Inverse this please (1 Viewer)

[currysauce]

y = (x-4)/(x-1)

 

[acmilan]

It is its own inverse, ie f^-1 = f

 

[currysauce]

are u sure, because the answer is

(2x-4)/(x-1)??

 

[gordo]

plot it, draw a y=x line and find the intercepts of the inverse

 

[gordo]

although doing it algebraicly don;t u get:

x = (y-4)(y-1)
x = y^2 - 5y +4
y^2 - 5y = x - 4
y^2 - 5y + (5/2)^2 = x - 4 - (5/2)^2 (complete the square)
(y - (5/2) )^2 = x -41/4
y = sqrt(x - 41/4) + 5/2

u get x intercepts of 16.5 , 0 and 1 , 0

correct me if i'm wrong?

 

[rama_v]

y = (x-4)/(x-1)

y = (x-4)/(x-1)
x = (y-4)/(y-1)
x(y-1) = y - 4
xy - x = y - 4
xy - x - y + 4 =0
y(x-1)=x-4
y = (x-4)/(x-1)

 

[KFunk]

y = (x-4)/(x-1) so switch the x's and the y's ...

x = (y-4)/(y-1)

yx - x = y - 4

y(x - 1) = x - 4

∴ y = (x - 4)/(x - 1) so f(x) = f^-1(x)

As acmilan said, it is it's own inverse.

 

[rama_v]

although doing it algebraicly don;t u get:

x = (y-4)(y-1)
x = y^2 - 5y +4
y^2 - 5y = x - 4
y^2 - 5y + (5/2)^2 = x - 4 - (5/2)^2 (complete the square)
(y - (5/2) )^2 = x -41/4
y = sqrt(x - 41/4) + 5/2

u get x intercepts of 16.5 , 0 and 1 , 0

correct me if i'm wrong?

You multiplied them - its:

x = (y-4)/(y-1) not x = (y-4)(y-1)

 

[currysauce]

ok guys, i'm going to go with the assumption that the answer in the back of the book is incorrect. Thanks for your time!

 

[香港!]

didn't u ask that in dat other "Inverse" thread already