The inverse of a one to one function

**vaironxxrd** · November 6th 2012, 01:12 PM

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
$f(x) = \frac{1}{x-2}$

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

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

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

**Plato** · November 6th 2012, 01:50 PM

Originally Posted by vaironxxrd

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

Note that 2 is not in the domain of $f$ .
Try $g(x)=\frac{2x+1}{x}$ .

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

**vaironxxrd** · November 6th 2012, 02:13 PM

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

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 ^_^

**Plato** · November 6th 2012, 02:18 PM

[QUOTE=vaironxxrd;749609]

Originally Posted by Plato

Note that 2 is not in the domain of $f$ .
Try $g(x)=\frac{2x+1}{x}$ .
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 $x=\frac{1}{y-2}$ for $y$ .

**vaironxxrd** · November 7th 2012, 04:17 AM

[QUOTE=Plato;749610]

Originally Posted by vaironxxrd

Here is the process.
Solve $x=\frac{1}{y-2}$ for $y$ .

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.

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

$x(y-2) = 1$

$xy - 2x = 1$

$xy = 1 + 2x$

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

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

**vaironxxrd** · November 7th 2012, 04:29 AM

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

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

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

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

$\frac{x}{1}$

$= x$

**HallsofIvy** · November 7th 2012, 04:37 AM

[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.

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

$x(y-2) = 1$

$xy - 2x = 1$

$xy = 1 + 2x$

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

Wouldn't the two x cancel out?

What is $\frac{1+ 2(3)}{3}$ ? Is that the same thing as $\frac{1+2}{1}$ ?

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 $\frac{x(a+ b)}{x(c+ d)}$ but not in $\frac{x+ b}{x(c+ d)}$ or $\frac{x+ b}{x+ d}$ .

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