Mixing
Finding domain of implicit functions.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Suppose $f(x,y)=k$ where k is a constant. I am supposed to find the domain of $ f(x,y)=1$ so should I check for the value of $x $where $f(x,y)$ is defined or first write $ y=f(x) $and then find the values of x for which f(x) is defined. Consider $x^2y=1$. The domain will be all real x I think but the answer in my book is given as only $x>0 $or $ x=0.$
functions
asked Jul 24 at 13:43
Jasmine
322111
add a comment |Â
up vote
1
down vote
favorite
Suppose $f(x,y)=k$ where k is a constant. I am supposed to find the domain of $ f(x,y)=1$ so should I check for the value of $x $where $f(x,y)$ is defined or first write $ y=f(x) $and then find the values of x for which f(x) is defined. Consider $x^2y=1$. The domain will be all real x I think but the answer in my book is given as only $x>0 $or $ x=0.$
functions
asked Jul 24 at 13:43
Jasmine
322111
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $f(x,y)=k$ where k is a constant. I am supposed to find the domain of $ f(x,y)=1$ so should I check for the value of $x $where $f(x,y)$ is defined or first write $ y=f(x) $and then find the values of x for which f(x) is defined. Consider $x^2y=1$. The domain will be all real x I think but the answer in my book is given as only $x>0 $or $ x=0.$
functions
asked Jul 24 at 13:43
Jasmine
322111
Suppose $f(x,y)=k$ where k is a constant. I am supposed to find the domain of $ f(x,y)=1$ so should I check for the value of $x $where $f(x,y)$ is defined or first write $ y=f(x) $and then find the values of x for which f(x) is defined. Consider $x^2y=1$. The domain will be all real x I think but the answer in my book is given as only $x>0 $or $ x=0.$
functions
asked Jul 24 at 13:43
Jasmine
322111
asked Jul 24 at 13:43
Jasmine
322111
asked Jul 24 at 13:43
Jasmine
322111
asked Jul 24 at 13:43
Jasmine
322111
322111
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
The given function
$$f(x,y):=x^2y$$
has as domain the right half plane $$H:=bigl x>0, -infty<y<inftybigr ,$$
because as soon as you consider powers $b^t$ with variable noninteger exponents the base $b$ is automatically restricted to $mathbb R_>0$.
Let's see whether the equation $f(x,y)=1$, i.e., $x^2y=1$, "defines $y$ as a function of $x$" in some way. Note that by definition $$1=x^2y:=exp(2ylog x) ,$$ and as $exp$ is monotonically increasing this is equivalent with $2ylog x=0$. It follows that
$$y=0quadveequad log x=0,quadrm resp.,quad y=0quadveequad x=1 .$$
In other words the solution set $$bigl f(x,y)=1bigr$$
consists of the horizontal half line $y=0$ together with the vertical line $x=1$.
We therefore can say the following: The given equation defines implicitly the function $y=phi(x):equiv0$ separately in the $x$-intervals $(0,1)$ and $(1,infty)$.
answered Jul 24 at 14:45
Christian Blatter
163k7107306
add a comment |Â
up vote
1
down vote
Take $x=-1,;y=1/2.$ Then $x^2y=-1$ and not $1.$ Given any negative $x$-value you can choose a $y$ which makes the equality fail.
edited Jul 24 at 19:34
amWhy
189k25219431
answered Jul 24 at 14:04
Jimmy Mixco
2013
But there can be certain y for which it satisfies the equation right. I think that at x=-1 y should be equal to any integer right?
– Jasmine
Jul 24 at 14:06
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The given function
$$f(x,y):=x^2y$$
has as domain the right half plane $$H:=bigl x>0, -infty<y<inftybigr ,$$
because as soon as you consider powers $b^t$ with variable noninteger exponents the base $b$ is automatically restricted to $mathbb R_>0$.
Let's see whether the equation $f(x,y)=1$, i.e., $x^2y=1$, "defines $y$ as a function of $x$" in some way. Note that by definition $$1=x^2y:=exp(2ylog x) ,$$ and as $exp$ is monotonically increasing this is equivalent with $2ylog x=0$. It follows that
$$y=0quadveequad log x=0,quadrm resp.,quad y=0quadveequad x=1 .$$
In other words the solution set $$bigl f(x,y)=1bigr$$
consists of the horizontal half line $y=0$ together with the vertical line $x=1$.
We therefore can say the following: The given equation defines implicitly the function $y=phi(x):equiv0$ separately in the $x$-intervals $(0,1)$ and $(1,infty)$.
answered Jul 24 at 14:45
Christian Blatter
163k7107306
add a comment |Â
up vote
0
down vote
accepted
The given function
$$f(x,y):=x^2y$$
has as domain the right half plane $$H:=bigl x>0, -infty<y<inftybigr ,$$
because as soon as you consider powers $b^t$ with variable noninteger exponents the base $b$ is automatically restricted to $mathbb R_>0$.
Let's see whether the equation $f(x,y)=1$, i.e., $x^2y=1$, "defines $y$ as a function of $x$" in some way. Note that by definition $$1=x^2y:=exp(2ylog x) ,$$ and as $exp$ is monotonically increasing this is equivalent with $2ylog x=0$. It follows that
$$y=0quadveequad log x=0,quadrm resp.,quad y=0quadveequad x=1 .$$
In other words the solution set $$bigl f(x,y)=1bigr$$
consists of the horizontal half line $y=0$ together with the vertical line $x=1$.
