# For what values of the constant k will the following function NOT be its own inverse.?

f(x) = (x-k) / (x-1)

Relevance

Let (x - k) / (x - 1) = y

x - k = yx - y

x = yx - y + k

x - yx = y + k

x(1 - y) = y + k

x = (-y + k) / (1 - y)

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

It would seem that the function is generally its own inverse. However...

f(x) = (x - k) / (x - 1)

Then f^-1(f(x)) should be equal to x.

f^-1(f(x)) = ( [(x - k) / (x - 1)] - k) / ( [(x - k) / (x - 1)] - 1)

We can see that this expression will generally be equal to x after some manipulation, except when the denominator is equal to 0, i.e. when k = 1. We can surmise that the function is not its own inverse when k = 1.

This is easily verifiable.

For example, for k = 1, f(2) = 1 but f^-1(1) is not even defined.

edit:

Here are the manipulations that make sure that:

( [(x - k) / (x - 1)] - k) / ( [(x - k) / (x - 1)] - 1) = x

( [(x - k) / (x - 1)] - k) / ( [(x - k) / (x - 1)] - 1)

= [x - k - k(x - 1)] / [x - k - x - 1]

= (x - k - kx + k) / (1 - k)

= (x - kx) / (1 - k)

= [x (1 - k)] / (1 - k)

= x

• 4 years ago

First discover the inverse of f(x). y = (x - ok)/(x - a million) Swop x and y and resolve for y. x = (y - ok))/(y - a million) ----> y = (x - ok)/(x - a million) are you able to fill interior the intermediate steps? This shows that f(x) is its very own inverse for any fee of ok so there is no fee for which it is not its very own inverse.