# Inverse function

What probably could be the inverse of following function?

y = (x^(1/t)-a^(1/t))*(x^(1/t)-c*b^(1/t))/(x^(1/t)-a^(1/t))*(x^(1/t)-c*b^(1/t))+
(x^(1/t)-   b^(1/t))*x^(1/t)-c*a^(1/t)


a,b,c & t are the parameters. Any help is going to be highly appreciated.

-
Please, refer to the help section in order to improve the fomatting of your future answers/questions. –  Sektor Aug 11 '13 at 18:00

You can use Solve on an equation (== instead of =). It is essentially a quadratic equation, so there are two solutions. Either there is no inverse or there are two, depending whether you restrict the domain appropriately.

eqn = y == (x^(1/t) - a^(1/t))*(x^(1/t) - c*b^(1/t))/(x^(1/t) - a^(1/t))*(x^(1/t) -
c*b^(1/t)) + (x^(1/t) - b^(1/t))*x^(1/t) - c*a^(1/t);
soln = Solve[eqn, x]

(* {{x -> 4^-t (-b^((1/t)) (-1 - 2 c) -
Sqrt[b^(2/t) + 8 a^(1/t) c + 4 b^(2/t) c - 4 b^(2/t) c^2 + 8 y])^t},
{x -> 4^-t (-b^((1/t)) (-1 - 2 c) +
Sqrt[b^(2/t) + 8 a^(1/t) c + 4 b^(2/t) c - 4 b^(2/t) c^2 + 8 y])^t}} *)


Check both solutions:

eqn /. soln // PowerExpand // Expand
(* {True, True} *)

-
Really grateful :) –  Arshad Aug 11 '13 at 18:33

Inverse of a function is calculated using InverseFunction. So converting your equation to some function,

g[a_] = (x^(1/t) -
a^(1/t))*(x^(1/t) - c*b^(1/t))/(x^(1/t) - a^(1/t))*(x^(1/t) -
c*b^(1/t)) + (x^(1/t) - b^(1/t))*x^(1/t) - c*a^(1/t)

gft = InverseFunction[g]


Now checking it,

gft[4]


gives ($\left(\frac{c^2 b^{2/t}-2 c b^{1/t} x^{1/t}-b^{1/t} x^{1/t}+2 x^{2/t}-4}{c}\right)^t$) which is probably the inverse function of your function. Repeat this with simple equation (say x^2), you shall get result as -Sqrt[2] for 4 as inverse.

-