How to find domain of a function in the implicit function theorem?
$begingroup$
So i am working with an excercise reguarding the implicit function theorem .
$$F(x, y) = y(e^y + x) − log(x).$$
I know $(x_0,y_0)=(1,0)$ such that $F(x_0, y_0) = 0$.
I want to check that there exist a strictly positive real number $delta > 0$ and a function $y =f(x)$ such that $$y_0 = f(x_0)quad text{ and }quad F(x, f(x)) = 0 qquad forall x ∈ (x_0 − delta, x_0 + delta).$$
Is it fair to say that , in order to solve the question, we have to check if it possible to apply the first part of the Implicit Function Theorem?
Which would imply, as the answer sheet says, that the point $(1, 0)$ belongs to $$DF = {(x, y) in mathbb{R}^2∶ x > 0, y in mathbb{R}}$$
the domain of $F$.
I have dyscalculia and it appears confusing how to solve the domain of this function, where $x$ is greater than $0$ and $y$ exists in real numbers..
Can somebody please show me, in the simplest way possible, how to find the domain so I can proceed with the excercise? Thanks!
implicit-function-theorem
edited Dec 8 '18 at 10:15
smcc
4,317517
asked Dec 8 '18 at 9:19
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So i am working with an excercise reguarding the implicit function theorem .
$$F(x, y) = y(e^y + x) − log(x).$$
I know $(x_0,y_0)=(1,0)$ such that $F(x_0, y_0) = 0$.
I want to check that there exist a strictly positive real number $delta > 0$ and a function $y =f(x)$ such that $$y_0 = f(x_0)quad text{ and }quad F(x, f(x)) = 0 qquad forall x ∈ (x_0 − delta, x_0 + delta).$$
Is it fair to say that , in order to solve the question, we have to check if it possible to apply the first part of the Implicit Function Theorem?
Which would imply, as the answer sheet says, that the point $(1, 0)$ belongs to $$DF = {(x, y) in mathbb{R}^2∶ x > 0, y in mathbb{R}}$$
the domain of $F$.
I have dyscalculia and it appears confusing how to solve the domain of this function, where $x$ is greater than $0$ and $y$ exists in real numbers..
Can somebody please show me, in the simplest way possible, how to find the domain so I can proceed with the excercise? Thanks!
implicit-function-theorem
edited Dec 8 '18 at 10:15
smcc
4,317517
asked Dec 8 '18 at 9:19
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So i am working with an excercise reguarding the implicit function theorem .
$$F(x, y) = y(e^y + x) − log(x).$$
I know $(x_0,y_0)=(1,0)$ such that $F(x_0, y_0) = 0$.
I want to check that there exist a strictly positive real number $delta > 0$ and a function $y =f(x)$ such that $$y_0 = f(x_0)quad text{ and }quad F(x, f(x)) = 0 qquad forall x ∈ (x_0 − delta, x_0 + delta).$$
Is it fair to say that , in order to solve the question, we have to check if it possible to apply the first part of the Implicit Function Theorem?
Which would imply, as the answer sheet says, that the point $(1, 0)$ belongs to $$DF = {(x, y) in mathbb{R}^2∶ x > 0, y in mathbb{R}}$$
the domain of $F$.
I have dyscalculia and it appears confusing how to solve the domain of this function, where $x$ is greater than $0$ and $y$ exists in real numbers..
Can somebody please show me, in the simplest way possible, how to find the domain so I can proceed with the excercise? Thanks!
implicit-function-theorem
edited Dec 8 '18 at 10:15
smcc
4,317517
asked Dec 8 '18 at 9:19
BM97BM97
758
$endgroup$
So i am working with an excercise reguarding the implicit function theorem .
$$F(x, y) = y(e^y + x) − log(x).$$
I know $(x_0,y_0)=(1,0)$ such that $F(x_0, y_0) = 0$.
