I just need some clarifications. I'm given a function of two variables $f(x,y)=2-x^2-y^2+2x-4y$ and I'm asked to find the domain and range of it. Now I know that the domain of this is all real numbers but would it be correct to say that the range would also be the set of all real numbers?
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
2
Not neccesarily.
\begin{align} 2-x^2-y^2+2x-4y&=-(x-1)^2+1-(y-2)^2+4+2\\&=-(x-1)^2-(y-2)^2+7\\&\le7 \text{ for all } x,y\in{\Bbb{R}} \end{align}
answered May 18, 2015 at 1:33
-
$\begingroup$ oh right it's R x R thank you so much for the help $\endgroup$– madisonCommented May 18, 2015 at 1:36
-
$\begingroup$ You're welcomed. And the others also have contributed. $\endgroup$ Commented May 18, 2015 at 1:37
$\begingroup$
$\endgroup$
0
The domain is NOT $\mathbb{R}$. The domain is $\mathbb{R}\times\mathbb{R}$, as it is a function of two variables.
The range is $\{z\in \mathbb{R} \colon z\leq 7\}$. You can see this by noting that the parabola $p(x)=-x^2+2x$ has range $\{z\in \mathbb{R} \colon z\leq 1\}$ (simply find the vertex) and the parabola $q(y)=-y^2-4y$ has range $\{z\in \mathbb{R} \colon z\leq 4\}$.
So the sum of these two functions (and 2) has range $\{z\in \mathbb{R} \colon z\leq 1+4+2=7\}$.
$\begingroup$
$\endgroup$
1
The domain is actually the set $\mathbb{R} \times \mathbb{R}$, which is the set of pairs of real numbers, since this is a function of two variables. The range is not the entire set $\mathbb{R}$. Far away from the origin it behaves like $-x^2 - y^2$, so there is no way it can achieve all positive values.
-
$\begingroup$ yes i completely forgot that it's supposed to be R x R thanks for the help though $\endgroup$– madisonCommented May 18, 2015 at 1:37