We therefore can say the following: The given equation defines implicitly the function $y=phi(x):equiv0$ separately in the $x$-intervals $(0,1)$ and $(1,infty)$.
answered Jul 24 at 14:45
Christian Blatter
163k7107306
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The given function
$$f(x,y):=x^2y$$
has as domain the right half plane $$H:=bigl x>0, -infty<y<inftybigr ,$$
because as soon as you consider powers $b^t$ with variable noninteger exponents the base $b$ is automatically restricted to $mathbb R_>0$.
Let's see whether the equation $f(x,y)=1$, i.e., $x^2y=1$, "defines $y$ as a function of $x$" in some way. Note that by definition $$1=x^2y:=exp(2ylog x) ,$$ and as $exp$ is monotonically increasing this is equivalent with $2ylog x=0$. It follows that
$$y=0quadveequad log x=0,quadrm resp.,quad y=0quadveequad x=1 .$$
In other words the solution set $$bigl f(x,y)=1bigr$$
consists of the horizontal half line $y=0$ together with the vertical line $x=1$.
We therefore can say the following: The given equation defines implicitly the function $y=phi(x):equiv0$ separately in the $x$-intervals $(0,1)$ and $(1,infty)$.
answered Jul 24 at 14:45
Christian Blatter
163k7107306
The given function
$$f(x,y):=x^2y$$
has as domain the right half plane $$H:=bigl x>0, -infty<y<inftybigr ,$$
because as soon as you consider powers $b^t$ with variable noninteger exponents the base $b$ is automatically restricted to $mathbb R_>0$.
Let's see whether the equation $f(x,y)=1$, i.e., $x^2y=1$, "defines $y$ as a function of $x$" in some way. Note that by definition $$1=x^2y:=exp(2ylog x) ,$$ and as $exp$ is monotonically increasing this is equivalent with $2ylog x=0$. It follows that
$$y=0quadveequad log x=0,quadrm resp.,quad y=0quadveequad x=1 .$$
In other words the solution set $$bigl f(x,y)=1bigr$$
consists of the horizontal half line $y=0$ together with the vertical line $x=1$.
We therefore can say the following: The given equation defines implicitly the function $y=phi(x):equiv0$ separately in the $x$-intervals $(0,1)$ and $(1,infty)$.
answered Jul 24 at 14:45
Christian Blatter
163k7107306
answered Jul 24 at 14:45
Christian Blatter
163k7107306
answered Jul 24 at 14:45
Christian Blatter
163k7107306
answered Jul 24 at 14:45
Christian Blatter
163k7107306
163k7107306
add a comment |Â
add a comment |Â
up vote
1
down vote
Take $x=-1,;y=1/2.$ Then $x^2y=-1$ and not $1.$ Given any negative $x$-value you can choose a $y$ which makes the equality fail.
edited Jul 24 at 19:34
amWhy
189k25219431
answered Jul 24 at 14:04
Jimmy Mixco
2013
But there can be certain y for which it satisfies the equation right. I think that at x=-1 y should be equal to any integer right?
– Jasmine
Jul 24 at 14:06
add a comment |Â
up vote
1
down vote
Take $x=-1,;y=1/2.$ Then $x^2y=-1$ and not $1.$ Given any negative $x$-value you can choose a $y$ which makes the equality fail.
edited Jul 24 at 19:34
amWhy
189k25219431
answered Jul 24 at 14:04
Jimmy Mixco
2013
But there can be certain y for which it satisfies the equation right. I think that at x=-1 y should be equal to any integer right?
– Jasmine
Jul 24 at 14:06
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Take $x=-1,;y=1/2.$ Then $x^2y=-1$ and not $1.$ Given any negative $x$-value you can choose a $y$ which makes the equality fail.
edited Jul 24 at 19:34
amWhy
189k25219431
answered Jul 24 at 14:04
Jimmy Mixco
2013
Take $x=-1,;y=1/2.$ Then $x^2y=-1$ and not $1.$ Given any negative $x$-value you can choose a $y$ which makes the equality fail.
edited Jul 24 at 19:34
amWhy
189k25219431
answered Jul 24 at 14:04
Jimmy Mixco
2013
edited Jul 24 at 19:34
amWhy
189k25219431
edited Jul 24 at 19:34
amWhy
189k25219431
edited Jul 24 at 19:34
amWhy
189k25219431
189k25219431
answered Jul 24 at 14:04
Jimmy Mixco
2013
answered Jul 24 at 14:04
Jimmy Mixco
2013
answered Jul 24 at 14:04
Jimmy Mixco
2013
2013
But there can be certain y for which it satisfies the equation right. I think that at x=-1 y should be equal to any integer right?
– Jasmine
Jul 24 at 14:06
add a comment |Â
But there can be certain y for which it satisfies the equation right. I think that at x=-1 y should be equal to any integer right?
– Jasmine
Jul 24 at 14:06
But there can be certain y for which it satisfies the equation right. I think that at x=-1 y should be equal to any integer right?
– Jasmine
Jul 24 at 14:06
But there can be certain y for which it satisfies the equation right. I think that at x=-1 y should be equal to any integer right?
– Jasmine
Jul 24 at 14:06
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861348%2ffinding-domain-of-implicit-functions%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password