Given that the inverse function of the function f (x) = 3x passes through the point (18, a + 2), let the definition domain of G (x) = 3ax-4x be the interval [- 1,1], and obtain the analytic expression of G (x); 3x: 3 to the power of X, 3ax: 3 to the power of ax, 4x: 4 to the power of X

Given that the inverse function of the function f (x) = 3x passes through the point (18, a + 2), let the definition domain of G (x) = 3ax-4x be the interval [- 1,1], and obtain the analytic expression of G (x); 3x: 3 to the power of X, 3ax: 3 to the power of ax, 4x: 4 to the power of X

The inverse function of F (x) = 3 ^ x is y = log3 (x)
A + 2 = log3 (18) is obtained by substituting point (18, a + 2)
So a = log3 (18) - 2 = log3 (18) - log3 (9) = log3 (18 / 9) = log3 (2)
So g (x) = 3 ^ (AX) - 4 ^ x = (3 ^ a) ^ X - 4 ^ x = 2 ^ X - 4 ^ X

Function y = 2 ^ x + 1 / 2 ^ X-1 (x

The definition domain of inverse function is the value range of function
y=2^x+1/2^x-1 =1+2/(2^x-1)
X 0 inverse function y = log (2) [(x + 1) / (x-1)] (x)

Let y = - x (x + 2) (x > = 0), then find the definition domain of its inverse function?

The definition domain of inverse function is the range of original function
So the definition domain of inverse function is (negative infinite, 0]

Find the inverse function definition domain of y = x ^ 2 + 2x (x > = 0) ·········

The domain of inverse function is the range of Y
y=(x+1)^2-1
x>=0
When x = 0, y min = 0
So the domain of the inverse function is [0, positive infinity]

Y = (2-x) \ (2 + x) is the inverse function, and the definition domain is obtained

The inverse function is y = (2-2x) / (x + 1), and the definition field is x ≠ 1

Find the inverse function of y = x + 2 / x, and point out its definition domain

Y = x / (x + 2) = (x + 2) / (x + 2) = 1-2 / (x + 2) and 2 / (x + 2) is any value that is not 0, so y is any value that is not 1, y = x / (x + 2) XY + 2Y = XX (1-y) = 2Y. Therefore, x = 2Y / (1-y) is the inverse function: y = 2x / (1-x), where the definition domain of inverse function is the range of original function

Given that f (x) = ((x-1) / (x + 1)) ^ 2 (x ≥ 1), find the inverse function of F (x) and its definition domain, judge and prove the monotonicity of this inverse function (1) Finding the definition domain of inverse function and inverse function (2) Judge and prove the monotonicity of inverse function

The definition domain of inverse function is the value of the original function (domain is the domain of original function definition
Original function = 1-4 / (x + 1 / x + 2) > = 0, so the definition domain of inverse function is 0 infinite left closed right open
The range is 1 infinite
Because it is an inverse function defined by F (y) = X
The solution y = - (radical x + 1) / (radical x-1)

It is known that the function y = f (x) decreases in the definition domain D, and there is an inverse function F-1 (x). It is proved that the function decreases in the definition domain Y = F-1 (x) is the inverse function

Use definition directly!
For any X1 > X2, Y1 = F-1 (x1), y2 = F-1 (x2)
Then (F = 1), YX2 = 1;
Because y = f (x) is a decreasing function, Y1

If y = f (x) has an inverse function in the domain of definition (- ∞, 0), and f (x-1) = x ^ 2-2x, then the value of F-1 (- 1 / 4) is obtained Why f (x) = x ^ 2-1

For example, y = x + 1, then its inverse function is: x = Y-1. Usually, the independent variable is represented by X, and the dependent variable is expressed by Y. then the inverse function of y = x + 1 is y = X-1
OK. This is the basic knowledge. This problem uses the method of variable substitution
Let X-1 = t, then x = t + 1, bring in the formula of the question stem to get: F (T) = (T + 1) ^ 2-2 (T + 1) = T ^ 2 + 2T + 1-2t-2 = T ^ 2-1
However, f (x) is a corresponding relation, i.e. functional relationship. No matter what letter the independent variable in brackets is expressed, its corresponding relationship remains unchanged. From the above, we can get: F (T) = T ^ 2-1, or we can write: F (s) = s ^ 2-1, Now that you know the expression of F (x), then, I think, you also know the expression of its inverse function? The next thing to do is to find the inverse function of a function with known expression, which is not difficult. Y = x ^ 2-1, y + 1 = x ^ 2, x = - √ (y + 1), so f ^ - 1 (x) = - √ (y + 1) takes the value in and gets: f ^ - 1 (- 1 / 4) = - √ (3 / 4) = - (√ 3) / 2

It is known that there exists an inverse function F-1 (x) for the function f (x) whose domain is R. for any x ∈ R, there is always f (x) + F (- x) = 1, then F-1 + F-1 (x-2009) = () A. 0 B. 2 C. 3 D. It's about X

∵f（x）+f（-x）=1，
Let 2010-x = m, x-2009 = n, M + n = 1,
ν Let f (T) = m, f (- t) = n, which is known by the definition of inverse function,
∴t=f-1（m），-t=f-1（n）
∴f′（m）+f′（n）=0，
That is, the value of F-1 (2010-x) + F-1 (x-2009) is 0,
Therefore, a