2
$\begingroup$

Restrict the domain and codomain of $f$ to make it one-to-one on new domain, onto on new codomain

The function $f(x,y)=3x^2+2y^2-5$

$z=3x^2+2y^2-5$

The domain = $\{(x,y)\in\mathbb{R}^2|x,y\in(-\infty,\infty)\}$

The range = $\{z\in\mathbb{R}|z\in[2,\infty)\}$


One to one means that if $f(x_1,y_1)=f(x_2,y_2)$, then $x_1=x_2,y_1=y_2$

Clearly that is not the case here, take $x=-1,y=-1$, then $f(x,y)=3+2-5=0$, take $x=1,y=1$, then $f(x,y)=3+2-5=0$. Therefore, I would restrict the domain to not include negative values of $x,y$

New domain = $\{(x,y)\in\mathbb{R}^2|x,y\in[0,\infty]\}$


Onto means that every $y\in Y$, where $Y$ is the codomain, has a $x\in X$ that maps to it, where $X$ is the domain.

The range = $\{z\in\mathbb{R}|z\in[2,\infty)\}$

The codomain is just $\{z\in\mathbb{R}\}$

I am having trouble restricting this. I could just set the new codomain to be the range, but is this correct?

$\endgroup$
  • $\begingroup$ Setting the codomain to the range will make the function onto. What about one to one? $\endgroup$ – Ethan Bolker Sep 21 '17 at 14:05
  • $\begingroup$ For one to one I argue that the new domain should be $\{(x,y)\in\mathbb{R})^2|x,y\in[0\infty)\}$ $\endgroup$ – K Split X Sep 21 '17 at 14:14
  • $\begingroup$ I think your answer is correct (I didn't see the part where you set the domain). Note: if the problem doesn't ask for the largest domain you could be sneaky, set the domain to the single point $(0,0)$ and the codomain to the single number $-5%. $\endgroup$ – Ethan Bolker Sep 21 '17 at 14:22
0
$\begingroup$

If you take $D=\{(x,y)\in\mathbb{R}^2\mid x,y\in[0,+\infty]\}=\{(x,y)\in\mathbb{R}^2\mid x,y\ge0\}$, i.e. the entire first quadrant, as the domain, the function is still not one-to-one. For example, consider $z=10$, which is an element of the range. Then $$3x^2+2y^2-5=10 \iff 3x^2+2y^2=15,$$ which, considering $D$ as the domain, still is a quarter of an ellipse, not a single point. So there are infinitely many points $(x,y)\in D$ satisfying $f(x,y)=10$, and the function is not one-to-one. The same holds for any $z\in\text{your range}$.

So you're right that we have to restrict $x$ because of "$x^2$" in the formula, and the same goes for $y$. But that is not sufficient. This question doesn't have a single answer. Some possible answers are to define the domain as $\{(x,y)\in\mathbb{R}^2\mid x\ge0,y=0\}$ or $\{(x,y)\in\mathbb{R}^2\mid x=0,y\ge0\}$ or $\{(x,y)\in\mathbb{R}^2\mid x\ge0,y=x\}$ or …

I'm also not sure about the codomain that you gave. The codomain could be $\mathbb{R}$, but with this codomain the function is obviously not onto. You're right when you say that to make the function onto the codomain must be precisely the range. But the range of the original function is $[-5,+\infty)$, not $[2,+\infty)$.

Depending on how we interpret the question, you probably are allowed to restrict the codomain even further. But then you'll have to adjust the domain accordingly too. You can't consider $f(x,y)=3x^2+2y^2-5$ as a function from $\{(x,y)\in\mathbb{R}^2\mid x,y\ge0\}$ to $[2,+\infty)$, because in this case some values in the domain, such as $(x,y)=(1,1)$, land outside of the codomain.

$\endgroup$
  • $\begingroup$ Thanks for this answer. What about this. For the new domain, I could let $D=\{(x,y)\in\mathbb{R}^2|x\geq 0, y=0\}$, and the new codomain could be the original range, which was $\{z\in\mathbb{R}|z\in[5,+\infty]\}$ $\endgroup$ – K Split X Sep 21 '17 at 22:33
  • $\begingroup$ Then for every $z\in\mathbb{R}|z\in[5,+\infty]$, there are $(x,y)$ that satisfy every solution? $\endgroup$ – K Split X Sep 21 '17 at 22:34
  • $\begingroup$ @KSplitX: Yes. Except that the original range is $[-5,+\infty)$, not "$[5,+\infty]$". $\endgroup$ – zipirovich Sep 21 '17 at 23:56
0
$\begingroup$

This work is mostly correct, but it seems to me that you have miss-identified the range (assuming "range" means "the largest set of points for which the function attains each point in the set"). $f(0,0)=-5$ which doesn't fall into your range. This doesn't stop you from being right that the function is onto on $[2,\infty)$ however.

If you define the image of a function to be the points that are achieved by the function on the domain, then it follows that in every case restricting the co-domain to be the image yields an onto function. As I said, it looks like this is what you mean by "range," though I've more commonly seen "range" and "co-domain" used interchangeably. "Image" is a widely used and unambiguous term for "the largest set of points that the function actually achieves on some input."

One other point of note is that I have read this question a little different from you. By listing the two conditions together, that is, by saying "to make it one-to-one on new domain, onto on new co-domain" I read the question as asking you to do both, simultaneously. So you need to first restrict the domain to make the function one-to-one, and then restrict the co-domain to make the function onto.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.