Given that the function f (x) = loga (X-2 / x + 2), the definition field [α, β], the range [logaa (β - 1), logaa (α - 1)] {* subscript will not be marked, a (β - 1) and a (α - 1) in the range are true numbers}, and f (x) is a subtraction function on [α, β], so we can find the range of real number a, It's OK to give you some ideas

Given that the function f (x) = loga (X-2 / x + 2), the definition field [α, β], the range [logaa (β - 1), logaa (α - 1)] {* subscript will not be marked, a (β - 1) and a (α - 1) in the range are true numbers}, and f (x) is a subtraction function on [α, β], so we can find the range of real number a, It's OK to give you some ideas

f(x)=loga((x-2)/(x+2))
=loga((x+2-4)/(x+2))
=loga(1-4/(x+2))
1-4 / (x + 2) is an increasing function
F (x) is a decreasing function on [α, β]
Therefore, 0 (X-2) / (x + 2) > 0 - 2-2 < α < β < 2
F (x) is a decreasing function on [α, β]
So f (β) = log (a) a (β - 1)
log(a)(1-4/(β+2))=log(a)a(β-1)
1-4/(β+2)=a(β-1)
β-2=a(β-1)(β+2)
In the same way, we can see that
α-2=a(α-1)(α+2)
The equation
X-2 = a (x-1) (x + 2) has two solutions - 2aX ^ 2 + AX-2 = X-2
ax^2+(a-1)x=0
ax(x+(a-1))=0
x1=0 x2=1-a
So - 2 < 1 - a < 2
-1 to sum up, 0
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The set a = {x ｜ x-a = 4}. Set B = {1.2. B} 1. Is there a real number a such that for any real number B, there is a ≤ B? If so, find the corresponding a; if not, explain the reason 2 if a ≤ B holds, find the corresponding pair of real numbers (a, b)

1) If a contains B, x = a ± 4 ∈ Ba + 4 = 1, a = - 3, B = A-4 = - 3-4 = - 7a + 4 = 4 = 2, a = - 3, B = A-4 = - 3-4 = - 7a + 4 = 2, a = - 2, B = A-4 = - 2-4 = - 6a-4 = 1, a = 5, B = A-4 = 5-4 = 1a-4 = 1, a = 6, B = A-4 = 5-4 = 1a-4 = 1, a = 6, B = A-4 = 6-4 = 6-4 = 2, a = 6, B = A-4 = 6-4 = 2, B = a, B = A-4 = 6 = 6-4 = 2, corresponding real number pairs (a, b) is: (- 3, - 3, - -

Let the sum of the first n terms of the sequence {an} be Sn, and an + Sn = 1 (n belongs to N +) 1. Find the same term formula of {an} 2. If the sequence {BN} satisfies B1 = 1 and B (n + 1) = BN + an, find the general term formula of sequence {BN}

The following can be obtained: a|n = a|n-1-a|n-1-a|n-1-a n n n = a n n = 1 / 2 a|n-1 a n n-1 series {an} is a common ratio Q 1 / 2 equal proportion sequence, let n = 1, then 2 A1 = 1, A1 = 1, A1 = 1 / 2 according to the formula of equiproportion series: general term {an} = [1 / 2 (1-1-1 / 2)] / [1 / 2 (1-1 / 2)] / [1 / 2 (1-1 / 2)] / [1 / [1 / 2 (1-1 / 2)] / [1 / 2 (1-1 / 2)] / [1 / 2 (1-1 / 2)] / 1 - (1 / 2) ^ n]

Why 3 ^ (a + 2) = 18.3 ^ a = 2?

