Inverse function of y = 2x parts x + 1

Inverse function of y = 2x parts x + 1

Y = (x + 1) / 2x = 1 / 2 + 1 / 2x, definition domain: {x ≠ 0}, range: {y ≠ 1 / 2}
So 1 / 2x = Y-1 / 2
So 1 / x = 2y-1
So x = 1 / (2y-1)
If "X" and "Y" are interchanged, y = 1 / (2x-1), the definition domain is {x ≠ 1 / 2}, and the range is {y ≠ 0}
So the inverse function of y = (x + 1) / 2x is:
y=1/(2x-1)

Find the inverse function of y = (2x + 1) / (x + 3)

2x+1=y(x+3)
x(2-y)=3y-1
x=(3y-1)/(2-y)
So the inverse function is y = (3x-1) / (2-x)

Inverse function of y = 2x-1 / X

y=x/(2x-1)
be
y(2x-1)=x
2yx-y=x
(2y-1)x=y
x=y/(2y-1)
therefore
The inverse function is as follows:
Y = x / (2x-1), X is not equal to 1 / 2

The inverse function of y = 4x + 2 / 2x-3

Y(4X+2)=2X-3 4Y*X+2Y=2X-3 (4Y-2)X=-3-2Y X=(-3-2Y)/(4Y-2)
So the inverse function y = (- 3-2x) / (4x-2)

Inverse function of y = x + 1 / 2x – 3

x=3y+1/2y-1

Find the inverse function of y = - 3 ^ X-1 / 3 ^ x + 1

In order to calculate easily, let 3 ^ x = t, then t > 0
y=-3^x-1/3^x+1=-（t+1/t)+1.
Because t + 1 / T ≥ 2, y ≤ - 1
T = (1-y) + √ (y ^ 2-2y-3) / 2. The minus sign is omitted because t > 0
3^x=(1-y)+√（y^2-2y-3)/2
X = log3 [(1-y) + √ (y ^ 2-2y-3) / 2], 3 is the base number
Change the letter Y = log3 [(1-x) + √ (x ^ 2-2x-3) / 2],
The definition domain is (- ∞, - 3]

Inverse function of y = 3 ^ x + 1 My approach is: Because 3 ^ x = Y-1, log3 ^ (Y-1) = x, so the inverse function is: y = log3 ^ (x-1) But why is the answer y = ln (x-1) / Ln3

These two are the same
Log3 M = ln M / Ln3 change to natural logarithm base

Inverse function of y = 3 ^ (x-1) (x > 1)

X>1
x-1>0
3^(x-1)>1
Y>1
log3(y)=x-1
x=log3(y)+1
Inverse function
y=1+log3(x),x>1

How to find the inverse function of the function y = x + 9 / x?

The domain of y = x + 9 / X is x0, and the range is y ≥ 6 or Y ≤ - 6
y=x+9/x=(x²+9)/x
x²-yx+9=0
To solve the quadratic equation with respect to X
x=[y±√(y²-36)]/2
Then the inverse function is: y = [x ±√ (x? - 36)] / 2 (x ≥ 6 or X ≤ - 6)

If the function f (x) = 1 + 2x and the inverse function y = F-1 (x), then F-1 (9)=______ .

Let f (1) = 1,
∴f（a）=1+2a=9，
A = 3, that is, F-1 (9) = 3
The inverse function of the function f (x) = 1 + 2x is y = F-1 (x) = log2 (x-1)
∴f-1（9）=log28=3
So the answer is: 3