
X − 1 monotonically increases on [10, + ∞) and is positive,
‡ k-1 ＜ 0 and 10K − 1
10−1＞0，
∴1
10＜k＜1．

Known function f (x) = a • 2x + a − 2 2X + 1 (x ∈ R), if f (x) satisfies f (- x) = - f (x) (1) Find the value of real number a; (2) It is proved that f (x) is an increasing function on R; (3) Find the value range of function f (x)

(1) The definition domain of function f (x) is r, and f (x) satisfies f (- x) = - f (x),
So f (- 0) = - f (0), that is, f (0) = 0. So 2A − 2
2 = 0, a = 1,... (3 points)
At this time, f (x) = 2x − 1
2X + 1, f (x) meets the meaning of the question, so a = 1            … (4 points)
(2) Let x1 ＜ X2,
Then f (x2) − f (x1) = 2x2 − 1
1+2x2−2x1−1
1+2x1＝2(2x2−2x1)
(1+2x1)(1+2x2)
∵x1＜x2，
∴0＜2x1＜2x2，
∴2x2−2x1＞0，(1+2x1)(1+2x2)＞0
∴f（ x2）-f（ x1）＞0
f（ x2）＞f（ x1）
So f (x) is an increasing function on the definition field r.... (8 points)
（3）f(x)＝2x−1
2x+1=1−2
2X + 1,... (11 points)
Since 2x + 1 > 1, 0 < 2
2X + 1 ＜ 2, that is, the value range of F (x) is (- 1, 1).... (12 points)