The definition domain and value domain of y = log2 (- x ^ 2 + 2x) are obtained

The definition domain and value domain of y = log2 (- x ^ 2 + 2x) are obtained

∵-x^2+2x>0
∴(x-1)^2

Given the inverse function F-1 (x) = log2 (1 + X / 1-x) of F (x), find the analytic formula of F (x)

It is to find the inverse function y = F-1 (x)
2^y=(1+x)/(1-x)=-(x-1+2)/(x-1)
=-1-2/(x-1)
2/(x-1)=-2^y-1
x-1=-2/(2^y+1)
x=1-2/(2^y-1)=(2^y-1)/(2^y+1)
So f (x) = (2 ^ x-1) / (2 ^ x + 1)

F (x) = log2 (1 + 1 / x) (x is greater than 0),

Change the position of X and y
x=log2（1+1/y）
1/y=2^x-1
y=1/(2^x-1).

F (x) = 2sinx + 1, X ∈ [π / 2,3 π / 2], find the definition and range of inverse function and inverse function f(x)=2sinx+1,x∈[π/2,3π/2]. Find the definition and range of inverse function and inverse function

Y = 2sinx + 1, the domain is [π / 2,3 π / 2],
Zero

The definition domain of F (x) = Log1 / 2 as the base (8-2 ^ x) is the value domain and inverse function of (- ∞, 2] The more detailed the train of thought is, the more you will get, and the full score will be 100 What you say is what you say

Because the domain is (- ∞, 2], so 4 =

There are two ways to determine the definition domain of inverse function: 1. Using the value range of the original function to determine; 2. Finding the natural domain of inverse function directly, which is more accurate?

Even if you don't agree, but I want to say that they are equally accurate. Only in specific problems, the difficulty of doing is different. Reason: because the y of the original function and the X of the inverse function are actually the same thing, which is called equivalent transformation in mathematics, so method 1 is absolutely correct

On inverse functions When f (x) = (AX + b) / (Cx + D), ABCD ≠ 0, when a, B, C, D satisfy any condition, the inverse function of F (x) is itself The answer is that when BC ad ≠ 0 and a + D = 0, the inverse function of F (x) is itself A + D = 0 understand. BC ad ≠ 0 is how to find out It is mainly to ask how to deduce the answer without knowing the answer, rather than relying on the answer to deduce whether it is correct or not

Suppose ad = BC
Then a / C = B / d
Then (AX + b) / (Cx + D) is a constant, = A / C = B / d
So f (x) = constant, of course, there is no inverse function

Seeking the higher inverse function Find the function f (x) = [x2 + 1 (x > = 0), x + 1 (x)

When x > = 0, f (x) > = 1
The inverse function is: y = √ X-1
When x

Given the set a = {x} - 2 ≤ x ≤ 2}, B = {X - 1 ≤ x ≤ 1}. Corresponding to F: X → y = ax. If the mapping from a to B can be established under the action of F, the value range of real number can be obtained The last sentence is to find the value range of real number A. how can we get that? I can see it at a glance. How do you want me to write about it?

That is to say - 2 ≤ x ≤ 2 in y = ax has - 1 ≤ ax ≤ 1 for all x values
So - 1 / 2

If the function f (x) defined on [- 20132013] satisfies: for any x1, X2 ∈ [- 20132013], there is f (x1 + x2) = f (x1) + F (x2) - 2012, and when x > 0, there is f (x) > 2012, and the maximum and minimum values of F (x) are m and N respectively, then the value of M + n is () A. 2011 B. 2012 C. 4022 D. 4024

Let X1 = x2 = 0, then f (0 + 0) = f (0) + F (0) - 2012,  f (0) = 2012, let - 2013 ≤ X1 < x2 ≤ 2013, and x2-x1 = t > 0, then f (x1) - f (x2) = f (x1) - f (x1 + T) = f (x1) - f (x1) - f (T) + 2012 = 2012-f (T) ∵ t ﹥ 0, ﹥ f (T) ﹥ 2012, ﹥ 2012, ﹥ 2012, ﹥ 2012 -