Function: $f(x,y) = (x^2+4y^2-5)(x-1)$. Find where $f(x,y)=0; f(x,y)>0; f(x,y)<0$ [closed]
$begingroup$
I just begin with the 3- dimension function. I dont really understand how to begin with this problem.
Function $$f(x,y)= (x^2+4y^2-5)(x-1)$$
Find where $$f(x,y)=0; f(x,y)>0; f(x,y)<0.$$
To find where $$f(x,y)=0$$ i already have the ellipse function $$(x^2)/5 +(y^2)/(5/4)=1$$ and the straight line $$x=1$$ but the other two i dont know how to solve.
Thank you very much.
functions
edited Dec 16 '18 at 23:18
Brian Tung
25.8k32554
asked Dec 16 '18 at 23:13
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
closed as off-topic by amWhy, Eevee Trainer, user10354138, Erick Wong, Leucippus Dec 17 '18 at 8:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, user10354138, Erick Wong, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I just begin with the 3- dimension function. I dont really understand how to begin with this problem.
Function $$f(x,y)= (x^2+4y^2-5)(x-1)$$
Find where $$f(x,y)=0; f(x,y)>0; f(x,y)<0.$$
To find where $$f(x,y)=0$$ i already have the ellipse function $$(x^2)/5 +(y^2)/(5/4)=1$$ and the straight line $$x=1$$ but the other two i dont know how to solve.
Thank you very much.
functions
edited Dec 16 '18 at 23:18
Brian Tung
25.8k32554
asked Dec 16 '18 at 23:13
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
closed as off-topic by amWhy, Eevee Trainer, user10354138, Erick Wong, Leucippus Dec 17 '18 at 8:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, user10354138, Erick Wong, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
if you are on the ellipse, it's clear that those inequalities occur outside the ellipse or inside of it.
$endgroup$
– valer
Dec 16 '18 at 23:23
add a comment |
$begingroup$
I just begin with the 3- dimension function. I dont really understand how to begin with this problem.
Function $$f(x,y)= (x^2+4y^2-5)(x-1)$$
Find where $$f(x,y)=0; f(x,y)>0; f(x,y)<0.$$
To find where $$f(x,y)=0$$ i already have the ellipse function $$(x^2)/5 +(y^2)/(5/4)=1$$ and the straight line $$x=1$$ but the other two i dont know how to solve.
Thank you very much.
functions
edited Dec 16 '18 at 23:18
Brian Tung
25.8k32554
asked Dec 16 '18 at 23:13
Vu Thanh PhanVu Thanh Phan
347
$endgroup$
I just begin with the 3- dimension function. I dont really understand how to begin with this problem.
Function $$f(x,y)= (x^2+4y^2-5)(x-1)$$
Find where $$f(x,y)=0; f(x,y)>0; f(x,y)<0.$$
To find where $$f(x,y)=0$$ i already have the ellipse function $$(x^2)/5 +(y^2)/(5/4)=1$$ and the straight line $$x=1$$ but the other two i dont know how to solve.
Thank you very much.
functions
functions
edited Dec 16 '18 at 23:18
Brian Tung
25.8k32554
asked Dec 16 '18 at 23:13
Vu Thanh PhanVu Thanh Phan
347
edited Dec 16 '18 at 23:18
Brian Tung
25.8k32554
asked Dec 16 '18 at 23:13
Vu Thanh PhanVu Thanh Phan
347
edited Dec 16 '18 at 23:18
Brian Tung
25.8k32554
edited Dec 16 '18 at 23:18
Brian Tung
25.8k32554
edited Dec 16 '18 at 23:18
Brian Tung
25.8k32554
25.8k32554
asked Dec 16 '18 at 23:13
Vu Thanh PhanVu Thanh Phan
347
asked Dec 16 '18 at 23:13
Vu Thanh PhanVu Thanh Phan
347
asked Dec 16 '18 at 23:13
Vu Thanh PhanVu Thanh Phan
347
347
closed as off-topic by amWhy, Eevee Trainer, user10354138, Erick Wong, Leucippus Dec 17 '18 at 8:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, user10354138, Erick Wong, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Eevee Trainer, user10354138, Erick Wong, Leucippus Dec 17 '18 at 8:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, user10354138, Erick Wong, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
if you are on the ellipse, it's clear that those inequalities occur outside the ellipse or inside of it.
$endgroup$
– valer
Dec 16 '18 at 23:23
add a comment |
$begingroup$
if you are on the ellipse, it's clear that those inequalities occur outside the ellipse or inside of it.
