Describe the preimage of any point $(b,c) in mathbb{R}^2$ for $F(x, y) = (x+y,xy)$
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I plotted the line and the hyperbola implied by the first and second coordinates respectively. I realize that the preimage can be
Empty for $c > 0 land b < c$
Size one for $c = 0 lor (b = c land c > 0)$
Size two for $c < 0 lor b > c$.
But I need help actually writing down the expressions for each scenario. I would also appreciate knowing what course/area I should look into to get more practice with this type of problems.
functions inverse-function
asked Nov 18 at 18:35
zipzapboing
907
add a comment |
up vote
0
down vote
favorite
I plotted the line and the hyperbola implied by the first and second coordinates respectively. I realize that the preimage can be
Empty for $c > 0 land b < c$
Size one for $c = 0 lor (b = c land c > 0)$
Size two for $c < 0 lor b > c$.
But I need help actually writing down the expressions for each scenario. I would also appreciate knowing what course/area I should look into to get more practice with this type of problems.
functions inverse-function
asked Nov 18 at 18:35
zipzapboing
907
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I plotted the line and the hyperbola implied by the first and second coordinates respectively. I realize that the preimage can be
Empty for $c > 0 land b < c$
Size one for $c = 0 lor (b = c land c > 0)$
Size two for $c < 0 lor b > c$.
But I need help actually writing down the expressions for each scenario. I would also appreciate knowing what course/area I should look into to get more practice with this type of problems.
functions inverse-function
asked Nov 18 at 18:35
zipzapboing
907
I plotted the line and the hyperbola implied by the first and second coordinates respectively. I realize that the preimage can be
Empty for $c > 0 land b < c$
Size one for $c = 0 lor (b = c land c > 0)$
Size two for $c < 0 lor b > c$.
But I need help actually writing down the expressions for each scenario. I would also appreciate knowing what course/area I should look into to get more practice with this type of problems.
functions inverse-function
functions inverse-function
asked Nov 18 at 18:35
zipzapboing
907
asked Nov 18 at 18:35
zipzapboing
907
asked Nov 18 at 18:35
zipzapboing
907
asked Nov 18 at 18:35
zipzapboing
907
asked Nov 18 at 18:35
zipzapboing
907
907
add a comment |
add a comment |
1 Answer
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1
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You are looking for solutions to the equation $f(x,y)=(b,c)$, i.e.
$$left{begin{matrix}
x+y=b \
xy=c
end{matrix}right.$$
Consider two cases
(1) $c=0$. Then $xy=0$ meaning either $x=0$ or $y=0$. The first equation then implies that either $y=b$ or $x=b$. Meaning for $c=0$ there are two solutions: $(0, b)$ and $(b,0)$.
(2) $cneq 0$. Then $xy=c$ implies neither $x=0$ nor $y=0$. Therefore $y=c/x$ makes total sense. So in the first equation we get
$$x+y=b$$
$$x+c/x=b$$
$$x^2 + c=bx$$
$$x^2-bx+c=0$$
So all that is left is to solve this equation, which can have at most two solutions depending on $b,c$. Can you complete the calculations?
answered Nov 18 at 18:44
freakish
10.6k1527
You can summarize both cases as the pre-image is the set $$F^{-1}((b,c))={(x,y) , | , x text{ and } y text{ are roots of } t^2-bt+c=0}.$$
– Anurag A
Nov 18 at 19:26
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are looking for solutions to the equation $f(x,y)=(b,c)$, i.e.
$$left{begin{matrix}
x+y=b \
xy=c
end{matrix}right.$$
Consider two cases
(1) $c=0$. Then $xy=0$ meaning either $x=0$ or $y=0$. The first equation then implies that either $y=b$ or $x=b$. Meaning for $c=0$ there are two solutions: $(0, b)$ and $(b,0)$.
