If the image of the inverse function of the function f (x) = loga x passes through the point (- 1,2), then a = ()

If the image of the inverse function of the function f (x) = loga x passes through the point (- 1,2), then a = ()

Because the image of the inverse function passes through the point (- 1,2)
So the original function image goes through (2, - 1)
Substituting (2, - 1) into:
-1 = loga 2
a = 1/2

F (x) = x 2-2x + 4, (x > = 1), find the definition domain of inverse function and inverse function

y=x^2-2x+4
In other words, y = (x-1) ^ 2 + 3, and the range is y ≥ 3;
y-3=(x-1)^2；
Because x ≥ 1, so: X-1 = √ (Y-3)
That is, x = 1 + √ (Y-3)
Therefore, the inverse function is: y = f (x) = 1 + √ (x-3), and the definition domain is: X ≥ 3;
If you don't understand, please hi me,

If the inverse function of function f (x) = x 2, X ∈ [1,2] is F-1 (x), find the value range of function y = [F-1 (x)] 2 + F-1 (2x) Thank you for your trouble

From F (x) = x ^ 2, X ∈ [1,2], we get: F (x) ∈ [1,4]
Inverse function F-1 (x) = √ x, X ∈ [1,4]
F-1 (2x) = √ (2x), whose domain is 1=

It is known that the function f (x) = x ^ 2-2x + 2 x belongs to [a, b] (a > = 1) if its inverse function domain [a, b] finds a B As the title

Because f (x) = x ^ 2-2x + 2 = (x-1) ^ 2 + 1, it is an increasing function on x > = 1. Therefore, when x belongs to [a, b] (a > = 1), the minimum value of the function is f (a) = (A-1) ^ 2 + 1, and the maximum value is f (b) = (B-1) ^ 2 + 1 because of the definition domain of inverse function [a, b], the value range of function f (x) is [a, b], so (A-1) ^ 2 + 1 = a (B-1) ^ 2 + 1 = B

Given the function f (x) = 2x, write the inverse function g (x) of function f (x) and its definition domain

The inverse function g (x) = log2x is defined as (0, + ∞)

What is the inverse function and domain of the function f (x) = 5-2x, X ∈ [0, + ∞)?

The inverse function is y = 2.5-0.5x
The domain is (- infinite, 5]
Brother integral ah, urgent need integral! Urgent!
(the inverse function is to change the position of variable and independent variable, such as f (x) = 5-2x
(x = 2.5-0.5f (x). Then rewrite it to y = 2.5-0.5x. As for the definition field, it is the value range of the original function

Given that the function y = f (x) has an inverse function in its definition domain (- ∞, 0), and f (x-1) = x-2x, find the negative first power (- 1 / 2) of F Is to find the value of the negative first power (- 1 / 2) of F!

F (x-1) = x-2x
Let X-1 = t, x = t + 1
f(t)=(t+1)^2-2(t+1)=t^2-1
So f (x) = x ^ 2-1
When x ^ 2-1 = - 1 / 2,
x^2=1/2,
Because the domain is defined (- ∞, 0)
x=-√2/2
So [f (- 1 / 2)] ^ (- 1) = - √ 2 / 2

Given that the function y = f (x) has an inverse function on its domain (- ∞, 0], and f (x-1) = x2-2x, find f − 1 (− 1) 2) Value of

∵f（x-1）=x2-2x=（x-1）2-1，
ν f (x) = x2-1, and X ∈ (- ∞, 0],
Let f (x) = - 1
2, get: X=-
Two
2 (negative value discarded)
∴f−1(−1
2)=-
Two
2．

Inverse function of F (x) = 2x? Is it 1 / 2x or (1 / 2) x I think it should be the first one? But in that case, X can't be equal to 0. Isn't that consistent with F (x) = 2x? By the way, tell me the concept of inverse function~

First of all, you should make clear the concept of inverse function
F (x) = 2x is y = 2x, and the inverse function is x = 2Y
That is, y = x / 2 or F (x) = x / 2

The inverse function f '(x) of function f (x), the image of F (x) passes (1,2) and f' (2x + 1) = 1, find X

Because the inverse function of F (x) is f '(x), and the image of F (x) passes (1,2)
So the image of F '(x) passes (2,1), that is, f' (2) = 1
Because f '(2x + 1) = 1
So we have 2x + 1 = 2
The solution is x = 1 / 2