$endgroup$
– valer
Dec 16 '18 at 23:23
$begingroup$
if you are on the ellipse, it's clear that those inequalities occur outside the ellipse or inside of it.
$endgroup$
– valer
Dec 16 '18 at 23:23
$begingroup$
if you are on the ellipse, it's clear that those inequalities occur outside the ellipse or inside of it.
$endgroup$
– valer
Dec 16 '18 at 23:23
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As Brian Tung explained there are 4 regions to check. I will explain why this works.
Since $fcolonmathbb R^2tomathbb R$ is continuous the images of connected subsets connected. In $mathbb R$ an open subset is connected if and only if it is an open interval. So as our regions are connected open subsets of $mathbb R^2$ the images of them under $f$ are open intervals.
Let $R$ be one of the regions. We know that $forall (x,y)in R: f(x,y)ne0$. So as $f(X)$ is an Interval $f(X)subset mathbb R_{>0}$ or $f(X) subsetmathbb R_{<0}$. In words: all Points $(x,y)$ in a region are either positive or negative.
Just calculate one of the values in each region and you know if the whole region is positive or negative.
answered Dec 16 '18 at 23:57
quiliupquiliup
1799
$endgroup$
$begingroup$
+1. Thanks for filling in the gaps. Though I must say, I'm mildly surprised that this particular aspect was the OP's gap.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:42
add a comment |
$begingroup$
HINT:
2.$f(x,y)gt 0 =(x^2+4y^2-5)(x-1)gt 0$is only possible when (x-1) and(x^2+4y^2-5) both are positive or both are negative.
so,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
solve these 4 inequality conditions and you will get your second answer.
Now,
3.$f(x,y)lt 0 =(x^2+4y^2-5)(x-1)lt 0$is only possible when between (x-1) and(x^2+4y^2-5) one is positive and the other one is negative.
So,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
solve these 4 inequality conditions and you will get your third answer.
answered Dec 16 '18 at 23:29
Rakibul Islam PrinceRakibul Islam Prince
988211
$endgroup$
add a comment |
$begingroup$
You've already correctly identified the boundaries of the regions you need to identify:
These two curves divide the plane into four regions:
- inside the ellipse and to the left of the line
- inside the ellipse and to the right of the line
- outside the ellipse and to the left of the line
- outside the ellipse and to the right of the line
In each region, simply decide whether the first term $x^2+4y^2-5$ is positive or negative, and whether the second term $x-1$ is positive or negative. Then keep in mind how positive and negative terms multiply, and you should have your answer.
answered Dec 16 '18 at 23:28
Brian TungBrian Tung
25.8k32554
$endgroup$
$begingroup$
Hi. Thanks for your answer. Can you explain alittle bit more, base on what we have those divided regions and how can they determine the positive or negative value of f(x,y)?
$endgroup$
– Vu Thanh Phan
Dec 16 '18 at 23:44
$begingroup$
It's no different from seeing that for a function of one variable—say, $x^2+x-2$—the function equals $0$ at $x = -2$ and $x = 1$, and realizing that it must be positive or negative in the regions bounded by those points. (In fact, it is positive for $x < -2$, negative for $-2 < x < 1$, and positive again for $x > 1$.) Here, you have already correctly identified the boundaries of the regions in question; I have plotted them in my answer. Within those regions, are those two terms positive or negative?
$endgroup$
– Brian Tung
Dec 17 '18 at 0:09
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As Brian Tung explained there are 4 regions to check. I will explain why this works.
Since $fcolonmathbb R^2tomathbb R$ is continuous the images of connected subsets connected. In $mathbb R$ an open subset is connected if and only if it is an open interval. So as our regions are connected open subsets of $mathbb R^2$ the images of them under $f$ are open intervals.
Let $R$ be one of the regions. We know that $forall (x,y)in R: f(x,y)ne0$. So as $f(X)$ is an Interval $f(X)subset mathbb R_{>0}$ or $f(X) subsetmathbb R_{<0}$. In words: all Points $(x,y)$ in a region are either positive or negative.
Just calculate one of the values in each region and you know if the whole region is positive or negative.
answered Dec 16 '18 at 23:57
quiliupquiliup
1799
$endgroup$
$begingroup$
+1. Thanks for filling in the gaps. Though I must say, I'm mildly surprised that this particular aspect was the OP's gap.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:42
add a comment |
$begingroup$
As Brian Tung explained there are 4 regions to check. I will explain why this works.
Since $fcolonmathbb R^2tomathbb R$ is continuous the images of connected subsets connected. In $mathbb R$ an open subset is connected if and only if it is an open interval. So as our regions are connected open subsets of $mathbb R^2$ the images of them under $f$ are open intervals.
