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Domain of multivariable function
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I have a function of two real variables which is given by the transformation rule
$$f(x,y)=fracy1+x^2+y^2.$$
I have to find the domain of $f$ which consists of all points $(x,y)$.
When I examine the function I would say the domain is $$|x,y in BbbR^2:yneq0, x text are real numbers$$, but looking at the results-list it says that both $x$ and $y$ are real numbers. How come that is?
This might be straightforward for some of you, but I can't seem to wrap my head around this on my own and hope some of you can help. Thanks in advance
multivariable-calculus functions
asked Aug 10 at 10:00
JayFreddy
344
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up vote
2
down vote
favorite
I have a function of two real variables which is given by the transformation rule
$$f(x,y)=fracy1+x^2+y^2.$$
I have to find the domain of $f$ which consists of all points $(x,y)$.
When I examine the function I would say the domain is $$|x,y in BbbR^2:yneq0, x text are real numbers$$, but looking at the results-list it says that both $x$ and $y$ are real numbers. How come that is?
This might be straightforward for some of you, but I can't seem to wrap my head around this on my own and hope some of you can help. Thanks in advance
multivariable-calculus functions
asked Aug 10 at 10:00
JayFreddy
344
1
The domain of a function is the set of values which the function can take as input. You're probably tasked with finding the maximal domain in $mathbbR^2$. So you have to ask yourself, for which $(x,y)inmathbbR^2$ is the function $f$ you stated (not) defined.
– zzuussee
Aug 10 at 10:03
Why do you think $f$ should not take the value $0$. As long as the denominator in non-zero the functions is well defined.
– Kavi Rama Murthy
Aug 10 at 10:03
However, if $x,y in BbbC$, things will be a bit different.
– twalberg
Aug 10 at 15:52
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a function of two real variables which is given by the transformation rule
$$f(x,y)=fracy1+x^2+y^2.$$
I have to find the domain of $f$ which consists of all points $(x,y)$.
When I examine the function I would say the domain is $$|x,y in BbbR^2:yneq0, x text are real numbers$$, but looking at the results-list it says that both $x$ and $y$ are real numbers. How come that is?
This might be straightforward for some of you, but I can't seem to wrap my head around this on my own and hope some of you can help. Thanks in advance
multivariable-calculus functions
asked Aug 10 at 10:00
JayFreddy
344
I have a function of two real variables which is given by the transformation rule
$$f(x,y)=fracy1+x^2+y^2.$$
I have to find the domain of $f$ which consists of all points $(x,y)$.
When I examine the function I would say the domain is $$|x,y in BbbR^2:yneq0, x text are real numbers$$, but looking at the results-list it says that both $x$ and $y$ are real numbers. How come that is?
This might be straightforward for some of you, but I can't seem to wrap my head around this on my own and hope some of you can help. Thanks in advance
multivariable-calculus functions
asked Aug 10 at 10:00
JayFreddy
344
asked Aug 10 at 10:00
JayFreddy
344
asked Aug 10 at 10:00
JayFreddy
344
asked Aug 10 at 10:00
JayFreddy
344
344
1
The domain of a function is the set of values which the function can take as input. You're probably tasked with finding the maximal domain in $mathbbR^2$. So you have to ask yourself, for which $(x,y)inmathbbR^2$ is the function $f$ you stated (not) defined.
– zzuussee
Aug 10 at 10:03
Why do you think $f$ should not take the value $0$. As long as the denominator in non-zero the functions is well defined.
– Kavi Rama Murthy
Aug 10 at 10:03
However, if $x,y in BbbC$, things will be a bit different.
– twalberg
Aug 10 at 15:52
add a comment |Â
1
The domain of a function is the set of values which the function can take as input. You're probably tasked with finding the maximal domain in $mathbbR^2$. So you have to ask yourself, for which $(x,y)inmathbbR^2$ is the function $f$ you stated (not) defined.
– zzuussee
Aug 10 at 10:03
Why do you think $f$ should not take the value $0$. As long as the denominator in non-zero the functions is well defined.
– Kavi Rama Murthy
Aug 10 at 10:03
However, if $x,y in BbbC$, things will be a bit different.
– twalberg
Aug 10 at 15:52
1
1
The domain of a function is the set of values which the function can take as input. You're probably tasked with finding the maximal domain in $mathbbR^2$. So you have to ask yourself, for which $(x,y)inmathbbR^2$ is the function $f$ you stated (not) defined.
– zzuussee
Aug 10 at 10:03
The domain of a function is the set of values which the function can take as input. You're probably tasked with finding the maximal domain in $mathbbR^2$. So you have to ask yourself, for which $(x,y)inmathbbR^2$ is the function $f$ you stated (not) defined.
– zzuussee
Aug 10 at 10:03
Why do you think $f$ should not take the value $0$. As long as the denominator in non-zero the functions is well defined.
– Kavi Rama Murthy
Aug 10 at 10:03
Why do you think $f$ should not take the value $0$. As long as the denominator in non-zero the functions is well defined.
– Kavi Rama Murthy
Aug 10 at 10:03
However, if $x,y in BbbC$, things will be a bit different.
– twalberg
Aug 10 at 15:52
However, if $x,y in BbbC$, things will be a bit different.
