If the symmetric center of the inverse function image of the function f (x) = (A-X) / (x-a-1) is (- 1,4), then the real number a =? Isn't the original function over (4,

If the symmetric center of the inverse function image of the function f (x) = (A-X) / (x-a-1) is (- 1,4), then the real number a =? Isn't the original function over (4,

∵ the image symmetry center of the inverse function f (x) = (A-X) / (x-a-1) is (- 1,4)
The image symmetry center of the function f (x) = (A-X) / (x-a-1) is (4, - 1)
∴a+1=4, a=3
Methods Summary: the coordinates of the point of symmetry center of the function of y = (AX + m) / (BX + n) are (C, d), C = - N / B.D = A / b

Given the function f (x) = (A-X) / (x-a-1), the symmetry center of the image of the inverse function of F (x) is (m, 3), then a is equal to several

So the center of symmetry of the function is (x, y). The midpoint of the line between the two centers of symmetry must be on the line y = x, and the line must be perpendicular to y = x, so (x + m) / 2 = (3 + y) / 2 (Y

The function f (x) = a − x If the symmetry center of the image of the inverse function F-1 (x) of X − a − 1 is (- 1,3), then the real number a=______ .

∵ the symmetry center of the image of the inverse function F-1 (x) of the function f (x) = a − XX − a − 1 is (- 1,3),  f (x) is (3, - 1), y = f (x) = a − XX − a − 1 = - x − a − 1 + 1x − 1 = - 1x − (a + 1) − 1,

Given that f (x) = loga ((1 + x) / (1-x)), (a > 0 and a is not equal to 1), find the definition domain. Find the value range of X when f (x) > 0

One
Definition domain (1 + x) / (1-x) > 0
That is (x + 1) (x-1) < 0
x∈(-1.1)
Two
f(x)＞0
If a > 1, then (1 + x) / (1-x) > 1
2x/(x-1)＜0
x∈（0,1）
If 0 < a < 1, then (1 + x) / (1-x) < 1
2x/(x-1)＞0
X > 1 or x < 0
X ∈ (-1,0) is obtained by combining the domain of definition

We know that f (x) = loga (x + 1) - loga (1-x), a > 0 and a is not equal to 1, (1) find the domain of F (x); (2) when a > 1, find x with F (x) > 0 Given the culvert number f (x) = loga (x + 1) - loga (1-x), a > 0 and a is not equal to 1, (1) find the domain of F (x); (2) when a > 1, find the range of X values for f (x) > 0

1、
If the true number is greater than 0, x + 1 > 0, 1-x > 0
So define the domain (-1,1)
2、
f(x)=loga[(x+1)/(1-x)]>0
a> Then loga (x) is an increasing function
And 0 = loga (1)
So (x + 1) / (1-x) > 1
(x+1)/(1-x)-1>0
2x(x-1)

Given the function f (x) = loga root 2 x power minus 1, (a > 0 and a ≠ 1) ① find the definition domain of function; ② find the value range of X with F (x) > 0

f(x)=loga(√(2^x-1))
√(2^x-1)>0
2^x-1>0
2^x>1=2^0
X>0
f(x)>0
√(2^x-1)>1
2^x-1>1
2^x>2^1
X>1

Let f (x) = loga (AX-1) (a > 0 and a ≠ 1) (I) find the domain of F (x); (II) when x is a value, f (x) > 1?

(1) From the meaning of the title, AX-1 > 0, that is, ax > 1 = A0,
When 0 < a < 1, then x < 0 is defined as (- ∞, 0),
When a > 1, then x > 0, then the definition domain is (0, + ∞);
(II) from the meaning of the title, loga (AX-1) > 1 = logaa,
When 0 < a < 1, 0 < AX-1 < A, then 1 < ax < A + 1,
That is, A0 < ax < alg
A+1
A
Log
A+1
A
＜x＜0，
When a > 1, AX-1 > A, that is, ax > A + 1 = analog
A+1
A
,
The solution is x > log
A+1
A
,
To sum up, when 0 < a < 1, log
A+1
A
＜x＜0，
When a > 1, x > log
A+1
A
.

Let f (x) = loga (x power-1 of a) (a is greater than 0 and a is not equal to 1) 1. Find the definition domain of F (x) 2. Discuss the monotonicity of function f (x) 3. The inverse function of solving the equation f (2x) = f (x) (i.e. F-1 (x))

It is known that f (x) = loga (x-1 of a) (a is greater than 0 and a is not equal to 1) 1. Find the definition domain of F (x). Discuss the monotonicity of function f (x). 3. Inverse function of solving equation f (2x) = f (x) (i.e. F-1 (x)) (1) analysis: ∵ f (x) = log (a, a ^ x-1) (a > 0 and a ≠ 1) a ^ X-1 > 0 = = > A ^ x > A ^ 0

Given the function f (x) = loga (a ^ x-1), (a > 0, and a is not equal to 1), find the domain of F (x) and monotonicity of F (x)

When a > 1,
A ^ x increases monotonically,
If a ^ x > 1, x > 0, that is, the domain of F is (0, + infinity)
A ^ x monotonically increases, loga (x) monotonically increases
Then f is a composite function, monotonically increasing
When 0

Inverse function of function y = 1 / X-2 (x is not equal to 2)

Let y = x, x = y be substituted into the original function to obtain x = 1 / Y-2, that is, y = 1 / (x + 2), that is, the inverse function is y = 1 / (x + 2)