Mixing
Applying the implicit function theorem on $e^y+x^2ye^{-x}=3$
1
$begingroup$
I know the implicit function theorem but no clue how to show this..
Show that $e^y+x^2ye^{-x}=3$ implicitly defines a unique function $y=f(x)$ defined on all real numbers.
multivariable-calculus implicit-function-theorem
edited Jan 20 at 11:18
Git Gud
28.9k1050100
asked Jan 20 at 10:50
ArchimedessArchimedess
236
$endgroup$
add a comment |
1
$begingroup$
I know the implicit function theorem but no clue how to show this..
Show that $e^y+x^2ye^{-x}=3$ implicitly defines a unique function $y=f(x)$ defined on all real numbers.
multivariable-calculus implicit-function-theorem
edited Jan 20 at 11:18
Git Gud
28.9k1050100
asked Jan 20 at 10:50
ArchimedessArchimedess
236
$endgroup$
$begingroup$
Can you state the version of the implicit function theorem you're familiar with? That will help people fit an answer to your needs.
$endgroup$
– Git Gud
Jan 20 at 11:02
add a comment |
1
1
1
$begingroup$
I know the implicit function theorem but no clue how to show this..
Show that $e^y+x^2ye^{-x}=3$ implicitly defines a unique function $y=f(x)$ defined on all real numbers.
multivariable-calculus implicit-function-theorem
edited Jan 20 at 11:18
Git Gud
28.9k1050100
asked Jan 20 at 10:50
ArchimedessArchimedess
236
$endgroup$
I know the implicit function theorem but no clue how to show this..
Show that $e^y+x^2ye^{-x}=3$ implicitly defines a unique function $y=f(x)$ defined on all real numbers.
multivariable-calculus implicit-function-theorem
multivariable-calculus implicit-function-theorem
edited Jan 20 at 11:18
Git Gud
28.9k1050100
asked Jan 20 at 10:50
ArchimedessArchimedess
236
edited Jan 20 at 11:18
Git Gud
28.9k1050100
asked Jan 20 at 10:50
ArchimedessArchimedess
236
edited Jan 20 at 11:18
Git Gud
28.9k1050100
edited Jan 20 at 11:18
Git Gud
28.9k1050100
edited Jan 20 at 11:18
Git Gud
28.9k1050100
28.9k1050100
asked Jan 20 at 10:50
ArchimedessArchimedess
236
asked Jan 20 at 10:50
ArchimedessArchimedess
236
asked Jan 20 at 10:50
ArchimedessArchimedess
236
236
$begingroup$
Can you state the version of the implicit function theorem you're familiar with? That will help people fit an answer to your needs.
$endgroup$
– Git Gud
Jan 20 at 11:02
add a comment |
$begingroup$
Can you state the version of the implicit function theorem you're familiar with? That will help people fit an answer to your needs.
$endgroup$
– Git Gud
Jan 20 at 11:02
$begingroup$
Can you state the version of the implicit function theorem you're familiar with? That will help people fit an answer to your needs.
$endgroup$
– Git Gud
Jan 20 at 11:02
$begingroup$
Can you state the version of the implicit function theorem you're familiar with? That will help people fit an answer to your needs.
$endgroup$
– Git Gud
Jan 20 at 11:02
add a comment |
1 Answer
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Hint: note that $x^2 e^{-x} ge 0$, and so for each $x$, the function $y mapsto e^y + x^2e^{-x}y$ is strictly increasing.
answered Jan 20 at 10:54
Theo BenditTheo Bendit
19.2k12353
$endgroup$
$begingroup$
F(x,y) Strictly increasing means that $forall xinmathbb{R}exists!yinmathbb{R}$ such that $F(x,y)=0$ $implies$ $y=f(x)$ is unique and defined on all real numbers right?
$endgroup$
– Archimedess
Jan 20 at 10:59
$begingroup$
@Archimedess Unfortunately not (e.g. $arctan$ is strictly increasing and has a range only of $(-pi/2, pi/2)$), but it does mean the function is injective, which means if there exists a $y$, then there exists a unique $y$. To get existence, try using the intermediate value theorem, and the fact that the function tends to $pm infty$ as $x to pm infty$ respectively.
$endgroup$
– Theo Bendit
Jan 20 at 11:01
$begingroup$
Thanks Theo, i got it now.
$endgroup$
– Archimedess
Jan 20 at 11:29
add a comment |
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$begingroup$
Hint: note that $x^2 e^{-x} ge 0$, and so for each $x$, the function $y mapsto e^y + x^2e^{-x}y$ is strictly increasing.
answered Jan 20 at 10:54
Theo BenditTheo Bendit
19.2k12353
$endgroup$
$begingroup$
F(x,y) Strictly increasing means that $forall xinmathbb{R}exists!yinmathbb{R}$ such that $F(x,y)=0$ $implies$ $y=f(x)$ is unique and defined on all real numbers right?
$endgroup$
– Archimedess
Jan 20 at 10:59
$begingroup$
@Archimedess Unfortunately not (e.g. $arctan$ is strictly increasing and has a range only of $(-pi/2, pi/2)$), but it does mean the function is injective, which means if there exists a $y$, then there exists a unique $y$. To get existence, try using the intermediate value theorem, and the fact that the function tends to $pm infty$ as $x to pm infty$ respectively.
$endgroup$
– Theo Bendit
Jan 20 at 11:01
$begingroup$
Thanks Theo, i got it now.