Let $R$ be one of the regions. We know that $forall (x,y)in R: f(x,y)ne0$. So as $f(X)$ is an Interval $f(X)subset mathbb R_{>0}$ or $f(X) subsetmathbb R_{<0}$. In words: all Points $(x,y)$ in a region are either positive or negative.
Just calculate one of the values in each region and you know if the whole region is positive or negative.
answered Dec 16 '18 at 23:57
quiliupquiliup
1799
$endgroup$
$begingroup$
+1. Thanks for filling in the gaps. Though I must say, I'm mildly surprised that this particular aspect was the OP's gap.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:42
add a comment |
$begingroup$
As Brian Tung explained there are 4 regions to check. I will explain why this works.
Since $fcolonmathbb R^2tomathbb R$ is continuous the images of connected subsets connected. In $mathbb R$ an open subset is connected if and only if it is an open interval. So as our regions are connected open subsets of $mathbb R^2$ the images of them under $f$ are open intervals.
Let $R$ be one of the regions. We know that $forall (x,y)in R: f(x,y)ne0$. So as $f(X)$ is an Interval $f(X)subset mathbb R_{>0}$ or $f(X) subsetmathbb R_{<0}$. In words: all Points $(x,y)$ in a region are either positive or negative.
Just calculate one of the values in each region and you know if the whole region is positive or negative.
answered Dec 16 '18 at 23:57
quiliupquiliup
1799
$endgroup$
As Brian Tung explained there are 4 regions to check. I will explain why this works.
Since $fcolonmathbb R^2tomathbb R$ is continuous the images of connected subsets connected. In $mathbb R$ an open subset is connected if and only if it is an open interval. So as our regions are connected open subsets of $mathbb R^2$ the images of them under $f$ are open intervals.
Let $R$ be one of the regions. We know that $forall (x,y)in R: f(x,y)ne0$. So as $f(X)$ is an Interval $f(X)subset mathbb R_{>0}$ or $f(X) subsetmathbb R_{<0}$. In words: all Points $(x,y)$ in a region are either positive or negative.
Just calculate one of the values in each region and you know if the whole region is positive or negative.
answered Dec 16 '18 at 23:57
quiliupquiliup
1799
edited Dec 17 '18 at 11:49
answered Dec 16 '18 at 23:57
quiliupquiliup
1799
answered Dec 16 '18 at 23:57
quiliupquiliup
1799
answered Dec 16 '18 at 23:57
quiliupquiliup
1799
1799
$begingroup$
+1. Thanks for filling in the gaps. Though I must say, I'm mildly surprised that this particular aspect was the OP's gap.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:42
add a comment |
$begingroup$
+1. Thanks for filling in the gaps. Though I must say, I'm mildly surprised that this particular aspect was the OP's gap.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:42
$begingroup$
+1. Thanks for filling in the gaps. Though I must say, I'm mildly surprised that this particular aspect was the OP's gap.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:42
$begingroup$
+1. Thanks for filling in the gaps. Though I must say, I'm mildly surprised that this particular aspect was the OP's gap.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:42
add a comment |
$begingroup$
HINT:
2.$f(x,y)gt 0 =(x^2+4y^2-5)(x-1)gt 0$is only possible when (x-1) and(x^2+4y^2-5) both are positive or both are negative.
so,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
solve these 4 inequality conditions and you will get your second answer.
Now,
3.$f(x,y)lt 0 =(x^2+4y^2-5)(x-1)lt 0$is only possible when between (x-1) and(x^2+4y^2-5) one is positive and the other one is negative.
So,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
solve these 4 inequality conditions and you will get your third answer.
answered Dec 16 '18 at 23:29
Rakibul Islam PrinceRakibul Islam Prince
988211
$endgroup$
add a comment |
$begingroup$
HINT:
2.$f(x,y)gt 0 =(x^2+4y^2-5)(x-1)gt 0$is only possible when (x-1) and(x^2+4y^2-5) both are positive or both are negative.
so,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
solve these 4 inequality conditions and you will get your second answer.
Now,
3.$f(x,y)lt 0 =(x^2+4y^2-5)(x-1)lt 0$is only possible when between (x-1) and(x^2+4y^2-5) one is positive and the other one is negative.
So,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
solve these 4 inequality conditions and you will get your third answer.
answered Dec 16 '18 at 23:29
Rakibul Islam PrinceRakibul Islam Prince
988211
$endgroup$
add a comment |
$begingroup$
HINT:
2.$f(x,y)gt 0 =(x^2+4y^2-5)(x-1)gt 0$is only possible when (x-1) and(x^2+4y^2-5) both are positive or both are negative.
so,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
solve these 4 inequality conditions and you will get your second answer.
