MHB Inverse of the function and find if is surjective/injective

[theakdad]

For the given function i have to find if is surjective/injective and find the inverse of the function:

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

I now that for inverse i have to express $x$ somehow,but i don't know how to do it...
Thank you for the help!

 

[topsquark]

For the given function i have to find if is surjective/injective and find the inverse of the function:

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

I now that for inverse i have to express $x$ somehow,but i don't know how to do it...
Thank you for the help!

Let y = f(x) so you have
$$y=\frac{3x-2}{x+2}$$

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

Now solve for y.

-Dan

 

[theakdad]

What about surjective and injective?

 

[I like Serena]

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?

 

[theakdad]

I understand this,but how to prove it mathematicly?

 

[I like Serena]

I understand this,but how to prove it mathematicly?

Can you write down the inverse function first?

 

[theakdad]

Can you write down the inverse function first?

I have calculated that inverse is : $$-\frac{2-2x}{x-3}$$

 

[I like Serena]

I have calculated that inverse is : $$-\frac{2-2x}{x-3}$$

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.

 

[theakdad]

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...

 

[I like Serena]

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 $$y=f(x)=\frac{3x-2}{x+2}$$, then $$x = -\frac{2y+2}{y-3}$$.Anyway, I guess you want it a bit more formal?Well, $f(x)=3$ has no solution, which we can see from the formula $$x = -\frac{2y+2}{y-3}$$, 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 $$x_1 = -\frac{2y+2}{y-3}$$, but also that $$x_2 = -\frac{2y+2}{y-3}$$.
These are both the same value, which is a contradiction.
Therefore f is injective (proof by contradiction).

 

[theakdad]

First off, I just noticed that your inverse is incorrect.
If you have $$y=f(x)=\frac{3x-2}{x+2}$$, then $$x = -\frac{2y+2}{y-3}$$.Anyway, I guess you want it a bit more formal?Well, $f(x)=3$ has no solution, which we can see from the formula $$x = -\frac{2y+2}{y-3}$$, 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 $$x_1 = -\frac{2y+2}{y-3}$$, but also that $$x_2 = -\frac{2y+2}{y-3}$$.
These are both the same value, which is a contradiction.
Therefore f is injective (proof by contradiction).

i don't know...seems inverse is correct,checked by many programs...

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

 

[I like Serena]

i don't know...seems inverse is correct,checked by many programs...

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

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

 

[theakdad]

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

And here's my result:

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

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

 

[I like Serena]

I have calculated that inverse is : $$-\frac{2-2x}{x-3}$$

And here's my result:

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

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.

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

Good! :)

 

[theakdad]

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)

 

[I like Serena]

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.

 

[theakdad]

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 $$-\frac{1}{3}$$ or $$\frac{-1}{3}$$ the same thing?
But yes,i believe it was my mistake,i don't want to be a smart guy here,just want to learn proper math ;)

 

[I like Serena]

Isnt $$-\frac{1}{3}$$ or $$\frac{-1}{3}$$ the same thing?
But yes,i believe it was my mistake,i don't want to be a smart guy here,just want to learn proper math ;)

Yes, that is the same thing.

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

 

[Petrus]

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:
$$-\frac{2-2x}{x-3} =\frac{-(2-2x)}{x-3} $$
so $$-\frac{1}{3}=\frac{-(1)}{3}$$
does this make it easy to see?

Regards,
$$|\pi\rangle$$