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#### wishmaster

##### Active member

- Oct 11, 2013

- 211

\(\displaystyle f(x)=\frac{3x-2}{x+2}\)

I now that for inverse i have to express $x$ somehow,but i dont know how to do it....

Thank you for the help!

- Thread starter wishmaster
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- Thread starter
- #1

- Oct 11, 2013

- 211

\(\displaystyle f(x)=\frac{3x-2}{x+2}\)

I now that for inverse i have to express $x$ somehow,but i dont know how to do it....

Thank you for the help!

- Aug 30, 2012

- 1,123

Let y = f(x) so you have

\(\displaystyle f(x)=\frac{3x-2}{x+2}\)

I now that for inverse i have to express $x$ somehow,but i dont know how to do it....

Thank you for the help!

\(\displaystyle y=\frac{3x-2}{x+2}\)

To find the inverse exchange the roles of x and y:

\(\displaystyle x=\frac{3y-2}{y+2}\)

Now solve for y.

-Dan

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- #3

- Oct 11, 2013

- 211

What about surjective and injective?

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- #4

- Mar 5, 2012

- 8,774

You will need the inverse to answer those questions.What about surjective and injective?

Does every value in the codomain have at least one original?

Does every value in the codomain have at most one original?

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- #5

- Oct 11, 2013

- 211

I understand this,but how to prove it mathematicly?

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- #6

- Mar 5, 2012

- 8,774

Can you write down the inverse function first?I understand this,but how to prove it mathematicly?

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- #7

- Oct 11, 2013

- 211

I have calculated that inverse is : \(\displaystyle -\frac{2-2x}{x-3}\)Can you write down the inverse function first?

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- #8

- Mar 5, 2012

- 8,774

Good.I have calculated that inverse is : \(\displaystyle -\frac{2-2x}{x-3}\)

So pick any element in $\mathbb R$.

Your formula gives a unique original of that value for the original function f.

There is 1 exception: the value 3 does not have an original.

So each element in the codomain $\mathbb R$ of f has either 1 or 0 originals.

As such the function f is injective, since each element has at most 1 original.

But the function f is not surjective, since

Therefore, f is not bijective either.

If we restrict the codomain to $\mathbb R \backslash \{ 3 \}$, it

- Thread starter
- #9

- Oct 11, 2013

- 211

Thank you very much! But how do i prove this mathematicly? This is theory,i have to calculate it somehow......Good.

So pick any element in $\mathbb R$.

Your formula gives a unique original of that value for the original function f.

There is 1 exception: the value 3 does not have an original.

So each element in the codomain $\mathbb R$ of f has either 1 or 0 originals.

As such the function f is injective, since each element has at most 1 original.

But the function f is not surjective, sincenotevery element hasat least1 original.

Therefore, f is not bijective either.

If we restrict the codomain to $\mathbb R \backslash \{ 3 \}$, itisboth surjective and bijective.

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- #10

- Mar 5, 2012

- 8,774

First off, I just noticed that your inverse is incorrect.Thank you very much! But how do i prove this mathematicly? This is theory,i have to calculate it somehow......

If you have \(\displaystyle y=f(x)=\frac{3x-2}{x+2}\), then \(\displaystyle x = -\frac{2y+2}{y-3}\).

Anyway, I guess you want it a bit more formal?

Well, $f(x)=3$ has no solution, which we can see from the formula \(\displaystyle x = -\frac{2y+2}{y-3}\), which we deduced from $f(x)=y$, therefore f is not surjective (proof by counter example).

Then suppose f is

That means that \(\displaystyle x_1 = -\frac{2y+2}{y-3}\), but also that \(\displaystyle x_2 = -\frac{2y+2}{y-3}\).

These are both the same value, which is a contradiction.

Therefore f is injective (proof by contradiction).

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- #11

- Oct 11, 2013

- 211

i dont know.....seems inverse is correct,checked by many programs.....First off, I just noticed that your inverse is incorrect.

If you have \(\displaystyle y=f(x)=\frac{3x-2}{x+2}\), then \(\displaystyle x = -\frac{2y+2}{y-3}\).

Anyway, I guess you want it a bit more formal?

Well, $f(x)=3$ has no solution, which we can see from the formula \(\displaystyle x = -\frac{2y+2}{y-3}\), which we deduced from $f(x)=y$, therefore f is not surjective (proof by counter example).

Then suppose f isnotinjective, then there are $x_1, x_2, y \in \mathbb R$, with $x_1 \ne x_2$, such that $y=f(x_1)=f(x_2)$.