$endgroup$
– Archimedess
Jan 20 at 11:29
add a comment |
1
$begingroup$
Hint: note that $x^2 e^{-x} ge 0$, and so for each $x$, the function $y mapsto e^y + x^2e^{-x}y$ is strictly increasing.
answered Jan 20 at 10:54
Theo BenditTheo Bendit
19.2k12353
$endgroup$
$begingroup$
F(x,y) Strictly increasing means that $forall xinmathbb{R}exists!yinmathbb{R}$ such that $F(x,y)=0$ $implies$ $y=f(x)$ is unique and defined on all real numbers right?
$endgroup$
– Archimedess
Jan 20 at 10:59
$begingroup$
@Archimedess Unfortunately not (e.g. $arctan$ is strictly increasing and has a range only of $(-pi/2, pi/2)$), but it does mean the function is injective, which means if there exists a $y$, then there exists a unique $y$. To get existence, try using the intermediate value theorem, and the fact that the function tends to $pm infty$ as $x to pm infty$ respectively.
$endgroup$
– Theo Bendit
Jan 20 at 11:01
$begingroup$
Thanks Theo, i got it now.
$endgroup$
– Archimedess
Jan 20 at 11:29
add a comment |
1
1
1
$begingroup$
Hint: note that $x^2 e^{-x} ge 0$, and so for each $x$, the function $y mapsto e^y + x^2e^{-x}y$ is strictly increasing.
answered Jan 20 at 10:54
Theo BenditTheo Bendit
19.2k12353
$endgroup$
Hint: note that $x^2 e^{-x} ge 0$, and so for each $x$, the function $y mapsto e^y + x^2e^{-x}y$ is strictly increasing.
answered Jan 20 at 10:54
Theo BenditTheo Bendit
19.2k12353
answered Jan 20 at 10:54
Theo BenditTheo Bendit
19.2k12353
answered Jan 20 at 10:54
Theo BenditTheo Bendit
19.2k12353
answered Jan 20 at 10:54
Theo BenditTheo Bendit
19.2k12353
19.2k12353
$begingroup$
F(x,y) Strictly increasing means that $forall xinmathbb{R}exists!yinmathbb{R}$ such that $F(x,y)=0$ $implies$ $y=f(x)$ is unique and defined on all real numbers right?
$endgroup$
– Archimedess
Jan 20 at 10:59
$begingroup$
@Archimedess Unfortunately not (e.g. $arctan$ is strictly increasing and has a range only of $(-pi/2, pi/2)$), but it does mean the function is injective, which means if there exists a $y$, then there exists a unique $y$. To get existence, try using the intermediate value theorem, and the fact that the function tends to $pm infty$ as $x to pm infty$ respectively.
$endgroup$
– Theo Bendit
Jan 20 at 11:01
$begingroup$
Thanks Theo, i got it now.
$endgroup$
– Archimedess
Jan 20 at 11:29
add a comment |
$begingroup$
F(x,y) Strictly increasing means that $forall xinmathbb{R}exists!yinmathbb{R}$ such that $F(x,y)=0$ $implies$ $y=f(x)$ is unique and defined on all real numbers right?
$endgroup$
– Archimedess
Jan 20 at 10:59
$begingroup$
@Archimedess Unfortunately not (e.g. $arctan$ is strictly increasing and has a range only of $(-pi/2, pi/2)$), but it does mean the function is injective, which means if there exists a $y$, then there exists a unique $y$. To get existence, try using the intermediate value theorem, and the fact that the function tends to $pm infty$ as $x to pm infty$ respectively.
$endgroup$
– Theo Bendit
Jan 20 at 11:01
$begingroup$
Thanks Theo, i got it now.
$endgroup$
– Archimedess
Jan 20 at 11:29
$begingroup$
F(x,y) Strictly increasing means that $forall xinmathbb{R}exists!yinmathbb{R}$ such that $F(x,y)=0$ $implies$ $y=f(x)$ is unique and defined on all real numbers right?
$endgroup$
– Archimedess
Jan 20 at 10:59
$begingroup$
F(x,y) Strictly increasing means that $forall xinmathbb{R}exists!yinmathbb{R}$ such that $F(x,y)=0$ $implies$ $y=f(x)$ is unique and defined on all real numbers right?
$endgroup$
– Archimedess
Jan 20 at 10:59
$begingroup$
@Archimedess Unfortunately not (e.g. $arctan$ is strictly increasing and has a range only of $(-pi/2, pi/2)$), but it does mean the function is injective, which means if there exists a $y$, then there exists a unique $y$. To get existence, try using the intermediate value theorem, and the fact that the function tends to $pm infty$ as $x to pm infty$ respectively.
$endgroup$
– Theo Bendit
Jan 20 at 11:01
$begingroup$
@Archimedess Unfortunately not (e.g. $arctan$ is strictly increasing and has a range only of $(-pi/2, pi/2)$), but it does mean the function is injective, which means if there exists a $y$, then there exists a unique $y$. To get existence, try using the intermediate value theorem, and the fact that the function tends to $pm infty$ as $x to pm infty$ respectively.
$endgroup$
– Theo Bendit
Jan 20 at 11:01
$begingroup$
Thanks Theo, i got it now.
$endgroup$
– Archimedess
Jan 20 at 11:29
$begingroup$
Thanks Theo, i got it now.
$endgroup$
– Archimedess
Jan 20 at 11:29
add a comment |
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$begingroup$
Can you state the version of the implicit function theorem you're familiar with? That will help people fit an answer to your needs.
$endgroup$
– Git Gud
Jan 20 at 11:02