Misuse of inverse function theorem
$begingroup$
Let $U={binom{u}{v}in mathbb{R}^2|0<v<u}$. Let $f:Utomathbb{R}^2, f(u,v)=(u+v,uv)$. Show that $f$ has a global reverse function, find $g=f^{-1}$ and its domain.
Not a valid solution: For any $(u,v)inmathbb{R}^2$,
$$
\ Df(u,v)=begin{pmatrix}
1& 1\
v& u
end{pmatrix} ne0 Leftrightarrow une v
$$
then $Df(u,v)$ is reversable for any $(u,v)in U$. Thus, $forall (u,v)in U$,
$$
\ g(u,v)=frac{operatorname{adj}(Df(u,v))}{operatorname{det}(Df(u,v))}=frac{1}{u-v}cdot begin{pmatrix}
u & -1 \
-v & 1 end{pmatrix}
$$ I understand this is not the solution, but I don't understand where is the mistake.
algebra-precalculus multivariable-calculus inverse-function-theorem
asked Dec 2 '18 at 19:37
J. DoeJ. Doe
1637
$endgroup$
add a comment |
$begingroup$
Let $U={binom{u}{v}in mathbb{R}^2|0<v<u}$. Let $f:Utomathbb{R}^2, f(u,v)=(u+v,uv)$. Show that $f$ has a global reverse function, find $g=f^{-1}$ and its domain.
Not a valid solution: For any $(u,v)inmathbb{R}^2$,
$$
\ Df(u,v)=begin{pmatrix}
1& 1\
v& u
end{pmatrix} ne0 Leftrightarrow une v
$$
then $Df(u,v)$ is reversable for any $(u,v)in U$. Thus, $forall (u,v)in U$,
$$
\ g(u,v)=frac{operatorname{adj}(Df(u,v))}{operatorname{det}(Df(u,v))}=frac{1}{u-v}cdot begin{pmatrix}
u & -1 \
-v & 1 end{pmatrix}
$$ I understand this is not the solution, but I don't understand where is the mistake.
algebra-precalculus multivariable-calculus inverse-function-theorem
asked Dec 2 '18 at 19:37
J. DoeJ. Doe
1637
$endgroup$
add a comment |
$begingroup$
Let $U={binom{u}{v}in mathbb{R}^2|0<v<u}$. Let $f:Utomathbb{R}^2, f(u,v)=(u+v,uv)$. Show that $f$ has a global reverse function, find $g=f^{-1}$ and its domain.
Not a valid solution: For any $(u,v)inmathbb{R}^2$,
$$
\ Df(u,v)=begin{pmatrix}
1& 1\
v& u
end{pmatrix} ne0 Leftrightarrow une v
$$
then $Df(u,v)$ is reversable for any $(u,v)in U$. Thus, $forall (u,v)in U$,
$$
\ g(u,v)=frac{operatorname{adj}(Df(u,v))}{operatorname{det}(Df(u,v))}=frac{1}{u-v}cdot begin{pmatrix}
u & -1 \
-v & 1 end{pmatrix}
$$ I understand this is not the solution, but I don't understand where is the mistake.
algebra-precalculus multivariable-calculus inverse-function-theorem
asked Dec 2 '18 at 19:37
J. DoeJ. Doe
1637
$endgroup$
Let $U={binom{u}{v}in mathbb{R}^2|0<v<u}$. Let $f:Utomathbb{R}^2, f(u,v)=(u+v,uv)$. Show that $f$ has a global reverse function, find $g=f^{-1}$ and its domain.
Not a valid solution: For any $(u,v)inmathbb{R}^2$,
$$
\ Df(u,v)=begin{pmatrix}
1& 1\
v& u
end{pmatrix} ne0 Leftrightarrow une v
$$
then $Df(u,v)$ is reversable for any $(u,v)in U$. Thus, $forall (u,v)in U$,
$$
\ g(u,v)=frac{operatorname{adj}(Df(u,v))}{operatorname{det}(Df(u,v))}=frac{1}{u-v}cdot begin{pmatrix}
u & -1 \
-v & 1 end{pmatrix}
$$ I understand this is not the solution, but I don't understand where is the mistake.
algebra-precalculus multivariable-calculus inverse-function-theorem
algebra-precalculus multivariable-calculus inverse-function-theorem
asked Dec 2 '18 at 19:37
J. DoeJ. Doe
1637
asked Dec 2 '18 at 19:37
J. DoeJ. Doe
1637
asked Dec 2 '18 at 19:37
J. DoeJ. Doe
1637
asked Dec 2 '18 at 19:37
J. DoeJ. Doe
1637
asked Dec 2 '18 at 19:37
J. DoeJ. Doe
1637
1637
add a comment |
add a comment |
1 Answer
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$begingroup$
You've found the inverse of $Df$, not the inverse of $f$.
answered Dec 2 '18 at 19:41
TedTed
21.5k13260
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You've found the inverse of $Df$, not the inverse of $f$.
answered Dec 2 '18 at 19:41
TedTed
21.5k13260
$endgroup$
add a comment |
$begingroup$
You've found the inverse of $Df$, not the inverse of $f$.
answered Dec 2 '18 at 19:41
TedTed
21.5k13260
$endgroup$
add a comment |
$begingroup$
You've found the inverse of $Df$, not the inverse of $f$.
answered Dec 2 '18 at 19:41
TedTed
21.5k13260
$endgroup$
You've found the inverse of $Df$, not the inverse of $f$.
answered Dec 2 '18 at 19:41
TedTed
21.5k13260
answered Dec 2 '18 at 19:41
TedTed
21.5k13260
answered Dec 2 '18 at 19:41
TedTed
21.5k13260
answered Dec 2 '18 at 19:41
TedTed
21.5k13260
21.5k13260
add a comment |
add a comment |
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