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Determine function's domain type
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I'm trying to solve this two excercise:
1) Determine the domain of $f(x,y)=sqrt3-xy$, then say if it is open, close, not open neither close; say if it is bounded or not.
To find the solution I solve $3-xy ge 0$ and I find $y le frac3x$. In my opinion the domain is not open neither close, and it is not bounded.
2) Determine the domain of $f(x,y)=log(3-xy)$, then say if it is open, close, not open neither close; say if it is bounded or not.
To find the solution I solve $3-xy gt 0$ and I find $y lt frac3x$. In my opinion the domain is open, and it is not bounded.
I'm not sure of my answers can someone help me?
analysis multivariable-calculus functions
asked Jul 16 at 14:59
Marco
186
add a comment |Â
up vote
1
down vote
favorite
I'm trying to solve this two excercise:
1) Determine the domain of $f(x,y)=sqrt3-xy$, then say if it is open, close, not open neither close; say if it is bounded or not.
To find the solution I solve $3-xy ge 0$ and I find $y le frac3x$. In my opinion the domain is not open neither close, and it is not bounded.
2) Determine the domain of $f(x,y)=log(3-xy)$, then say if it is open, close, not open neither close; say if it is bounded or not.
To find the solution I solve $3-xy gt 0$ and I find $y lt frac3x$. In my opinion the domain is open, and it is not bounded.
I'm not sure of my answers can someone help me?
analysis multivariable-calculus functions
asked Jul 16 at 14:59
Marco
186
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to solve this two excercise:
1) Determine the domain of $f(x,y)=sqrt3-xy$, then say if it is open, close, not open neither close; say if it is bounded or not.
To find the solution I solve $3-xy ge 0$ and I find $y le frac3x$. In my opinion the domain is not open neither close, and it is not bounded.
2) Determine the domain of $f(x,y)=log(3-xy)$, then say if it is open, close, not open neither close; say if it is bounded or not.
To find the solution I solve $3-xy gt 0$ and I find $y lt frac3x$. In my opinion the domain is open, and it is not bounded.
I'm not sure of my answers can someone help me?
analysis multivariable-calculus functions
asked Jul 16 at 14:59
Marco
186
I'm trying to solve this two excercise:
1) Determine the domain of $f(x,y)=sqrt3-xy$, then say if it is open, close, not open neither close; say if it is bounded or not.
To find the solution I solve $3-xy ge 0$ and I find $y le frac3x$. In my opinion the domain is not open neither close, and it is not bounded.
2) Determine the domain of $f(x,y)=log(3-xy)$, then say if it is open, close, not open neither close; say if it is bounded or not.
To find the solution I solve $3-xy gt 0$ and I find $y lt frac3x$. In my opinion the domain is open, and it is not bounded.
I'm not sure of my answers can someone help me?
analysis multivariable-calculus functions
asked Jul 16 at 14:59
Marco
186
asked Jul 16 at 14:59
Marco
186
asked Jul 16 at 14:59
Marco
186
asked Jul 16 at 14:59
Marco
186
186
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
votes
up vote
1
down vote
You want to solve $xy le 3$.
$x = 0$ is a solution for all $y$.
If $x > 0$, then $y le dfrac 3x$.
If $x < 0$, then $y ge dfrac 3x$
Below is a graph of $xy le 3$.
clearly its complement is an open set. So it must be a closed set.
answered Jul 17 at 0:08
steven gregory
16.5k22055
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0
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1) Be careful when dividing by $x$. If $x$ is negative, the inequality switches directions. If $x$ is $0$, you can't divide by it, but $x=0$ is still a solution to your inequality. The domain is slightly more complicated than what you have expressed.
2) Ditto for the dividing by $x$ problem. Also, depending on what you come up with for an answer, it might not be open EVEN IF you have strict inequalities $<$ running around everywhere. That gives you a good hunch that it might be open, but look carefully at the boundary. Sets don't necessarily have to be open or closed.
In either case, it's important to justify why the set is open or closed. Go back to whichever definition your book uses and see if these domains have those properties.
answered Jul 16 at 15:08
Hans Musgrave
1,393111
Ok so do the domains became for 1) $y le 3/x$ and $y ge -3/x$ And for 2) $y lt 3/x$ and $y gt -3/x$?
– Marco
Jul 16 at 15:15
See @StevenGregory's answer. Handling the $x=0$ case is important, and graphing the solution set makes it easier to understand.
– Hans Musgrave
Jul 17 at 0:13
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You want to solve $xy le 3$.
$x = 0$ is a solution for all $y$.
If $x > 0$, then $y le dfrac 3x$.
If $x < 0$, then $y ge dfrac 3x$
Below is a graph of $xy le 3$.
clearly its complement is an open set. So it must be a closed set.
answered Jul 17 at 0:08
steven gregory
16.5k22055
add a comment |Â
up vote
1
down vote
You want to solve $xy le 3$.
$x = 0$ is a solution for all $y$.
If $x > 0$, then $y le dfrac 3x$.
If $x < 0$, then $y ge dfrac 3x$
Below is a graph of $xy le 3$.
clearly its complement is an open set. So it must be a closed set.
answered Jul 17 at 0:08
steven gregory
16.5k22055
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You want to solve $xy le 3$.
$x = 0$ is a solution for all $y$.
If $x > 0$, then $y le dfrac 3x$.
If $x < 0$, then $y ge dfrac 3x$
Below is a graph of $xy le 3$.
clearly its complement is an open set. So it must be a closed set.
answered Jul 17 at 0:08
steven gregory
16.5k22055
You want to solve $xy le 3$.
$x = 0$ is a solution for all $y$.
If $x > 0$, then $y le dfrac 3x$.
