The inverse function is used to express y = x-x, 2 x belongs to [1,2]

The inverse function is used to express y = x-x, 2 x belongs to [1,2]

Y = 2 parts of x-x, X belongs to [1,2], which is an increasing function with the range of [- 1,1]
Inverse solution equation y = X-2 / X
x^2-2-yx=0
X = 1 / 2 * (y + - radical (y ^ 2 + 8))
Because x is positive
X = 1 / 2 * (y + radical (y ^ 2 + 8))
So the inverse function is
Y = 1 / 2 * (x + radical (x ^ 2 + 8)) is defined as [- 1,1]

How to find the inverse function of y = half x + 2?

In fact, it is very simple to find the inverse function. Originally, the formula of X is used to express y. in the inverse function, the formula containing y is used to express X. finally, X is replaced by Y, and X is replaced
On this topic:
First of all, the form of function representation changes, from y to x, we get: 2y-4 = X
Letter change: y = 2x - 4 is the inverse function of the original function
It must be understood here that the representation of a function has nothing to do with letters. In fact, in the first part, y is used to denote x, which is already an inverse function

Inverse function of y = (1-x) / (1 + x)

y=(1-x)/(1+x)

How to find the inverse function of y = 3 + 2 ^ X-1

Let's swap X and y
x=3+2^(y-1)
2^(y-1)=x-3
y-1=log2(x-3)
So the inverse function is y = 1 + log2 (x-3)

Inverse function y = 1 + X + e ^ x If f (x) = 1 + X + e ^ x, find f − 1 (2)

The range of the original function is equal to the domain of the inverse function
This problem knows the original function and inverse function, so it is not necessary to calculate it
Let f (x) = 2 and the value x = 0
That is, f − 1 (2) is the value of X when f (x) = 2
f−1(2)=0

Inverse function of y = √ (x-1) + 1 (x > = 1)

Step 1: the solution x is represented by y
y-1=√(x-1)
(y-1)²=x-1
x=(y-1)²+1
Step 2: exchange x, y, the inverse function is
y=(x-1)²+1,x≥1

y=(1_ x) Inverse function of / (1 + x)

The inverse function is XY commutation
That is, x = (1-y) / (1 + y)
x+xy=1-y
xy+y=1-x
y(1+x)=1-x
So the inverse function is y = (1-x) / (1 + x)

Inverse function of y = X-1

Isn't it itself, symmetric about y = x

How to find the inverse function of y = 2x-3 / 5x + 1,

y=(2x-3)/(5x+1)
=[(2/5)(5x+1)-17/5]/(5x+1)
=2/5-17/(25x+5)
25X + 5 is an increasing function, so 17 / (25X + 5) is a decreasing function, so - 17 / (25X + 5) is an increasing function
So the inverse function exists
And - 17 / (25X + 5) is not equal to 0
So y is not equal to 2 / 5
So the domain of the inverse function is x, not equal to 2 / 5
y=(2x-3)/(5x+1)
5xy+y=2x-3
(5y-2)x=-y-3
x=-(y+3)/(5y-2)
So the inverse function is y = - (x + 3) / (5x-2), X is not equal to 2 / 5

2x-3 / 5x + 1 = y find its inverse function?

2X-3/5X+1=Y
2x-3=5xy+y
2x-5xy=3+y
x=3+y/(2-5y)
That is, the inverse function y = 3 + X / (2-5x) x is not equal to 0.4