3^（a+2)=18
3^a*3^2=18
3^a*9=18
3^a=2

[1 - (sin160 °) ^ 2] But there is no cos20 in the option A. Cos160 B. - cos160 C. soil cos160 D. cannot be determined

From (SiNx) ^ 2 + (cosx) ^ 2 = 1
1-(sin160)^2=(cos160)^2
cos160=cos(180-20)=-cos20
So (cos160) ^ 2 = (cos20) ^ 2
So the original formula = cos20

Find the inverse function of y = 1 / (x-1) ^ 2 + ln (x-1) I want to get y '= - 2 / (x-1) ^ 3 + 1 / (x-1), then get the original function after the reciprocal. The indefinite integral is 1 / 2 (x-1) ^ 2 + ln | 1 / 2 (x ^ 2-2x-1) | + root 2 * ln | (x-1-radical 2) / (x-1 + radical 2) | + C, but the verification is wrong

(1) Y = 1 / (x-1) 2 + ln (x-1) is a transcendental function. It is impossible to solve x as a function of Y, i.e. it is impossible to obtain x = f ֿ (y),
So you can't work out its inverse function!
(2) ∫ f ′ (x) DX = f (x) + C, that is to say, if you get y ′, and then integrate x, that is to test whether your derivative is correct, because you go back! Y ′ = - 2 / (x-1) 3 + 1 / (x-1), ∫ [- 2 / (x-1) 3 + 1 / (x-1)] DX = 1 / (x-1) 2 + ln (x-1) + C; that is to say, differential sum
Integrals are two kinds of inverse operations, and finding the inverse function of a function are two different things. It can be said that it is not the same thing! Integration is to find the original function, so that the derivative of this function is equal to the integrand function; and the inverse function is to inverse solve x as a function of Y, and then exchange X and y, so as to obtain y = f ֿ (x)
(3) You take the reciprocal of Y 'and then integrate it. What is this operation? The function integrated can only be the original function of 1 / y'. What do you want it to do?
Is full of nothing to do, do to play?

How to use Mathematica to solve the inverse function and draw the numerical distribution diagram y ^ 2 / (x ^ 2 + y ^ 2) + ln (sqrt (x ^ 2 + y ^ 2)) = 100 As the title Y ^ 2 / (x ^ 2 + y ^ 2) + ln (sqrt (x ^ 2 + y ^ 2)) = D the result of the equation is assumed to be d = 101001000 How to solve the inverse equation and draw the distribution map of D in X, y coordinate system in a certain result (101001000)

ContourPlot[
y^2/(x^2 + y^2) + Log[ Sqrt [x^2 + y^2]] == 10,{x,-50000,
50000},{y,-50000,50000}]
ContourPlot[
y^2/(x^2 + y^2) + Log[ Sqrt [x^2 + y^2]] == 100,{x,-10^44,
10^44},{y,-10^44,10^44}]
The image with y ^ 2 / (x ^ 2 + y ^ 2) + log [sqrt [x ^ 2 + y ^ 2] = = 1000 is too large to draw

The following is the solution to the function definition domain y = lnsinx y = ln (LNX). Another problem is that the inverse function of F = e to the power of X-1 is y = ln (x + 1), and its definition domain is y = ln (x + 1) But I don't know how to solve this domain?

The first two are logarithmic functions, which mainly consider the problem that the true number is greater than zero. SiNx > 0,2kp1. The second definition domain is greater than negative one, which can also be seen from the value range of exponential function

Definition domain of inverse function of y = 2 ^ X / (2 ^ x + 1)

The definition domain of inverse function is the range of Y
y=(2^x+1-1)/(2^x+1)=(2^x+1)/(2^x+1)-1/(2^x+1)=1-1/(2^x+1)
2^x>0
SO 2 ^ x + 1 > 1
Zero

Find the inverse function and its definition domain of the following functions: (1) y = 2x + 1; (2) y = 1-x / 1 + X

(1) The inverse function y = 2x + 1 y = x / 2-1 / 2 x belongs to R
(2)y=1-x/1+x x+1=y-xy x+xy=y-1 x(y+1)=(y-1) x=(y-1)/(y+1)
Inverse function y = (x-1) / (x + 1) x ≠ - 1
This kind of problem as long as you know the definition of the inverse function is the value range of the original function