Find the inverse of the function f(x)= 2x+1/x-2?
回答 (6)
2009-01-23 2:37 pm
✔ 最佳答案
put it in the form of y =y = 2x+1/x-2
I can interpret this in at least 3 different ways. I'll pick one.
y = (2x+1)/(x–2)
swap y and x
x = (2y+1)/(y–2)
solve for y
xy – 2x = 2y +1
xy – 2y = 2x +1
y(x–2) = (2x+1)
y = (2x+1) / (x–2)
.
2009-01-23 10:39 pm
First interchange Y with X
X= 2y+1/(y-2)
x(y-2)=2y+1
xy-2x = 2y+1
xy-2y =2x+1
y(x-2)= 2x+1
then
Y=((2x+1)/(x-2)
hope this help
X= 2y+1/(y-2)
x(y-2)=2y+1
xy-2x = 2y+1
xy-2y =2x+1
y(x-2)= 2x+1
then
Y=((2x+1)/(x-2)
hope this help
2009-01-23 11:00 pm
f(x)= 2x+1/x-2
y=2x+1/x-2
x=2y+1/y-2
x(y-2)=2y+1
xy-2x=2y+1
xy-2y=1+2x
y(x-2)=1+2x
y=(1+2x)/(x-2)
or f-1(x)=(2x+1)/(x-2) answer
y=2x+1/x-2
x=2y+1/y-2
x(y-2)=2y+1
xy-2x=2y+1
xy-2y=1+2x
y(x-2)=1+2x
y=(1+2x)/(x-2)
or f-1(x)=(2x+1)/(x-2) answer
參考: @(^_^)@
2009-01-23 10:48 pm
f(x) = (2x+1)/(x-2)
let...
f(x) = y
f^(-1)(y) = x
y = (2x + 1)/(x - 2)
y(x - 2) = 2x + 1
xy - 2y = 2x + 1
xy - 2x = 1 + 2y
x(y - 2) = 1 + 2y
x = (1+ 2y)/(y - 2)
f^(-1) (y) = (1+ 2y)/(y - 2)
therefore...
f^(-1) (x) = (1 + 2x)/(x - 2)
let...
f(x) = y
f^(-1)(y) = x
y = (2x + 1)/(x - 2)
y(x - 2) = 2x + 1
xy - 2y = 2x + 1
xy - 2x = 1 + 2y
x(y - 2) = 1 + 2y
x = (1+ 2y)/(y - 2)
f^(-1) (y) = (1+ 2y)/(y - 2)
therefore...
f^(-1) (x) = (1 + 2x)/(x - 2)
2009-01-23 10:47 pm
f(x) = (2x + 1)/(x - 2)
Change f(x) into y and solve for x:
f(x) = (2x + 1)/(x - 2)
y = (2x + 1)/(x - 2)
y(x - 2) = 2x + 1
xy - 2y = 2x + 1
xy - 2x = 2y + 1
x(y - 2) = 2y + 1
x = (2y + 1)/(y - 2)
Change x into f^-1(x) and y into x:
x = (2y + 1)/(y - 2)
f^-1(x) = (2y + 1)/(y - 2)
Change f(x) into y and solve for x:
f(x) = (2x + 1)/(x - 2)
y = (2x + 1)/(x - 2)
y(x - 2) = 2x + 1
xy - 2y = 2x + 1
xy - 2x = 2y + 1
x(y - 2) = 2y + 1
x = (2y + 1)/(y - 2)
Change x into f^-1(x) and y into x:
x = (2y + 1)/(y - 2)
f^-1(x) = (2y + 1)/(y - 2)
2009-01-23 10:43 pm
We can interpret the problem as finding the function such that
(2y+1)/(y-2) = x
(2y-4+5)/(y-2) = x
(2y-4)/(y-2) + 5/(y-2) = x
2 + 5/(y-2) = x
5/(y-2) = x-2
5/(x-2) = y - 2
y = 2 + 5/(x-2)
y = (2x-4)/(x-2) + 5/(x-2)
y = (2x+1)/(x-2)
So the function is the inverse of itself.
(2y+1)/(y-2) = x
(2y-4+5)/(y-2) = x
(2y-4)/(y-2) + 5/(y-2) = x
2 + 5/(y-2) = x
5/(y-2) = x-2
5/(x-2) = y - 2
y = 2 + 5/(x-2)
y = (2x-4)/(x-2) + 5/(x-2)
y = (2x+1)/(x-2)
So the function is the inverse of itself.
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