Inverse functions with restricted domain

I'm wondering how the domain affects the inverse function.

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

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

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?

Re: Inverse functions with restricted domain

Quote:

Originally Posted by

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

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

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

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

But $\displaystyle f(g(x))=\sqrt{(x^2-3)+3}=\sqrt{x^2}$. What does it take to insure that $\displaystyle \sqrt{x^2}=x~?$

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.

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 $\displaystyle \sqrt{x^2}=x$ it is necessary that $\displaystyle x\ge 0$.

Re: Inverse functions with restricted domain

Hm. I must be misunderstanding, as $\displaystyle \sqrt{(-1)^2}=1$

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!

Re: Inverse functions with restricted domain

What if we do it the other way around and take g(f(x))? Then we get $\displaystyle \sqrt{x+3}^2-3$ 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.

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 $\displaystyle \sqrt{x+3}^2-3$ 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 $\displaystyle x\ge 0$ then you know that $\displaystyle x\ge -3$. Right?

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).

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 $\displaystyle f(z)=\sqrt{x+3}~\&~g(x)=x^2-3$ are inverses of each other if $\displaystyle x\ge 0$.

Therefore, $\displaystyle -4$ has absolutely to do with any of this because $\displaystyle -4<0$.

Re: Inverse functions with restricted domain

Quote:

Originally Posted by

**Plato** I have no idea what you are talking about.

The functions $\displaystyle f(z)=\sqrt{x+3}~\&~g(x)=x^2-3$ are inverses of each other if $\displaystyle x\ge 0$.

Therefore, $\displaystyle -4$ has absolutely to do with any of this because $\displaystyle -4<0$.

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.