Now,
3.$f(x,y)lt 0 =(x^2+4y^2-5)(x-1)lt 0$is only possible when between (x-1) and(x^2+4y^2-5) one is positive and the other one is negative.
So,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
solve these 4 inequality conditions and you will get your third answer.
answered Dec 16 '18 at 23:29
Rakibul Islam PrinceRakibul Islam Prince
988211
$endgroup$
HINT:
2.$f(x,y)gt 0 =(x^2+4y^2-5)(x-1)gt 0$is only possible when (x-1) and(x^2+4y^2-5) both are positive or both are negative.
so,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
solve these 4 inequality conditions and you will get your second answer.
Now,
3.$f(x,y)lt 0 =(x^2+4y^2-5)(x-1)lt 0$is only possible when between (x-1) and(x^2+4y^2-5) one is positive and the other one is negative.
So,this yields,
$$(x-1)gt 0~~~~and~~~~(x^2+4y^2-5)lt 0$$
$$0r,$$
$$(x-1)lt 0~~~~and~~~~(x^2+4y^2-5)gt 0$$
solve these 4 inequality conditions and you will get your third answer.
answered Dec 16 '18 at 23:29
Rakibul Islam PrinceRakibul Islam Prince
988211
answered Dec 16 '18 at 23:29
Rakibul Islam PrinceRakibul Islam Prince
988211
answered Dec 16 '18 at 23:29
Rakibul Islam PrinceRakibul Islam Prince
988211
answered Dec 16 '18 at 23:29
Rakibul Islam PrinceRakibul Islam Prince
988211
988211
add a comment |
add a comment |
$begingroup$
You've already correctly identified the boundaries of the regions you need to identify:
These two curves divide the plane into four regions:
- inside the ellipse and to the left of the line
- inside the ellipse and to the right of the line
- outside the ellipse and to the left of the line
- outside the ellipse and to the right of the line
In each region, simply decide whether the first term $x^2+4y^2-5$ is positive or negative, and whether the second term $x-1$ is positive or negative. Then keep in mind how positive and negative terms multiply, and you should have your answer.
answered Dec 16 '18 at 23:28
Brian TungBrian Tung
25.8k32554
$endgroup$
$begingroup$
Hi. Thanks for your answer. Can you explain alittle bit more, base on what we have those divided regions and how can they determine the positive or negative value of f(x,y)?
$endgroup$
– Vu Thanh Phan
Dec 16 '18 at 23:44
$begingroup$
It's no different from seeing that for a function of one variable—say, $x^2+x-2$—the function equals $0$ at $x = -2$ and $x = 1$, and realizing that it must be positive or negative in the regions bounded by those points. (In fact, it is positive for $x < -2$, negative for $-2 < x < 1$, and positive again for $x > 1$.) Here, you have already correctly identified the boundaries of the regions in question; I have plotted them in my answer. Within those regions, are those two terms positive or negative?
$endgroup$
– Brian Tung
Dec 17 '18 at 0:09
add a comment |
$begingroup$
You've already correctly identified the boundaries of the regions you need to identify:
These two curves divide the plane into four regions:
- inside the ellipse and to the left of the line
- inside the ellipse and to the right of the line
- outside the ellipse and to the left of the line
- outside the ellipse and to the right of the line
In each region, simply decide whether the first term $x^2+4y^2-5$ is positive or negative, and whether the second term $x-1$ is positive or negative. Then keep in mind how positive and negative terms multiply, and you should have your answer.
answered Dec 16 '18 at 23:28
Brian TungBrian Tung
25.8k32554
$endgroup$
$begingroup$
Hi. Thanks for your answer. Can you explain alittle bit more, base on what we have those divided regions and how can they determine the positive or negative value of f(x,y)?
$endgroup$
– Vu Thanh Phan
Dec 16 '18 at 23:44
$begingroup$
It's no different from seeing that for a function of one variable—say, $x^2+x-2$—the function equals $0$ at $x = -2$ and $x = 1$, and realizing that it must be positive or negative in the regions bounded by those points. (In fact, it is positive for $x < -2$, negative for $-2 < x < 1$, and positive again for $x > 1$.) Here, you have already correctly identified the boundaries of the regions in question; I have plotted them in my answer. Within those regions, are those two terms positive or negative?
