# Thread: The inverse of a one to one function

1. ## The inverse of a one to one function

Hello forum,

I personally believe, though I'm not sure, that this topic should be in the Algebra forum. Though, because I'm taking this from a Precalculus class I wasn't entirely sure whether to place it there.

Find the inverse of
$\displaystyle f(x) = \frac{1}{x-2}$

To my understanding I must find $\displaystyle f^-^1$
I have the answers, but I'm interested in knowing how to find the inverse

$\displaystyle F^-^1(x) = How can I find this?$

The o stands for composed with
$\displaystyle f o f^-^1$
$\displaystyle f(f^-^1(x)) = \frac{1}{f^-^1(x)-2}$

2. ## Re: The inverse of a one to one function

Originally Posted by vaironxxrd
Find the inverse of
$\displaystyle f(x) = \frac{1}{x-2}$
Note that 2 is not in the domain of $\displaystyle f$.
Try $\displaystyle g(x)=\frac{2x+1}{x}$.

What is $\displaystyle f\circ g(x)~\&~g\circ f(x)=~?$

3. ## Re: The inverse of a one to one function

[QUOTE=Plato;749605]Note that 2 is not in the domain of $\displaystyle f$.
Try $\displaystyle g(x)=\frac{2x+1}{x}$.

Hello Plato,

I remember you are always actively helping me. I dont know if I portray the following well when I ask for help. But I'm eager to learn the process, the concept.

I'm on a rush to school I will solve it as soon as I have a break. Thanks for replying ^_^

4. ## Re: The inverse of a one to one function

[QUOTE=vaironxxrd;749609]
Originally Posted by Plato
Note that 2 is not in the domain of $\displaystyle f$.
Try $\displaystyle g(x)=\frac{2x+1}{x}$.
I dont know if I portray the following well when I ask for help. But I'm eager to learn the process, the concept.
Here is the process.
Solve $\displaystyle x=\frac{1}{y-2}$ for $\displaystyle y$.

5. ## Re: The inverse of a one to one function

[QUOTE=Plato;749610]
Originally Posted by vaironxxrd
Here is the process.
Solve $\displaystyle x=\frac{1}{y-2}$ for $\displaystyle y$.
Lets see if I can get it right this time. I forgot some of the rules when multiplying fractions with variables, fear not, there are many resources.

$\displaystyle x = \frac{1}{y-2}$

$\displaystyle x(y-2) = 1$

$\displaystyle xy - 2x = 1$

$\displaystyle xy = 1 + 2x$

$\displaystyle y = \frac{1+2x}{x}$

Wouldn't the two x cancel out?
My best answer is because we are adding 1 to it and therefore the values will change

6. ## Re: The inverse of a one to one function

When proving that
$\displaystyle F(x) ~o~ f^-^1(x)~~and~~F^-^1(x) o F(x)$ I had a little trouble.

$\displaystyle F(F^-^1(x)) = \frac{1}{\frac{1+2x}{x}-2}$

$\displaystyle \frac{1}{\frac{1+2x}{x}-2}*x$

$\displaystyle \frac{x}{1+2x-2x}$

$\displaystyle \frac{x}{1}$

$\displaystyle = x$

7. ## Re: The inverse of a one to one function

[QUOTE=vaironxxrd;749802]
Originally Posted by Plato

Lets see if I can get it right this time. I forgot some of the rules when multiplying fractions with variables, fear not, there are many resources.

$\displaystyle x = \frac{1}{y-2}$

$\displaystyle x(y-2) = 1$

$\displaystyle xy - 2x = 1$

$\displaystyle xy = 1 + 2x$

$\displaystyle y = \frac{1+2x}{x}$

Wouldn't the two x cancel out?
What is $\displaystyle \frac{1+ 2(3)}{3}$? Is that the same thing as $\displaystyle \frac{1+2}{1}$?

My best answer is because we are adding 1 to it and therefore the values will change
I'm not sure I can makes sense out of that sentence! Your best answer to what question? The question "Wouldn't the two x cancel out?" require a "yes" or "no" answer, not "because".

In any case, the only time something in numerator and denominator cancel is when they are multipied by the rest of the numerator and denominator: you can cancel the "x"s in $\displaystyle \frac{x(a+ b)}{x(c+ d)}$ but not in $\displaystyle \frac{x+ b}{x(c+ d)}$ or $\displaystyle \frac{x+ b}{x+ d}$.