Known function f (x) = LG (AX-1) - LG (x-1), in [10, positive infinity) is monotone increasing function, find the value range of real number a, thank you!

Known function f (x) = LG (AX-1) - LG (x-1), in [10, positive infinity) is monotone increasing function, find the value range of real number a, thank you!

f(x)=lg[(ax-1)/(x-1)]
Just look at the real number
Let u (x) = (AX-1) / (x-1) = [a (x-1) + A-1] / (x-1) = a + (A-1) / (x-1)
To increase on [10, + ∞), then A-10, a > 1 / 10
Therefore, the value range of real number a is (1 / 10,1)
Let f (x) = bracket x-3 (x is greater than or equal to 1) log 1 / 2 as the base of (1-x) (x0) solution set be?
It's a piecewise function
Train of thought: classified discussion! X1. Can you explain your question clearly!
The function defined on r y = f (x), f (0) ≠ 0, when x > 0, f (x) > 1, and for any real number x, y has f (x + y) = f (x) · f (y)
(1) . prove that when x
It is proved that: let x = 0, y = 0, then f (0 + 0) = f (0) & sup2; that is, f (0) - f (0) & sup2; = 0, because f (0) is not equal to 0, so f (0) = 1, and let y = - x, then f (x-x) = f (x) × f (- x), that is, f (0) = f (x) × f (- x), that is, 1 = f (x) × f (- x), so f (x) and f (- x) are reciprocal, that is
V
One
From: F (0 + 0) = f (0) * f (0)
That is: F (0) = [f (0)] ^ 2,
The solution of this equation is: F (0) = 1, or F (0) = 0
But it is known that f (0) is not equal to 0, so f (0) = 1
For X0
1=f(0)=f(x-x)=f(x)*f(-x)
That is (- F) = 1,
Since f (- x) > 1, the following results are obtained
X1, so in the above formula:
f(x2-x1)>1.
Therefore, the formula is more than 0
That is: F (x2) - f (x1) = f [(x2-x1) + X1] - f (x1)
=f(x2-x1) *f(x1)-f(x1)
=f(x1)*[f(x2-x1)-1]>0
That is: F (x2) - f (x1) > 0
That is, f (x 2) > F (x 1) holds when x 2 > x 1
That is, f (x) is an increasing function
3. If f (x ^ 2) * f (2x-x ^ 2 + 2) > 1
That is: F (x ^ 2 + 2x-x ^ 2 + 2) > 1
That is: x ^ 2 + 2x-x ^ 2 + 2) > 0
That is: 2x + 2 > 0, that is: x > - 1
F (x) is an increasing function defined on (0, positive infinity). For all positive real numbers x and y, f (XY) = f (x) + F (y), find the inequality f (㏒ 2x)
F (y) = f (y) + F (y)
Let x = y = 1
Then f (1) = f (1) + F (1)
So f (1) = 0
f(㏒2 X）0
log2(x)
If the function f (x) = log2a (x + 1) defined in the interval (- 1,0) satisfies f (x) > 0, then the value range of a is______ .
The value range of ∵ x ∈ (- 1,0), ∵ 0 ＜ x + 1 ＜ 1. And ∵ f (x) ＞ 0, ∵ 0 ＜ 2A ＜ 1, ∵ A is (0,12)
Let the definition field of function y = f (x) be x ≠ 0. For any real number x, y has f (XY) = f (x) + F (y), and when x > 1, f (x) > 0.1. Prove that function is even function
Let the definition field of function y = f (x) be x ≠ 0. For any real number x, y has f (XY) = f (x) + F (y), and when x > 1, f (x) > 0.1. Prove that function is even function
f(xy)=f(x)+f(y)
Let y = 1
f(x*1)=f(x)+f(1)
f(1)=0
Let x = - 1, y = - 1
f(1)=f(-1)+f(-1)
2*f(-1)=f(1)=0
f(-1)=0
Let y = - 1
f(x*(-1))=f(x)+f(-1)
f(-x)=f(x)+f(-1)=f(x)
So f (x) is even function
It is known that f (x) is an increasing function defined on [- 1,1], and f (x-1)
f(x-1)
LS is a positive solution
Zero
Let f (x) = x ^ 4 + ax ^ 3 + 2x ^ 2 + B (x ∈ R), a, B ∈ R. (1) for any a ∈ [- 2,2] inequality f (x) ≤ 1, it holds on [- 1,0],
Find the value range of B
f'(x)=4x^3+3ax^2+4x=4x(x^2+3ax/4+1)=4x[(x+3a/8)^2+1-(3a/8)^2]
Because a ∈ [- 2,2], 1 - (3a / 8) ^ 2 > 0
Therefore, f '(x) = 0 has only one extreme point x = 0 and is a minimum point
So when x ∈ [- 1,0], f (x) decreases monotonically
The maximum value of this interval is f (- 1) = 1-A + 2 + B = 3-A + B
There are 3-A + B
The function f (x) defined on the set of real numbers is a monotone decreasing function and satisfies f (x) + F (- x) = 0, if f (1-A) + F (1-A ^ 2)
It is easy to know that f (x) is a monotonically decreasing odd function, f (1-A) + F (1-A ^ 2)
Given the function f (x) = x & # 179; - 1 / 2x & # 178; + C, if for X ∈ [- 1,2] inequality f (x)
f'（x）=3x^2-x-2
Then 3x ^ 2-x-2 < C ^ 2 holds for X ∈ {- 1,2}
Then we only need the maximum value of the quadratic function y = 3x ^ 2-x-2 in the interval x ∈ {- 1,2}