– twalberg
Aug 10 at 15:52
add a comment |Â
2 Answers
2
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oldest
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For the domain of the given function, the denominator must be different than zero, but :
$$1+x^2+y^2 neq 0 Leftrightarrow 1 neq -x^2 - y^2$$
Note that $-x^2 -y^2 leq 0 ; forall ; x,y ; in mathbb R$ and since $1$ is a positive number, this can never equal it. There are no other constraints to check. Thus, the domain is $D_f = mathbb R^2$.
answered Aug 10 at 10:04
Rebellos
10.1k21039
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up vote
3
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We have $1+x^2+y^2 ge 1 >0$ for all $(x,y) in mathbb R^2$. Hence $1+x^2+y^2 ne 0$ for all $(x,y) in mathbb R^2$. This shows that $f$ is defined for all $(x,y) in mathbb R^2$.
answered Aug 10 at 10:04
Fred
38.2k1238
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
For the domain of the given function, the denominator must be different than zero, but :
$$1+x^2+y^2 neq 0 Leftrightarrow 1 neq -x^2 - y^2$$
Note that $-x^2 -y^2 leq 0 ; forall ; x,y ; in mathbb R$ and since $1$ is a positive number, this can never equal it. There are no other constraints to check. Thus, the domain is $D_f = mathbb R^2$.
answered Aug 10 at 10:04
Rebellos
10.1k21039
add a comment |Â
up vote
8
down vote
accepted
For the domain of the given function, the denominator must be different than zero, but :
$$1+x^2+y^2 neq 0 Leftrightarrow 1 neq -x^2 - y^2$$
Note that $-x^2 -y^2 leq 0 ; forall ; x,y ; in mathbb R$ and since $1$ is a positive number, this can never equal it. There are no other constraints to check. Thus, the domain is $D_f = mathbb R^2$.
answered Aug 10 at 10:04
Rebellos
10.1k21039
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
For the domain of the given function, the denominator must be different than zero, but :
$$1+x^2+y^2 neq 0 Leftrightarrow 1 neq -x^2 - y^2$$
Note that $-x^2 -y^2 leq 0 ; forall ; x,y ; in mathbb R$ and since $1$ is a positive number, this can never equal it. There are no other constraints to check. Thus, the domain is $D_f = mathbb R^2$.
answered Aug 10 at 10:04
Rebellos
10.1k21039
For the domain of the given function, the denominator must be different than zero, but :
$$1+x^2+y^2 neq 0 Leftrightarrow 1 neq -x^2 - y^2$$
Note that $-x^2 -y^2 leq 0 ; forall ; x,y ; in mathbb R$ and since $1$ is a positive number, this can never equal it. There are no other constraints to check. Thus, the domain is $D_f = mathbb R^2$.
answered Aug 10 at 10:04
Rebellos
10.1k21039
answered Aug 10 at 10:04
Rebellos
10.1k21039
answered Aug 10 at 10:04
Rebellos
10.1k21039
answered Aug 10 at 10:04
Rebellos
10.1k21039
10.1k21039
add a comment |Â
add a comment |Â
up vote
3
down vote
We have $1+x^2+y^2 ge 1 >0$ for all $(x,y) in mathbb R^2$. Hence $1+x^2+y^2 ne 0$ for all $(x,y) in mathbb R^2$. This shows that $f$ is defined for all $(x,y) in mathbb R^2$.
answered Aug 10 at 10:04
Fred
38.2k1238
add a comment |Â
up vote
3
down vote
We have $1+x^2+y^2 ge 1 >0$ for all $(x,y) in mathbb R^2$. Hence $1+x^2+y^2 ne 0$ for all $(x,y) in mathbb R^2$. This shows that $f$ is defined for all $(x,y) in mathbb R^2$.
answered Aug 10 at 10:04
Fred
38.2k1238
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We have $1+x^2+y^2 ge 1 >0$ for all $(x,y) in mathbb R^2$. Hence $1+x^2+y^2 ne 0$ for all $(x,y) in mathbb R^2$. This shows that $f$ is defined for all $(x,y) in mathbb R^2$.
answered Aug 10 at 10:04
Fred
38.2k1238
We have $1+x^2+y^2 ge 1 >0$ for all $(x,y) in mathbb R^2$. Hence $1+x^2+y^2 ne 0$ for all $(x,y) in mathbb R^2$. This shows that $f$ is defined for all $(x,y) in mathbb R^2$.
answered Aug 10 at 10:04
Fred
38.2k1238
answered Aug 10 at 10:04
Fred
38.2k1238
answered Aug 10 at 10:04
Fred
38.2k1238
answered Aug 10 at 10:04
Fred
38.2k1238
38.2k1238
add a comment |Â
add a comment |Â
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1
The domain of a function is the set of values which the function can take as input. You're probably tasked with finding the maximal domain in $mathbbR^2$. So you have to ask yourself, for which $(x,y)inmathbbR^2$ is the function $f$ you stated (not) defined.
– zzuussee
Aug 10 at 10:03
Why do you think $f$ should not take the value $0$. As long as the denominator in non-zero the functions is well defined.
– Kavi Rama Murthy
Aug 10 at 10:03
However, if $x,y in BbbC$, things will be a bit different.
– twalberg
Aug 10 at 15:52