I need to find the domain and range of $f(x,y)=\sqrt{1+x-y^2}$.

Can someone walk me through the proper reasonings in solving this problem?

My attempt

  1. Domain

From looking at the function I get:

$D=\{(x,y)\mid 1+x-y^2\geq 0\}$

This means that the domain can be sketched as some kind of parabola.

  1. Range

The range is

$\{z\mid z=\sqrt{1+x-y^2},(x,y)\in D\}$.

However, the textbook says that the range is $\mathbb{Z}^+$. How do I get that?

  • $\begingroup$ @pjs36 I had forgotten $\geq 0$ in the domain, updated now. $\endgroup$ – jukka.aalto Sep 1 '15 at 23:05
  • $\begingroup$ Is it possible to have a function without already having a domain? $\endgroup$ – zhw. Sep 2 '15 at 0:28

1) For the domain, you have the correct set; and it corresponds to the set of points to the right of, or on, the parabola $x=y^2-1$ (since $x\ge y^2-1$).

2) The range is given by $[0,\infty)$, since $z=\sqrt{1+x-y^2}\implies z\ge0$;

and if $z$ is any number with $z\ge0$, then $z$ is in the range of $f$ since $x=z^2-1, y=0$ gives

$f(x,y)=\sqrt{z^2}=|z|=z$ (since $z\ge0)$.


I am going to assume (you should really specify this) that $x,y\in\mathbb{R}$. The domain is rather straightforward since all you need is the expression within the root sign to be non-negative. That is, you need $1+x-y^2\geq0$ or "in more simple terms" $1+x\geq y^2$. So $$ \operatorname{Dom}\bigl(f(x,y)\bigr)=\{1+x\geq y^2\mid x,y\in\mathbb{R}\}. $$ The range is slightly more tricky but not so much. Since $x$ and $y$ are allowed to be anything in $\mathbb{R}$ (so long as they are in the domain), we can work out a way for all of non-negative $\mathbb{R}$ to be mapped to. That is, $$ \operatorname{Rng}\bigl(f(x,y)\bigr)=\mathbb{R}_{\geq0}. $$ For example, how could you get $f(x,y)=e$ for some values of $x$ and $y$? What if $x=y^2-1+e^2$? Then we have that $f(x,y)=\sqrt{1+x-y^2}=\sqrt{1+(y^2-1+e^2)-y^2}=\sqrt{e^2}=e$.

Can you figure it out from here?


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