$endgroup$
– Brian Tung
Dec 17 '18 at 0:09
add a comment |
$begingroup$
You've already correctly identified the boundaries of the regions you need to identify:
These two curves divide the plane into four regions:
- inside the ellipse and to the left of the line
- inside the ellipse and to the right of the line
- outside the ellipse and to the left of the line
- outside the ellipse and to the right of the line
In each region, simply decide whether the first term $x^2+4y^2-5$ is positive or negative, and whether the second term $x-1$ is positive or negative. Then keep in mind how positive and negative terms multiply, and you should have your answer.
answered Dec 16 '18 at 23:28
Brian TungBrian Tung
25.8k32554
$endgroup$
You've already correctly identified the boundaries of the regions you need to identify:
These two curves divide the plane into four regions:
- inside the ellipse and to the left of the line
- inside the ellipse and to the right of the line
- outside the ellipse and to the left of the line
- outside the ellipse and to the right of the line
In each region, simply decide whether the first term $x^2+4y^2-5$ is positive or negative, and whether the second term $x-1$ is positive or negative. Then keep in mind how positive and negative terms multiply, and you should have your answer.
answered Dec 16 '18 at 23:28
Brian TungBrian Tung
25.8k32554
edited Dec 16 '18 at 23:33
answered Dec 16 '18 at 23:28
Brian TungBrian Tung
25.8k32554
answered Dec 16 '18 at 23:28
Brian TungBrian Tung
25.8k32554
answered Dec 16 '18 at 23:28
Brian TungBrian Tung
25.8k32554
25.8k32554
$begingroup$
Hi. Thanks for your answer. Can you explain alittle bit more, base on what we have those divided regions and how can they determine the positive or negative value of f(x,y)?
$endgroup$
– Vu Thanh Phan
Dec 16 '18 at 23:44
$begingroup$
It's no different from seeing that for a function of one variable—say, $x^2+x-2$—the function equals $0$ at $x = -2$ and $x = 1$, and realizing that it must be positive or negative in the regions bounded by those points. (In fact, it is positive for $x < -2$, negative for $-2 < x < 1$, and positive again for $x > 1$.) Here, you have already correctly identified the boundaries of the regions in question; I have plotted them in my answer. Within those regions, are those two terms positive or negative?
$endgroup$
– Brian Tung
Dec 17 '18 at 0:09
add a comment |
$begingroup$
Hi. Thanks for your answer. Can you explain alittle bit more, base on what we have those divided regions and how can they determine the positive or negative value of f(x,y)?
$endgroup$
– Vu Thanh Phan
Dec 16 '18 at 23:44
$begingroup$
It's no different from seeing that for a function of one variable—say, $x^2+x-2$—the function equals $0$ at $x = -2$ and $x = 1$, and realizing that it must be positive or negative in the regions bounded by those points. (In fact, it is positive for $x < -2$, negative for $-2 < x < 1$, and positive again for $x > 1$.) Here, you have already correctly identified the boundaries of the regions in question; I have plotted them in my answer. Within those regions, are those two terms positive or negative?
$endgroup$
– Brian Tung
Dec 17 '18 at 0:09
$begingroup$
Hi. Thanks for your answer. Can you explain alittle bit more, base on what we have those divided regions and how can they determine the positive or negative value of f(x,y)?
$endgroup$
– Vu Thanh Phan
Dec 16 '18 at 23:44
$begingroup$
Hi. Thanks for your answer. Can you explain alittle bit more, base on what we have those divided regions and how can they determine the positive or negative value of f(x,y)?
$endgroup$
– Vu Thanh Phan
Dec 16 '18 at 23:44
$begingroup$
It's no different from seeing that for a function of one variable—say, $x^2+x-2$—the function equals $0$ at $x = -2$ and $x = 1$, and realizing that it must be positive or negative in the regions bounded by those points. (In fact, it is positive for $x < -2$, negative for $-2 < x < 1$, and positive again for $x > 1$.) Here, you have already correctly identified the boundaries of the regions in question; I have plotted them in my answer. Within those regions, are those two terms positive or negative?
$endgroup$
– Brian Tung
Dec 17 '18 at 0:09
$begingroup$
It's no different from seeing that for a function of one variable—say, $x^2+x-2$—the function equals $0$ at $x = -2$ and $x = 1$, and realizing that it must be positive or negative in the regions bounded by those points. (In fact, it is positive for $x < -2$, negative for $-2 < x < 1$, and positive again for $x > 1$.) Here, you have already correctly identified the boundaries of the regions in question; I have plotted them in my answer. Within those regions, are those two terms positive or negative?
$endgroup$
– Brian Tung
Dec 17 '18 at 0:09
add a comment |
$begingroup$
if you are on the ellipse, it's clear that those inequalities occur outside the ellipse or inside of it.
$endgroup$
– valer
Dec 16 '18 at 23:23