Mixing
How do I find the domain and range for a multivariable function
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I am aware of question : Domain and range of a multivariable function
However my question is slightly different, I have no idea where to start with a function such as :
$f(x,y) = logfrac1-xy-1$
The function is multivariable but the part that confuses me is the variation in variables is within the same fraction. I don't know where to start? How to exclude certain values or even sketch this graph for a rough idea?
Cheers
calculus multivariable-calculus
edited Jul 27 at 18:24
Math Lover
12.3k21232
asked Jul 27 at 18:20
K.M.
17712
add a comment |Â
up vote
1
down vote
favorite
I am aware of question : Domain and range of a multivariable function
However my question is slightly different, I have no idea where to start with a function such as :
$f(x,y) = logfrac1-xy-1$
The function is multivariable but the part that confuses me is the variation in variables is within the same fraction. I don't know where to start? How to exclude certain values or even sketch this graph for a rough idea?
Cheers
calculus multivariable-calculus
edited Jul 27 at 18:24
Math Lover
12.3k21232
asked Jul 27 at 18:20
K.M.
17712
1. $log(f(x, y))$ is only defined for positive values of $f(x, y)$. 2. y can't be 1.
– amWhy
Jul 27 at 18:23
When $(1-x)/(y-1)le 0$?
– Holo
Jul 27 at 18:24
@amWhy I get that y can't be 1 but what does your first point mean? I know it has to be greater than 0 since it's logs but how do I know the range after I find the domain?
– K.M.
Jul 27 at 18:26
For the range: set $y=2,x=1-z$ for $x>0$, what would you get?
– Holo
Jul 27 at 18:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am aware of question : Domain and range of a multivariable function
However my question is slightly different, I have no idea where to start with a function such as :
$f(x,y) = logfrac1-xy-1$
The function is multivariable but the part that confuses me is the variation in variables is within the same fraction. I don't know where to start? How to exclude certain values or even sketch this graph for a rough idea?
Cheers
calculus multivariable-calculus
edited Jul 27 at 18:24
Math Lover
12.3k21232
asked Jul 27 at 18:20
K.M.
17712
I am aware of question : Domain and range of a multivariable function
However my question is slightly different, I have no idea where to start with a function such as :
$f(x,y) = logfrac1-xy-1$
The function is multivariable but the part that confuses me is the variation in variables is within the same fraction. I don't know where to start? How to exclude certain values or even sketch this graph for a rough idea?
Cheers
calculus multivariable-calculus
edited Jul 27 at 18:24
Math Lover
12.3k21232
asked Jul 27 at 18:20
K.M.
17712
edited Jul 27 at 18:24
Math Lover
12.3k21232
edited Jul 27 at 18:24
Math Lover
12.3k21232
edited Jul 27 at 18:24
Math Lover
12.3k21232
12.3k21232
asked Jul 27 at 18:20
K.M.
17712
asked Jul 27 at 18:20
K.M.
17712
asked Jul 27 at 18:20
K.M.
17712
17712
1. $log(f(x, y))$ is only defined for positive values of $f(x, y)$. 2. y can't be 1.
– amWhy
Jul 27 at 18:23
When $(1-x)/(y-1)le 0$?
– Holo
Jul 27 at 18:24
@amWhy I get that y can't be 1 but what does your first point mean? I know it has to be greater than 0 since it's logs but how do I know the range after I find the domain?
– K.M.
Jul 27 at 18:26
For the range: set $y=2,x=1-z$ for $x>0$, what would you get?
– Holo
Jul 27 at 18:28
add a comment |Â
1. $log(f(x, y))$ is only defined for positive values of $f(x, y)$. 2. y can't be 1.
– amWhy
Jul 27 at 18:23
When $(1-x)/(y-1)le 0$?
– Holo
Jul 27 at 18:24
@amWhy I get that y can't be 1 but what does your first point mean? I know it has to be greater than 0 since it's logs but how do I know the range after I find the domain?
– K.M.
Jul 27 at 18:26
For the range: set $y=2,x=1-z$ for $x>0$, what would you get?
– Holo
Jul 27 at 18:28
1. $log(f(x, y))$ is only defined for positive values of $f(x, y)$. 2. y can't be 1.
– amWhy
Jul 27 at 18:23
1. $log(f(x, y))$ is only defined for positive values of $f(x, y)$. 2. y can't be 1.
– amWhy
Jul 27 at 18:23
When $(1-x)/(y-1)le 0$?
– Holo
Jul 27 at 18:24
When $(1-x)/(y-1)le 0$?
– Holo
Jul 27 at 18:24
@amWhy I get that y can't be 1 but what does your first point mean? I know it has to be greater than 0 since it's logs but how do I know the range after I find the domain?
