Inverse function of function y = 2 | x | (x ∈ [0, + ∞))) For example, is the answer y = x / 2 (x ∈ [0, + ∞)), or y = | x | / 2 (x ∈ [0, + ∞))

Inverse function of function y = 2 | x | (x ∈ [0, + ∞))) For example, is the answer y = x / 2 (x ∈ [0, + ∞)), or y = | x | / 2 (x ∈ [0, + ∞))

Because x ∈ [0, + ∞)
So y = 2 | x | = 2x, and Y > = 0, that is, x > = 0 in the domain of inverse function definition
From the solution of y = 2x, x = Y / 2 is obtained;
Exchange x, y to get y = x / 2;
So the inverse of the original function is
y=x/2 (x∈[0,+∞))

I want to ask a math question The inverse function of the function y = - (1 / x) + 3 (x ≠ 0) is? A.y = - 1 / X b.y = 1 / X - 3 C.Y = 1 / (x-3) D.Y = 1 / () A.y=-1/x b.y=1/x -3 c.y=1/(x-3) d.y=1/(3-x) What's the inverse function?

Y = - (1 / x) + 3 (x ≠ 0), 1 / x = - y + 3, x = 1 / (3-y), rewrite y = 1 / (3-x), so choose D
In general, let the value range of the function y = f (x) (x ∈ a) be c. according to the relationship between X and Y in this function, X is expressed by Y, and x = g (y). If for any value of Y in C, by x = g (Y), X in a has a unique value corresponding to it, then x = g (y) means that y is an independent variable and X is a function of dependent variable y. such function x = g (y) (Y ∈ C) is called the inverse function of function y = f (x) (x ∈ a), denoted as y = f ^ (- 1) (x) The definition domain and range of inverse function y = f ^ (- 1) (x) are the range and definition domain of function y = f (x)

··A mathematical problem about inverse function·· If the inverse function of the function f (x) defined on R is [F-1] (x), and for any x belonging to R, there is f (- x) + F (x) = 3, then [F-1] (x-1) + [F-1] (4-x)= Note: [F-1] is - 1 in the upper right corner of F

Let a = f (x), B = f (- x), so a + B = 3 from a = f (x), x = [F-1] (a) (1) From b = f (- x) - x = [F-1] (b) (2) (1) + (2) we get [F-1] (a) + [F-1] (b) = 0, and because a + B = 3, that is, for all a and B satisfying a + B = 3, there is [F-1] (...)

If the inverse function of the function y = f (x-1) is y = F-1 (x-1), then the following equation must hold is () A. f（x）=f（x-1） B. f（x）-f（x-1）=-1 C. f（x）-f（x-1）=1 D. f（x）=-f（x-1）

From y = F-1 (x-1), we can get the following results
x-1=f（y），
x=1+f（y），
The inverse function of y = F-1 (x-1) is obtained by exchanging X and y
y=1+f（x），
The inverse function of the function y = f (x-1) is y = F-1 (x-1),
/ / F (x-1) = 1 + F (x), that is: F (x) - f (x-1) = - 1
The answer is B

Solving equation (1) (2x + 1) ^ 2 = 3 (2) 2 (x-1) ^ 3 = 16 The first question should be simplified

（1）（2x+1）^2=3
2x+1=±√3
2x=±√3-1
X = (√ 3-1) / 2, or x = (- √ 3-1) / 2
Yes, the square root of both sides is for simplification
(2)2(x-1)^3=16
（x-1）³=8
x-1=2
X=3

The inverse function of the function y = (x + 1) 2 + 1 (x ≤ 1) is—————— Fill in the blanks, The inverse function of the function y = (x + 1) 2 + 1 (x ≤ 1) is——————

Y = (x + 1) ^ 2 + 1 (x ≤ 1) is not a monotone function,
There is no inverse function

A mathematical problem of finding inverse function If y = 1 / 5, x + B and y = - A + 5 are inverse functions of each other, then a + B =?

From y = x / 5 + B, we get the following result
x=5y-5b,
So the inverse function of y = x / 5 + B is y = 5x-5b,
Because y = x / 5 + B and y = - ax + 5 are inverse functions,
So y = 5x-5b is the same function as y = - ax + 5,
So - a = 5, - 5B = 5,
a=-5,b=-1,
So a + B = - 6

A mathematical problem about inverse function Inverse function of y = √ x + √ (x-1)?

In order to make the calculation process intuitive, we can set
a=√x
The above formula can be written as
y=a+√(a^2-1)
y-a=√(a^2-1)
Square the two sides to get:
y^2-2ay+a^2=a^2-1
y^2-2ay+1=0
√x=a=(y^2+1)/2y
x=(y^4+2y^2+1)/(4*y^2)
It can be seen from the meaning of the title
y≥1,x≥1

A mathematical problem about inverse function Given the function f (x) = (3x + 2) / (x + a) (x ≠ - A, a ≠ 2 / 3), (1) Find its inverse function; (2) Find the value of the real number a such that F-1 (x) = f (x); (3) When a = - 1, find F-1 (2)

X = f (x) = (3x + 2) / (x + a) (x ≠ - a a a ≠ 2 / 3), 1) y = f (x) = (3x + 2) / (x + a) = [(3x + 3a) + (2-3a)] / (x + a) = 3 + (2-3a) / (x + a) = 3 + (2-3a) / (x + a) so y is not equal to 3Y (x + a) = 3x + 2xy-3x = 3x + 2xy-3x = 2-ayx (Y-3) = (2-ay) x = (2-ay) / (Y-3) so the inverse function is F-1 (x) = (2-ax) / (x-3) (x-3) (x-3) (x-3) (x-3) (x-3 x ≠ 3) 2) F-1 (x

Given that y = f (x) is an odd function, if x is greater than or equal to 0, f (x) = 3 ^ X-1, let the inverse function of F (x) be y = g (x), find the value of G (- 8)

Y = f (x) is an odd function
There are:
f(x)=-f(-x)
f(x)=3^x-1,(x>=0)
Let x