# Finding the inverse of a function

• June 23rd 2011, 05:42 AM
IBstudent
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
• June 23rd 2011, 05:57 AM
e^(i*pi)
Re: Finding the inverse of a function
Expand $y(x-1) = x$

This gives $yx - y = x$. Now you can get all your x terms on the LHS and factor
• June 23rd 2011, 06:12 AM
Prove It
Re: Finding the inverse of a function
As an alternative, write $\displaystyle \frac{x}{x - 1} = \frac{x - 1 + 1}{x - 1} = 1 + \frac{1}{x - 1}$ so that your original function is

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

Now evaluating the inverse function should be much easier.
• June 23rd 2011, 06:53 AM
IBstudent
Re: Finding the inverse of a function
Quote:

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

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

Now evaluating the inverse function should be much easier.

how? why is (x+1-1)/x-1 = 1+ 1/x-1?
• June 23rd 2011, 07:19 AM
Prove It
Re: Finding the inverse of a function
Quote:

Originally Posted by IBstudent
how? why is (x+1-1)/x-1 = 1+ 1/x-1?

My method is perfectly clear if you understand that $\displaystyle \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$, but anyway, $\displaystyle \frac{x - 1 + 1}{x - 1} = \frac{x - 1}{x - 1} + \frac{1}{x - 1} = 1 + \frac{1}{x - 1}$
• June 23rd 2011, 07:51 AM
IBstudent
Re: Finding the inverse of a function
ahh I see
Thanks :-)
• June 23rd 2011, 08:38 AM
topsquark
Re: Finding the inverse of a function
Quote:

Originally Posted by IBstudent
how? why is (x+1-1)/x-1 = 1+ 1/x-1?

1. Another way to find the inverse function is to switch the roles of x and y. For example:
$y = \frac{x}{x - 1} \to x = \frac{y}{y - 1}$
Now solve for y and you get your inverse function.

2. Please be more careful in how you write you problem. For example:
$y = x/x - 1$
is really
$y = (x/x) - 1 = 1 - 1 = 0$

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

-Dan
• June 23rd 2011, 09:09 PM
IBstudent
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?
• June 23rd 2011, 09:55 PM
mr fantastic
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 = .....
• June 23rd 2011, 10:53 PM
IBstudent
Re: Finding the inverse of a function
I know that, so, the inverse of x/(x-1) is x/(x-1) ?!
thats weird
• June 24th 2011, 12:43 AM
Ackbeet
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.
• June 24th 2011, 03:36 AM
TKHunny
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 $f(x) = \frac{x}{x-1}$ Note, first that $x \ne 1$

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

Notation is important. Don't just skip it.
• June 24th 2011, 04:46 AM
mr fantastic
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.

$f(x) = \frac{ax + b}{cx - a}$ 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?