Question

is it always possible to find expression for the inverse of a function ? (im talking only about the expression and not failure cause of domain)

Answer 1

no. especially since it is not always possible to find the expression for a function

Answer 2

well when i have something like :x - (2/x) = (1/y) + 1
can i make x(y) ?

Answer 3

and for 2y = x^2 - (1/x^2) ?

Answer 4

cause i could not find..

Answer 5

the OS seems a bit buggy lately

Answer 6

are you asking "can i solve for x?"

Answer 7

yes .. i want x as a function of y

Answer 8

no i don't think you can in this case because if you think of y as a function of x it is not one to one. so you will not be able to use algebra to solve for x

Answer 9

you know i want to break something.. i wasted much time for this two .. cause i believed (since it was given in my home work and said : "find the inverse") that i actually can find an expression for it

Answer 10

the equations that i wrote you is my end point ..

Answer 11

ok what was the original function?

Answer 12

for example the real one they gave me (before i got where i got ) is :x/(x^2-x-2)
but this is only one of the two that i wrote

Answer 13

so you have
\[y=\frac{x}{x^2-x-2}\] right? and you want the inverse. first off lets graph
http://www.wolframalpha.com/input/?i=y%3Dx%2F%28x^2-x-2%29 and see that it is not one to one

Answer 14

then we could still try to solve, we just wont get a function.

Answer 15

i dont care.. i just need an expression but i cant find any..

Answer 16

ok so we start with
\[y=\frac{x}{x^2-x-2}\] and solve for x. first get
\[yx^2-yx-2y=x\] then
\[yx^2-yx-x-2y=0\] and then
\[yx^2-(y+1)x-2y=0\] then use the quadratic formula to get x wtih
\[a=y, b=-(x+1), c = -2y\]

Answer 17

grr i cant believe !!

Answer 18

lol thank you!