# Inverse functions with restricted domain

• Sep 8th 2013, 07:27 AM
Paze
Inverse functions with restricted domain
I'm wondering how the domain affects the inverse function.

Taking the example: $y=\sqrt{x+3}, x\geq 3$

The inverse is $y=x^2-3, x\geq 0$

Why is the inverse restricted to 0 while the other was restricted to -3? I see it on a graph and I understand that it must be because the function stops being a 1 on 1 at x=0 but I'm wondering how we figure that out with functions that are more complex such as a sinus function or something?
• Sep 8th 2013, 08:35 AM
Plato
Re: Inverse functions with restricted domain
Quote:

Originally Posted by Paze
Taking the example: $y=\sqrt{x+3}, x\geq 3$
The inverse is $y=x^2-3, x\geq 0$
Why is the inverse restricted to 0 while the other was restricted to -3? I see it on a graph and I understand that it must be because the function stops being a 1 on 1 at x=0 but I'm wondering how we figure that out with functions that are more complex such as a sinus function or something?

Use function notation: $f(z)=\sqrt{x+3}~\&~g(x)=x^2-3$.

For inverses we require $f(g(x))=g(f(x))=x$

But $f(g(x))=\sqrt{(x^2-3)+3}=\sqrt{x^2}$. What does it take to insure that $\sqrt{x^2}=x~?$
• Sep 8th 2013, 10:39 AM
Paze
Re: Inverse functions with restricted domain
Hm, it seems to me that the function can take on negative values as well? Seems to me that it equals x by definition? Can you please clarify.
• Sep 8th 2013, 11:45 AM
Plato
Re: Inverse functions with restricted domain
Quote:

Originally Posted by Paze
Hm, it seems to me that the function can take on negative values as well? Seems to me that it equals x by definition? Can you please clarify.

In order for $\sqrt{x^2}=x$ it is necessary that $x\ge 0$.
• Sep 8th 2013, 02:59 PM
Paze
Re: Inverse functions with restricted domain
Hm. I must be misunderstanding, as $\sqrt{(-1)^2}=1$
sqrt((-1)^2) - Wolfram|Alpha

What am I misunderstanding?

EDIT: Woops..Now I see what you mean. -1 is of course not the same as 1. Thank you!
• Sep 9th 2013, 10:50 AM
Paze
Re: Inverse functions with restricted domain
What if we do it the other way around and take g(f(x))? Then we get $\sqrt{x+3}^2-3$ which must mean that x has to be equal to or larger than -3 as well? Or doesn't that matter?

As far as I remember, f(g(x))=g(f(x))=x for it to be a valid inverse.
• Sep 9th 2013, 10:57 AM
Plato
Re: Inverse functions with restricted domain
Quote:

Originally Posted by Paze
What if we do it the other way around and take g(f(x))? Then we get $\sqrt{x+3}^2-3$ which must mean that x has to be equal to or larger than -3 as well? Or doesn't that matter?
As far as I remember, f(g(x))=g(f(x))=x for it to be a valid inverse.

Well if $x\ge 0$ then you know that $x\ge -3$. Right?
• Sep 9th 2013, 11:06 AM
Paze
Re: Inverse functions with restricted domain
Well...According to wolfram I'm wrong (sqrt(-4+3))^2-3 - Wolfram|Alpha

How is it that this is calculated? I would have thought that we started calculating inside the root to get -1, then attempt to take the root but we would end up with an imaginary number...But perhaps that imaginary number squared is equal to -1? (I'm not very familiar with complex numbers).
• Sep 9th 2013, 11:16 AM
Plato
Re: Inverse functions with restricted domain
Quote:

Originally Posted by Paze
Well...According to wolfram I'm wrong (sqrt(-4+3))^2-3 - Wolfram|Alpha
How is it that this is calculated? I would have thought that we started calculating inside the root to get -1, then attempt to take the root but we would end up with an imaginary number...But perhaps that imaginary number squared is equal to -1? (I'm not very familiar with complex numbers).

I have no idea what you are talking about.
The functions $f(z)=\sqrt{x+3}~\&~g(x)=x^2-3$ are inverses of each other if $x\ge 0$.
Therefore, $-4$ has absolutely to do with any of this because $-4<0$.
• Sep 9th 2013, 12:49 PM
Paze
Re: Inverse functions with restricted domain
Quote:

Originally Posted by Plato
I have no idea what you are talking about.
The functions $f(z)=\sqrt{x+3}~\&~g(x)=x^2-3$ are inverses of each other if $x\ge 0$.
Therefore, $-4$ has absolutely to do with any of this because $-4<0$.

No wonder. I'm not even sure what I am asking anymore. Thank you for your insight, it has clarified the matter greatly. I will keep studying the subject.