That means that \(\displaystyle x_1 = -\frac{2y+2}{y-3}\), but also that \(\displaystyle x_2 = -\frac{2y+2}{y-3}\).

These are both the same value, which is a contradiction.

Therefore f is injective (proof by contradiction).

Yes,i want it formal,and you have given me a great answer,for what im very thankful!

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- #12

- Mar 5, 2012

- 8,774

Here's Wolfram's result: solve x for y=(3x-2)/(x+2) - Wolfram|Alpha Resultsi dont know.....seems inverse is correct,checked by many programs.....

Yes,i want it formal,and you have given me a great answer,for what im very thankful!

- Thread starter
- #13

- Oct 11, 2013

- 211

And heres my result:Here's Wolfram's result: solve x for y=(3x-2)/(x+2) - Wolfram|Alpha Results

(3x-2)/(x+2) inverse - Wolfram|Alpha Results

Anyway,it doesnt matter,i did my work,and i got 100%,which i could not get without your help! Thanks!

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- #14

- Mar 5, 2012

- 8,774

I have calculated that inverse is : \(\displaystyle -\frac{2-2x}{x-3}\)

Wolfram's result from your link is:

$$- \frac{2(x+1)}{x-3}$$

Note that it is not what you originally gave.

There's a difference with a minus sign.

Ah well.

Good!Anyway,it doesnt matter,i did my work,and i got 100%,which i could not get without your help! Thanks!

- Thread starter
- #15

- Oct 11, 2013

- 211

Im not as close so smart as you,but arent those results the same?Wolfram's result from your link is:

$$- \frac{2(x+1)}{x-3}$$

Note that it is not what you originally gave.

There's a difference with a minus sign.

Ah well.

Good!

Yes,good for me!

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- #16

- Mar 5, 2012

- 8,774

I'm afraid not.Im not as close so smart as you,but arent those results the same?

Let me try and show it:

\begin{array}{}

- \frac{2(x+1)}{x-3}

& =- \frac{2+2x}{x-3} & \qquad (1)\\

&= \frac{-(2+2x)}{x-3} \\

&= \frac{-2-2x}{x-3} & \qquad (2) \\

&\ne -\frac{2-2x}{x-3} & \qquad (\text{your version})

\end{array}

I'm trying to show here that (1) and (2) are proper alternate forms of the inverse.

However, (your version) is different from either of them.

- Thread starter
- #17

- Oct 11, 2013

- 211

Isnt \(\displaystyle -\frac{1}{3}\) or \(\displaystyle \frac{-1}{3}\) the same thing?I'm afraid not.

Let me try and show it:

\begin{array}{}

- \frac{2(x+1)}{x-3}

& =- \frac{2+2x}{x-3} & \qquad (1)\\

&= \frac{-(2+2x)}{x-3} \\

&= \frac{-2-2x}{x-3} & \qquad (2) \\

&\ne -\frac{2-2x}{x-3} & \qquad (\text{your version})

\end{array}

I'm trying to show here that (1) and (2) are proper alternate forms of the inverse.

However, (your version) is different from either of them.

But yes,i believe it was my mistake,i dont want to be a smart guy here,just want to learn proper math

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- #18

- Mar 5, 2012

- 8,774

Yes, that is the same thing.Isnt \(\displaystyle -\frac{1}{3}\) or \(\displaystyle \frac{-1}{3}\) the same thing?

But yes,i believe it was my mistake,i dont want to be a smart guy here,just want to learn proper math

However, \(\displaystyle -\frac{1+1}{3}=- \frac 2 3\) and \(\displaystyle \frac{-1+1}{3} = 0\) are not the same thing.

- Feb 21, 2013

- 739

look like this:I'm afraid not.

Let me try and show it:

\begin{array}{}

- \frac{2(x+1)}{x-3}

& =- \frac{2+2x}{x-3} & \qquad (1)\\

&= \frac{-(2+2x)}{x-3} \\

&= \frac{-2-2x}{x-3} & \qquad (2) \\

&\ne -\frac{2-2x}{x-3} & \qquad (\text{your version})

\end{array}

I'm trying to show here that (1) and (2) are proper alternate forms of the inverse.

However, (your version) is different from either of them.

\(\displaystyle -\frac{2-2x}{x-3} =\frac{-(2-2x)}{x-3} \)

so \(\displaystyle -\frac{1}{3}=\frac{-(1)}{3}\)

does this make it easy to see?

Regards,

\(\displaystyle |\pi\rangle\)