If $x < 0$, then $y ge dfrac 3x$
Below is a graph of $xy le 3$.
clearly its complement is an open set. So it must be a closed set.
answered Jul 17 at 0:08
steven gregory
16.5k22055
answered Jul 17 at 0:08
steven gregory
16.5k22055
answered Jul 17 at 0:08
steven gregory
16.5k22055
answered Jul 17 at 0:08
steven gregory
16.5k22055
16.5k22055
add a comment |Â
add a comment |Â
up vote
0
down vote
1) Be careful when dividing by $x$. If $x$ is negative, the inequality switches directions. If $x$ is $0$, you can't divide by it, but $x=0$ is still a solution to your inequality. The domain is slightly more complicated than what you have expressed.
2) Ditto for the dividing by $x$ problem. Also, depending on what you come up with for an answer, it might not be open EVEN IF you have strict inequalities $<$ running around everywhere. That gives you a good hunch that it might be open, but look carefully at the boundary. Sets don't necessarily have to be open or closed.
In either case, it's important to justify why the set is open or closed. Go back to whichever definition your book uses and see if these domains have those properties.
answered Jul 16 at 15:08
Hans Musgrave
1,393111
Ok so do the domains became for 1) $y le 3/x$ and $y ge -3/x$ And for 2) $y lt 3/x$ and $y gt -3/x$?
– Marco
Jul 16 at 15:15
See @StevenGregory's answer. Handling the $x=0$ case is important, and graphing the solution set makes it easier to understand.
– Hans Musgrave
Jul 17 at 0:13
add a comment |Â
up vote
0
down vote
1) Be careful when dividing by $x$. If $x$ is negative, the inequality switches directions. If $x$ is $0$, you can't divide by it, but $x=0$ is still a solution to your inequality. The domain is slightly more complicated than what you have expressed.
2) Ditto for the dividing by $x$ problem. Also, depending on what you come up with for an answer, it might not be open EVEN IF you have strict inequalities $<$ running around everywhere. That gives you a good hunch that it might be open, but look carefully at the boundary. Sets don't necessarily have to be open or closed.
In either case, it's important to justify why the set is open or closed. Go back to whichever definition your book uses and see if these domains have those properties.
answered Jul 16 at 15:08
Hans Musgrave
1,393111
Ok so do the domains became for 1) $y le 3/x$ and $y ge -3/x$ And for 2) $y lt 3/x$ and $y gt -3/x$?
– Marco
Jul 16 at 15:15
See @StevenGregory's answer. Handling the $x=0$ case is important, and graphing the solution set makes it easier to understand.
– Hans Musgrave
Jul 17 at 0:13
add a comment |Â
up vote
0
down vote
up vote
0
down vote
1) Be careful when dividing by $x$. If $x$ is negative, the inequality switches directions. If $x$ is $0$, you can't divide by it, but $x=0$ is still a solution to your inequality. The domain is slightly more complicated than what you have expressed.
2) Ditto for the dividing by $x$ problem. Also, depending on what you come up with for an answer, it might not be open EVEN IF you have strict inequalities $<$ running around everywhere. That gives you a good hunch that it might be open, but look carefully at the boundary. Sets don't necessarily have to be open or closed.
In either case, it's important to justify why the set is open or closed. Go back to whichever definition your book uses and see if these domains have those properties.
answered Jul 16 at 15:08
Hans Musgrave
1,393111
1) Be careful when dividing by $x$. If $x$ is negative, the inequality switches directions. If $x$ is $0$, you can't divide by it, but $x=0$ is still a solution to your inequality. The domain is slightly more complicated than what you have expressed.
2) Ditto for the dividing by $x$ problem. Also, depending on what you come up with for an answer, it might not be open EVEN IF you have strict inequalities $<$ running around everywhere. That gives you a good hunch that it might be open, but look carefully at the boundary. Sets don't necessarily have to be open or closed.
In either case, it's important to justify why the set is open or closed. Go back to whichever definition your book uses and see if these domains have those properties.
answered Jul 16 at 15:08
Hans Musgrave
1,393111
answered Jul 16 at 15:08
Hans Musgrave
1,393111
answered Jul 16 at 15:08
Hans Musgrave
1,393111
answered Jul 16 at 15:08
Hans Musgrave
1,393111
1,393111
Ok so do the domains became for 1) $y le 3/x$ and $y ge -3/x$ And for 2) $y lt 3/x$ and $y gt -3/x$?
– Marco
Jul 16 at 15:15
See @StevenGregory's answer. Handling the $x=0$ case is important, and graphing the solution set makes it easier to understand.
– Hans Musgrave
Jul 17 at 0:13
add a comment |Â
Ok so do the domains became for 1) $y le 3/x$ and $y ge -3/x$ And for 2) $y lt 3/x$ and $y gt -3/x$?
– Marco
Jul 16 at 15:15
See @StevenGregory's answer. Handling the $x=0$ case is important, and graphing the solution set makes it easier to understand.
– Hans Musgrave
Jul 17 at 0:13
Ok so do the domains became for 1) $y le 3/x$ and $y ge -3/x$ And for 2) $y lt 3/x$ and $y gt -3/x$?
– Marco
Jul 16 at 15:15
Ok so do the domains became for 1) $y le 3/x$ and $y ge -3/x$ And for 2) $y lt 3/x$ and $y gt -3/x$?
– Marco
Jul 16 at 15:15
See @StevenGregory's answer. Handling the $x=0$ case is important, and graphing the solution set makes it easier to understand.
– Hans Musgrave
Jul 17 at 0:13
See @StevenGregory's answer. Handling the $x=0$ case is important, and graphing the solution set makes it easier to understand.
– Hans Musgrave
Jul 17 at 0:13
add a comment |Â
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