Given that the definition domain of function y = f (x) is R and there is an inverse function y = F-1 (x), then the inverse function of function y = 2F (x / 2-1) is

Given that the definition domain of function y = f (x) is R and there is an inverse function y = F-1 (x), then the inverse function of function y = 2F (x / 2-1) is

1 f [F-1 (x)] = x; 2 F-1 [f (x)] = X

If the function y = f (x) is r in definition domain and range, and there is inverse function. If f (x) increases on (- ∞, + ∞), it is proved that y = F-1 (x) is also an increasing function on (- ∞, + ∞)

It is proved that if y = F-1 (x), take two points (x1, Y1) and (X2, Y2), let x1

Given that the domain of y = f (x) is (- ∞, 0) and there is an inverse function, and f (x-1) = x ^ 2-2x, then the value of f ^ - 1 (- 1 / 4) is

f(x-1)=x^2-2x=(x-1)^2-1
So, f (x) = x ^ 2-1, (x)

Find the inverse function and inverse function definition domain of the function f (x) = (x-1) 2 square + 1

Let f (x) = y
Y-1 = (x-1) ^ 2 (^ 2 stands for square, press and hold shift and then hold number key 6 to get out ^)
Radical (Y-1) = X-1
X = 1 + radical (Y-1),
That is, the inverse function y = 1 + root sign (x-1), and the definition field is (1, positive infinity)
The inverse function is to express x with y, and then exchange y and X. The definition domain of inverse function x is the definition domain of original function y

Let f (x) = LG (3 + x) + LG (3-x) (1) Find the definition domain of function f (x); (2) Judge the parity of function f (x)

(1) According to the meaning of the title
3+x＞0
3 − x ＞ 0, the solution is - 3 ＜ x ＜ 3,
So the definition domain of function f (x) is {x | - 3 ＜ x ＜ 3}
(2) From (1) we know that f (x) is symmetric about the origin,
∵f（x）=lg（3+x）+lg（3-x）=lg（9-x2），
∴f（-x）=lg（9-（-x）2）=lg（9-x2）=f（x），
The function f (x) is an even function

Inverse function of y = 3 ^ 4-x ^ 2, X ∈ {- 2,0}

0

0

1y = - (cubic root) √ (x)
2 y=2x/(1-x)
3 y=-√（x-1）（x＞1）

The inverse function of the function y = - 1 / x + 3 (x ≠ 0) is

By y = - 1 / x + 3 (x ≠ 0, y ≠ 3)
1 / x = 3-y
∴x=1/(3-y)
x. Y transposition
So the inverse function of the original function is
y=1/(3-x) (x≠3)

Find the value range of the function y = (log24 / x) (log4x / 32) in the definition domain [1 / 2,8] Give another 100 points if you answer!

y=[log2(4)-log2(x)][log4(x)-log4(32)]
=[2-log2(x)](lgx/lg4-lg32/lg4)
=[2-log2(x)](lgx/2lg2-5lg2/2lg2)
=[2-log2(x)][1/2*log2(x)-5/2]
Loga (x = 2)
1/2

Definition domain and value domain of y = log2 (X-2)

Definition domain requirements:
X-2 > 0, so x > 2
The definition domain is: (2, + ∞)
Since the function is a logarithmic function with the base of 2, its range is: (- ∞, + ∞)