Finding the inverse of a function

What is the inverse of f(x) = x/x-1 ?

this is my attempt to solve it:

y = x / x-1

y(x-1) = x

hence x(x-1) is the inverse.....

however, substituting with x=5 gives......

5/4=1.25

so pluging this is the inverse function should give 5, right?

lets see:

1.25(1.25-1) = 1.25*.25= 0.3125. WRONG!!

can anyone tell me where I went wrong?

thanks

Re: Finding the inverse of a function

Expand $\displaystyle y(x-1) = x$

This gives $\displaystyle yx - y = x$. Now you can get all your x terms on the LHS and factor

Re: Finding the inverse of a function

As an alternative, write $\displaystyle \displaystyle \frac{x}{x - 1} = \frac{x - 1 + 1}{x - 1} = 1 + \frac{1}{x - 1}$ so that your original function is

$\displaystyle \displaystyle y = 1 + \frac{1}{x - 1} $.

Now evaluating the inverse function should be much easier.

Re: Finding the inverse of a function

Quote:

Originally Posted by

**Prove It** As an alternative, write $\displaystyle \displaystyle \frac{x}{x - 1} = \frac{x - 1 + 1}{x - 1} = 1 + \frac{1}{x - 1}$ so that your original function is

$\displaystyle \displaystyle y = 1 + \frac{1}{x - 1} $.

Now evaluating the inverse function should be much easier.

how? why is (x+1-1)/x-1 = 1+ 1/x-1?

can you please be more clear with your method? thanks

Re: Finding the inverse of a function

Quote:

Originally Posted by

**IBstudent** how? why is (x+1-1)/x-1 = 1+ 1/x-1?

can you please be more clear with your method? thanks

My method is perfectly clear if you understand that $\displaystyle \displaystyle \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$, but anyway, $\displaystyle \displaystyle \frac{x - 1 + 1}{x - 1} = \frac{x - 1}{x - 1} + \frac{1}{x - 1} = 1 + \frac{1}{x - 1}$

Re: Finding the inverse of a function

Re: Finding the inverse of a function

Quote:

Originally Posted by

**IBstudent** how? why is (x+1-1)/x-1 = 1+ 1/x-1?

can you please be more clear with your method? thanks

Two comments.

1. Another way to find the inverse function is to switch the roles of x and y. For example:

$\displaystyle y = \frac{x}{x - 1} \to x = \frac{y}{y - 1}$

Now solve for y and you get your inverse function.

2. Please be more careful in how you write you problem. For example:

$\displaystyle y = x/x - 1$

is really

$\displaystyle y = (x/x) - 1 = 1 - 1 = 0$

What you meant to write is y = x/(x - 1). Please use parentheses in the future.

-Dan

Re: Finding the inverse of a function

okay so 1/(x-1) = y-1

(y-1)(x-1) = 1

yx-y-x+1 = 1

y(x-1)=x

ahhhh I'm confused...:S

is it x/(x-1) or x^2-x that is the inverse?

Re: Finding the inverse of a function

Quote:

Originally Posted by

**IBstudent** okay so 1/(x-1) = y-1

(y-1)(x-1) = 1

yx-y-x+1 = 1

y(x-1)=x

ahhhh I'm confused...:S

is it x/(x-1) or x^2-x that is the inverse?

It is basic algebra.

1/(x-1) = y-1

=> x - 1 = 1/(y - 1) => x = .....

Re: Finding the inverse of a function

I know that, so, the inverse of x/(x-1) is x/(x-1) ?!

thats weird

Re: Finding the inverse of a function

Quote:

Originally Posted by

**IBstudent** I know that, so, the inverse of x/(x-1) is x/(x-1) ?!

thats weird

Any function that is symmetric with respect to a reflection about the line y = x will behave that way. Other examples: y = x and y = 1/x. I'm sure there are others, but they're not coming to mind.

Re: Finding the inverse of a function

Quote:

Originally Posted by

**IBstudent** I know that, so, the inverse of x/(x-1) is x/(x-1) ?!

thats weird

Not quite and not really.

If $\displaystyle f(x) = \frac{x}{x-1}$ Note, first that $\displaystyle x \ne 1$

Then, $\displaystyle f^{-1}(x) = \frac{x}{x-1}$, where again, $\displaystyle x \ne 1$

Notation is important. Don't just skip it.

Re: Finding the inverse of a function

Quote:

Originally Posted by

**Ackbeet** Any function that is symmetric with respect to a reflection about the line y = x will behave that way. Other examples: y = x and y = 1/x. I'm sure there are others, but they're not coming to mind.

To expand on this, functions that are their own inverses are called involutes. It is simple to show that the following functions are involutes:

f(x) = x.

f(x) = -x + b, where b is a real number.

$\displaystyle f(x) = \frac{ax + b}{cx - a}$ where a, b and c are real numbers. f(x) = 1/x is included as a special case.

The question I'll pose for the interested reader is: Are the above functions the only involutes?