I want to check that there exist a strictly positive real number $delta > 0$ and a function $y =f(x)$ such that $$y_0 = f(x_0)quad text{ and }quad F(x, f(x)) = 0 qquad forall x ∈ (x_0 − delta, x_0 + delta).$$
Is it fair to say that , in order to solve the question, we have to check if it possible to apply the first part of the Implicit Function Theorem?
Which would imply, as the answer sheet says, that the point $(1, 0)$ belongs to $$DF = {(x, y) in mathbb{R}^2∶ x > 0, y in mathbb{R}}$$
the domain of $F$.
I have dyscalculia and it appears confusing how to solve the domain of this function, where $x$ is greater than $0$ and $y$ exists in real numbers..
Can somebody please show me, in the simplest way possible, how to find the domain so I can proceed with the excercise? Thanks!
implicit-function-theorem
implicit-function-theorem
edited Dec 8 '18 at 10:15
smcc
4,317517
asked Dec 8 '18 at 9:19
BM97BM97
758
edited Dec 8 '18 at 10:15
smcc
4,317517
asked Dec 8 '18 at 9:19
BM97BM97
758
edited Dec 8 '18 at 10:15
smcc
4,317517
edited Dec 8 '18 at 10:15
smcc
4,317517
edited Dec 8 '18 at 10:15
smcc
4,317517
4,317517
asked Dec 8 '18 at 9:19
BM97BM97
758
asked Dec 8 '18 at 9:19
BM97BM97
758
asked Dec 8 '18 at 9:19
BM97BM97
758
758
add a comment |
add a comment |
1 Answer
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$begingroup$
The formula defining $F$ only makes sense for $x>0$ because pf the second term, $log(x)$. The largest possible domain of the logarithm function is all positive real numbers, so we need $x>0$. There is no restriction on $y$ because $y(e^y+x)$ makes sense for any real number $y$ (the largest possible domain of the exponential function is all real numbers).
answered Dec 8 '18 at 10:18
smccsmcc
4,317517
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add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
The formula defining $F$ only makes sense for $x>0$ because pf the second term, $log(x)$. The largest possible domain of the logarithm function is all positive real numbers, so we need $x>0$. There is no restriction on $y$ because $y(e^y+x)$ makes sense for any real number $y$ (the largest possible domain of the exponential function is all real numbers).
answered Dec 8 '18 at 10:18
smccsmcc
4,317517
$endgroup$
add a comment |
$begingroup$
The formula defining $F$ only makes sense for $x>0$ because pf the second term, $log(x)$. The largest possible domain of the logarithm function is all positive real numbers, so we need $x>0$. There is no restriction on $y$ because $y(e^y+x)$ makes sense for any real number $y$ (the largest possible domain of the exponential function is all real numbers).
answered Dec 8 '18 at 10:18
smccsmcc
4,317517
$endgroup$
add a comment |
$begingroup$
The formula defining $F$ only makes sense for $x>0$ because pf the second term, $log(x)$. The largest possible domain of the logarithm function is all positive real numbers, so we need $x>0$. There is no restriction on $y$ because $y(e^y+x)$ makes sense for any real number $y$ (the largest possible domain of the exponential function is all real numbers).
answered Dec 8 '18 at 10:18
smccsmcc
4,317517
$endgroup$
The formula defining $F$ only makes sense for $x>0$ because pf the second term, $log(x)$. The largest possible domain of the logarithm function is all positive real numbers, so we need $x>0$. There is no restriction on $y$ because $y(e^y+x)$ makes sense for any real number $y$ (the largest possible domain of the exponential function is all real numbers).
answered Dec 8 '18 at 10:18
smccsmcc
4,317517
answered Dec 8 '18 at 10:18
smccsmcc
4,317517
answered Dec 8 '18 at 10:18
smccsmcc
4,317517
answered Dec 8 '18 at 10:18
smccsmcc
4,317517
4,317517
add a comment |
add a comment |
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