– K.M.
Jul 27 at 18:26
@amWhy I get that y can't be 1 but what does your first point mean? I know it has to be greater than 0 since it's logs but how do I know the range after I find the domain?
– K.M.
Jul 27 at 18:26
For the range: set $y=2,x=1-z$ for $x>0$, what would you get?
– Holo
Jul 27 at 18:28
For the range: set $y=2,x=1-z$ for $x>0$, what would you get?
– Holo
Jul 27 at 18:28
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
$log(f(x, y))$ is only defined for positive values of $f(x, y)$.
$x, y$ can't be $1.$
In the event that both numerator and denominator are positive, we get $$logbig(f(x, y)big) = logleft(frac 1-xy-1right) = log(1-x) - log(y-1)$$
This is fulfilled when both $xlt 1$ AND $ygt 1$.
In the event that the numerator and denominator are both negative: i.e.
If both $x gt 1$ AND $ylt 1$, then $$f(x,y) = frac 1-xy-1 = left(frac-1cdot (1-x)-1cdot (y-1)right) = fracx-11-y.$$
So in this case, we have that $logleft(f(x, y)right) = log(x-1) - log(1-y)$.
answered Jul 27 at 18:27
amWhy
189k25219431
I'm guessing because you can't have log of 0 but for the $log(1-x)$ part you will get log of a negative number so shouldn't x have to be less than 0?
– K.M.
Jul 27 at 18:29
Not if $x lt 1$
– amWhy
Jul 27 at 18:29
Also note that $y$ needs to be such that $ygt 1$.
– amWhy
Jul 27 at 18:31
1
Okay I see that x can be less than 1 and y can be greater than 1 but lets say x is 5 and y is -3, we would get $log(1)$ which is a possibility so how do I define the domain and range?
– K.M.
Jul 27 at 18:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$log(f(x, y))$ is only defined for positive values of $f(x, y)$.
$x, y$ can't be $1.$
In the event that both numerator and denominator are positive, we get $$logbig(f(x, y)big) = logleft(frac 1-xy-1right) = log(1-x) - log(y-1)$$
This is fulfilled when both $xlt 1$ AND $ygt 1$.
In the event that the numerator and denominator are both negative: i.e.
If both $x gt 1$ AND $ylt 1$, then $$f(x,y) = frac 1-xy-1 = left(frac-1cdot (1-x)-1cdot (y-1)right) = fracx-11-y.$$
So in this case, we have that $logleft(f(x, y)right) = log(x-1) - log(1-y)$.
answered Jul 27 at 18:27
amWhy
189k25219431
I'm guessing because you can't have log of 0 but for the $log(1-x)$ part you will get log of a negative number so shouldn't x have to be less than 0?
– K.M.
Jul 27 at 18:29
Not if $x lt 1$
– amWhy
Jul 27 at 18:29
Also note that $y$ needs to be such that $ygt 1$.
– amWhy
Jul 27 at 18:31
1
Okay I see that x can be less than 1 and y can be greater than 1 but lets say x is 5 and y is -3, we would get $log(1)$ which is a possibility so how do I define the domain and range?
– K.M.
Jul 27 at 18:34
add a comment |Â
up vote
4
down vote
accepted
$log(f(x, y))$ is only defined for positive values of $f(x, y)$.
$x, y$ can't be $1.$
In the event that both numerator and denominator are positive, we get $$logbig(f(x, y)big) = logleft(frac 1-xy-1right) = log(1-x) - log(y-1)$$
This is fulfilled when both $xlt 1$ AND $ygt 1$.
In the event that the numerator and denominator are both negative: i.e.
If both $x gt 1$ AND $ylt 1$, then $$f(x,y) = frac 1-xy-1 = left(frac-1cdot (1-x)-1cdot (y-1)right) = fracx-11-y.$$
So in this case, we have that $logleft(f(x, y)right) = log(x-1) - log(1-y)$.
answered Jul 27 at 18:27
amWhy
189k25219431
I'm guessing because you can't have log of 0 but for the $log(1-x)$ part you will get log of a negative number so shouldn't x have to be less than 0?
– K.M.
Jul 27 at 18:29
Not if $x lt 1$
– amWhy
Jul 27 at 18:29
Also note that $y$ needs to be such that $ygt 1$.
– amWhy
Jul 27 at 18:31
1
Okay I see that x can be less than 1 and y can be greater than 1 but lets say x is 5 and y is -3, we would get $log(1)$ which is a possibility so how do I define the domain and range?
– K.M.
Jul 27 at 18:34
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$log(f(x, y))$ is only defined for positive values of $f(x, y)$.