(2) $cneq 0$. Then $xy=c$ implies neither $x=0$ nor $y=0$. Therefore $y=c/x$ makes total sense. So in the first equation we get
$$x+y=b$$
$$x+c/x=b$$
$$x^2 + c=bx$$
$$x^2-bx+c=0$$
So all that is left is to solve this equation, which can have at most two solutions depending on $b,c$. Can you complete the calculations?
answered Nov 18 at 18:44
freakish
10.6k1527
You can summarize both cases as the pre-image is the set $$F^{-1}((b,c))={(x,y) , | , x text{ and } y text{ are roots of } t^2-bt+c=0}.$$
– Anurag A
Nov 18 at 19:26
add a comment |
up vote
1
down vote
accepted
You are looking for solutions to the equation $f(x,y)=(b,c)$, i.e.
$$left{begin{matrix}
x+y=b \
xy=c
end{matrix}right.$$
Consider two cases
(1) $c=0$. Then $xy=0$ meaning either $x=0$ or $y=0$. The first equation then implies that either $y=b$ or $x=b$. Meaning for $c=0$ there are two solutions: $(0, b)$ and $(b,0)$.
(2) $cneq 0$. Then $xy=c$ implies neither $x=0$ nor $y=0$. Therefore $y=c/x$ makes total sense. So in the first equation we get
$$x+y=b$$
$$x+c/x=b$$
$$x^2 + c=bx$$
$$x^2-bx+c=0$$
So all that is left is to solve this equation, which can have at most two solutions depending on $b,c$. Can you complete the calculations?
answered Nov 18 at 18:44
freakish
10.6k1527
You can summarize both cases as the pre-image is the set $$F^{-1}((b,c))={(x,y) , | , x text{ and } y text{ are roots of } t^2-bt+c=0}.$$
– Anurag A
Nov 18 at 19:26
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are looking for solutions to the equation $f(x,y)=(b,c)$, i.e.
$$left{begin{matrix}
x+y=b \
xy=c
end{matrix}right.$$
Consider two cases
(1) $c=0$. Then $xy=0$ meaning either $x=0$ or $y=0$. The first equation then implies that either $y=b$ or $x=b$. Meaning for $c=0$ there are two solutions: $(0, b)$ and $(b,0)$.
(2) $cneq 0$. Then $xy=c$ implies neither $x=0$ nor $y=0$. Therefore $y=c/x$ makes total sense. So in the first equation we get
$$x+y=b$$
$$x+c/x=b$$
$$x^2 + c=bx$$
$$x^2-bx+c=0$$
So all that is left is to solve this equation, which can have at most two solutions depending on $b,c$. Can you complete the calculations?
answered Nov 18 at 18:44
freakish
10.6k1527
You are looking for solutions to the equation $f(x,y)=(b,c)$, i.e.
$$left{begin{matrix}
x+y=b \
xy=c
end{matrix}right.$$
Consider two cases
(1) $c=0$. Then $xy=0$ meaning either $x=0$ or $y=0$. The first equation then implies that either $y=b$ or $x=b$. Meaning for $c=0$ there are two solutions: $(0, b)$ and $(b,0)$.
(2) $cneq 0$. Then $xy=c$ implies neither $x=0$ nor $y=0$. Therefore $y=c/x$ makes total sense. So in the first equation we get
$$x+y=b$$
$$x+c/x=b$$
$$x^2 + c=bx$$
$$x^2-bx+c=0$$
So all that is left is to solve this equation, which can have at most two solutions depending on $b,c$. Can you complete the calculations?
answered Nov 18 at 18:44
freakish
10.6k1527
answered Nov 18 at 18:44
freakish
10.6k1527
answered Nov 18 at 18:44
freakish
10.6k1527
answered Nov 18 at 18:44
freakish
10.6k1527
10.6k1527
You can summarize both cases as the pre-image is the set $$F^{-1}((b,c))={(x,y) , | , x text{ and } y text{ are roots of } t^2-bt+c=0}.$$
– Anurag A
Nov 18 at 19:26
add a comment |
You can summarize both cases as the pre-image is the set $$F^{-1}((b,c))={(x,y) , | , x text{ and } y text{ are roots of } t^2-bt+c=0}.$$
– Anurag A
Nov 18 at 19:26
You can summarize both cases as the pre-image is the set $$F^{-1}((b,c))={(x,y) , | , x text{ and } y text{ are roots of } t^2-bt+c=0}.$$
– Anurag A
Nov 18 at 19:26
You can summarize both cases as the pre-image is the set $$F^{-1}((b,c))={(x,y) , | , x text{ and } y text{ are roots of } t^2-bt+c=0}.$$
– Anurag A
Nov 18 at 19:26
add a comment |
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