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[SOLVED] Inverse of the function and find if is surjective/injective

wishmaster

Active member
Oct 11, 2013
211
For the given function i have to find if is surjective/injective and find the inverse of the function:

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

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

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
For the given function i have to find if is surjective/injective and find the inverse of the function:

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

I now that for inverse i have to express $x$ somehow,but i dont know how to do it....
Thank you for the help!
Let y = f(x) so you have
\(\displaystyle y=\frac{3x-2}{x+2}\)

To find the inverse exchange the roles of x and y:
\(\displaystyle x=\frac{3y-2}{y+2}\)

Now solve for y.

-Dan
 

wishmaster

Active member
Oct 11, 2013
211
What about surjective and injective?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
What about surjective and injective?
You will need the inverse to answer those questions.

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

wishmaster

Active member
Oct 11, 2013
211
I understand this,but how to prove it mathematicly?
 

Klaas van Aarsen

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Mar 5, 2012
8,774

wishmaster

Active member
Oct 11, 2013
211

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,774
I have calculated that inverse is : \(\displaystyle -\frac{2-2x}{x-3}\)
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, since not every element has at least 1 original.
Therefore, f is not bijective either.

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

wishmaster

Active member
Oct 11, 2013
211
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, since not every element has at least 1 original.
Therefore, f is not bijective either.

If we restrict the codomain to $\mathbb R \backslash \{ 3 \}$, it is both surjective and bijective.
Thank you very much! But how do i prove this mathematicly? This is theory,i have to calculate it somehow......
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,774
Thank you very much! But how do i prove this mathematicly? This is theory,i have to calculate it somehow......
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 is not injective, 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).
 

wishmaster

Active member
Oct 11, 2013
211
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 is not injective, 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).
i 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!
 

Klaas van Aarsen

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Mar 5, 2012
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wishmaster

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Oct 11, 2013
211

Klaas van Aarsen

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Staff member
Mar 5, 2012
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wishmaster

Active member
Oct 11, 2013
211
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! :)
Im not as close so smart as you,but arent those results the same?

Yes,good for me! (Tongueout)
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,774
Im not as close so smart as you,but arent those results the same?
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.
 

wishmaster

Active member
Oct 11, 2013
211
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.
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 ;)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
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 ;)
Yes, that is the same thing.

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

Petrus

Well-known member
Feb 21, 2013
739
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.
look like this:
\(\displaystyle -\frac{2-2x}{x-3} =\frac{-(2-2x)}{x-3} \)
so \(\displaystyle -\frac{1}{3}=\frac{-(1)}{3}\)
does this make it easy to see?

Regards,
\(\displaystyle |\pi\rangle\)