$x, y$ can't be $1.$
In the event that both numerator and denominator are positive, we get $$logbig(f(x, y)big) = logleft(frac 1-xy-1right) = log(1-x) - log(y-1)$$
This is fulfilled when both $xlt 1$ AND $ygt 1$.
In the event that the numerator and denominator are both negative: i.e.
If both $x gt 1$ AND $ylt 1$, then $$f(x,y) = frac 1-xy-1 = left(frac-1cdot (1-x)-1cdot (y-1)right) = fracx-11-y.$$
So in this case, we have that $logleft(f(x, y)right) = log(x-1) - log(1-y)$.
answered Jul 27 at 18:27
amWhy
189k25219431
$log(f(x, y))$ is only defined for positive values of $f(x, y)$.
$x, y$ can't be $1.$
In the event that both numerator and denominator are positive, we get $$logbig(f(x, y)big) = logleft(frac 1-xy-1right) = log(1-x) - log(y-1)$$
This is fulfilled when both $xlt 1$ AND $ygt 1$.
In the event that the numerator and denominator are both negative: i.e.
If both $x gt 1$ AND $ylt 1$, then $$f(x,y) = frac 1-xy-1 = left(frac-1cdot (1-x)-1cdot (y-1)right) = fracx-11-y.$$
So in this case, we have that $logleft(f(x, y)right) = log(x-1) - log(1-y)$.
answered Jul 27 at 18:27
amWhy
189k25219431
edited Jul 27 at 21:02
answered Jul 27 at 18:27
amWhy
189k25219431
answered Jul 27 at 18:27
amWhy
189k25219431
answered Jul 27 at 18:27
amWhy
189k25219431
189k25219431
I'm guessing because you can't have log of 0 but for the $log(1-x)$ part you will get log of a negative number so shouldn't x have to be less than 0?
– K.M.
Jul 27 at 18:29
Not if $x lt 1$
– amWhy
Jul 27 at 18:29
Also note that $y$ needs to be such that $ygt 1$.
– amWhy
Jul 27 at 18:31
1
Okay I see that x can be less than 1 and y can be greater than 1 but lets say x is 5 and y is -3, we would get $log(1)$ which is a possibility so how do I define the domain and range?
– K.M.
Jul 27 at 18:34
add a comment |Â
I'm guessing because you can't have log of 0 but for the $log(1-x)$ part you will get log of a negative number so shouldn't x have to be less than 0?
– K.M.
Jul 27 at 18:29
Not if $x lt 1$
– amWhy
Jul 27 at 18:29
Also note that $y$ needs to be such that $ygt 1$.
– amWhy
Jul 27 at 18:31
1
Okay I see that x can be less than 1 and y can be greater than 1 but lets say x is 5 and y is -3, we would get $log(1)$ which is a possibility so how do I define the domain and range?
– K.M.
Jul 27 at 18:34
I'm guessing because you can't have log of 0 but for the $log(1-x)$ part you will get log of a negative number so shouldn't x have to be less than 0?
– K.M.
Jul 27 at 18:29
I'm guessing because you can't have log of 0 but for the $log(1-x)$ part you will get log of a negative number so shouldn't x have to be less than 0?
– K.M.
Jul 27 at 18:29
Not if $x lt 1$
– amWhy
Jul 27 at 18:29
Not if $x lt 1$
– amWhy
Jul 27 at 18:29
Also note that $y$ needs to be such that $ygt 1$.
– amWhy
Jul 27 at 18:31
Also note that $y$ needs to be such that $ygt 1$.
– amWhy
Jul 27 at 18:31
1
1
Okay I see that x can be less than 1 and y can be greater than 1 but lets say x is 5 and y is -3, we would get $log(1)$ which is a possibility so how do I define the domain and range?
– K.M.
Jul 27 at 18:34
Okay I see that x can be less than 1 and y can be greater than 1 but lets say x is 5 and y is -3, we would get $log(1)$ which is a possibility so how do I define the domain and range?
– K.M.
Jul 27 at 18:34
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%2f2864666%2fhow-do-i-find-the-domain-and-range-for-a-multivariable-function%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
1. $log(f(x, y))$ is only defined for positive values of $f(x, y)$. 2. y can't be 1.
– amWhy
Jul 27 at 18:23
When $(1-x)/(y-1)le 0$?
– Holo
Jul 27 at 18:24
@amWhy I get that y can't be 1 but what does your first point mean? I know it has to be greater than 0 since it's logs but how do I know the range after I find the domain?
– K.M.
Jul 27 at 18:26
For the range: set $y=2,x=1-z$ for $x>0$, what would you get?
– Holo
Jul 27 at 18:28