Inverse function of F (x) = x|x| + 2x

Inverse function of F (x) = x|x| + 2x

Obviously, if x ≥ 0, f (x) = x ^ 2 + 2x = (x + 1) ^ 2 - 1, so its inverse function f (x) = √ (x + 1) - 1 (x ≥ - 1); when x < 0, f (x) = - x ^ 2 + 2x = - (x-1) ^ 2 + 1, so its inverse function f (x) = √ (1-x) + 1 (x ≤ 1)

Inverse function of F (2x-1) = x + 1

f(2x-1)=x+1=(2x-1)/2+3/2
f(x)=x/2+3/2=y
x=2y-3
Inverse function 2x-3
Another one
Inverse function g (x)
g(x+1)=2x-1
g(x)=2(x-1)-1=2x-3

Inverse function F-1 (x) =? X ^ 2-2x + 1 (x > = 1)?

Let y = f (x) = x? - 2x + 1 = (x-1) 2
x≥1 x-1≥0 (x-1)²≥0 y≥0
x-1=√y
x=√y +1
Swap X and y
y=√x+1 (x≥0)
The inverse function is F-1 (x) = √ x + 1 (x ≥ 0)

The function f (x) = x ^ 2 + 2x + 2 (x

f(x)=x^2+2x+2
=(x+1)^2+1(x

For a function defined on R, f (x) has an inverse function F-1 (x), and f (x) passes through a point (2,1), and the inverse function of F (2x) is F-1 (2x), then F-1 (16) is (writing process)

Let t = 2x, y = f (T), then t = F-1 (y), so x = t / 2 = F-1 (y) / 2
That is, the inverse function of F (2x) should be F-1 (x) / 2
The condition in the question is given as F-1 (2x)
Thus F-1 (2x) = F-1 (x) / 2, thus F-1 (16) = F-1 (8) / 2 = F-1 (4) / 4 = F-1 (2) / 8
=F-1(1)/16
And f (x) passes through point (2,1), so F-1 (x) passes through point (1,2)
Thus F-1 (16) = 2 / 16 = 1 / 8

How to find the inverse function of function f (x) = 2x + 1

F (x) = 2x + 1, that is
y=2x+1
Move x to one side
x=(y-1)/2
The inverse function is y = (x-1) / 2

Inverse function f ^ - 1 (x) =? (x) = √ (1-2x)? Seeking process

y=√(1-2x)
1-2x=y²
2x=1-y²
x=(1-y²)/2
That is, f ^ - 1 (x) = (1-x 2) / 2 (x ≥ 0)

If the function f (x) = 2x + 1, then the inverse function of F (2x + 1) is

f-1(x)=(x-3)/4

Find the inverse function of F (x) = 2x-1 / 1-x

Let y = f (x) = (2x-1) / (1-x)
Then y (1-x) = 2x-1
y-yx=2x-1
(2+y)x=y+1
So x = (y + 1) / (y + 2) (Y ≠ - 2)
x. Y is converted into: y = (x + 1) / (x + 2) (x ≠ - 2)

It is known that the inverse function of function y = f (x) is y = f ^ - 1 (x), and the inverse function of function f (2x-1) + 1 is?

Let y = f (2x-1) + 1
The inverse function is XY transposition
x=f(2y-1)+1
x-1=f(2y-1)
So f ^ - 1 (x-1) = 2y-1
So the inverse function is y = [f ^ - 1 (x-1) + 1] / 2