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Show ${x : f(x) < g(x)}$ is bounded above
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I recently came across this problem:
Show that the set ${x : x^2 < 1-x}$ is bounded above.
How should I approach any similar problem of the form ${x : f(x) < g(x)}$?
Unfortunately, I cannot say what I have already tried; this is a very new concept to me, and I am not sure what things I can even try in the first place. I am not necessarily looking for a solution; I am wondering how I should begin to tackle such a problem.
analysis
asked Feb 1 at 15:25
NetherGraniteNetherGranite
1478
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add a comment |
$begingroup$
I recently came across this problem:
Show that the set ${x : x^2 < 1-x}$ is bounded above.
How should I approach any similar problem of the form ${x : f(x) < g(x)}$?
Unfortunately, I cannot say what I have already tried; this is a very new concept to me, and I am not sure what things I can even try in the first place. I am not necessarily looking for a solution; I am wondering how I should begin to tackle such a problem.
analysis
asked Feb 1 at 15:25
NetherGraniteNetherGranite
1478
$endgroup$
add a comment |
$begingroup$
I recently came across this problem:
Show that the set ${x : x^2 < 1-x}$ is bounded above.
How should I approach any similar problem of the form ${x : f(x) < g(x)}$?
Unfortunately, I cannot say what I have already tried; this is a very new concept to me, and I am not sure what things I can even try in the first place. I am not necessarily looking for a solution; I am wondering how I should begin to tackle such a problem.
analysis
asked Feb 1 at 15:25
NetherGraniteNetherGranite
1478
$endgroup$
I recently came across this problem:
Show that the set ${x : x^2 < 1-x}$ is bounded above.
How should I approach any similar problem of the form ${x : f(x) < g(x)}$?
Unfortunately, I cannot say what I have already tried; this is a very new concept to me, and I am not sure what things I can even try in the first place. I am not necessarily looking for a solution; I am wondering how I should begin to tackle such a problem.
analysis
analysis
asked Feb 1 at 15:25
NetherGraniteNetherGranite
1478
asked Feb 1 at 15:25
NetherGraniteNetherGranite
1478
edited Feb 1 at 15:26
NetherGranite
asked Feb 1 at 15:25
NetherGraniteNetherGranite
1478
asked Feb 1 at 15:25
NetherGraniteNetherGranite
1478
asked Feb 1 at 15:25
NetherGraniteNetherGranite
1478
1478
add a comment |
add a comment |
2 Answers
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You just need to prouve that there exists an $a>0$ such that $f(x)geq g(x)~~forall x>a$ or just prouve that $lim_{xto+infty} (f(x)-g(x))>0$. In your case $f(x)-g(x)=x^2+x-1to+infty$ when $xto+infty$ so you can deduce that the set is bounded from above.
edited Feb 1 at 15:56
NetherGranite
1478
answered Feb 1 at 15:32
MahranMahran
1235
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add a comment |
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The first thing I would do, and what Harnack suggested, is solve the given inequality. We are given that $x^2< 1- x$. That is the same as $x^2+ x- 1= 0$. Completing the square, $x^2+ x+ 1/4- 1/4- 1= (x- 1/2)^2- 5/4= 0$. $(x- 1/2)^2= 5/4$ so $x- 1/2= pmfrac{sqrt{5}}{2}$ so that $x= frac{1}{2}pmfrac{sqrt{5}}{2}$. $y= x^2+ x- 1$ is a parabola opening upward, crossing the x-axis at $x= frac{1- sqrt{5}}{2}$ and $x= frac{1+ sqrt{5}}{2}$. The parabola is below the x-axis, so the original inequality is true, between those two numbers.
answered Feb 1 at 15:40
user247327user247327
11.6k1516
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
You just need to prouve that there exists an $a>0$ such that $f(x)geq g(x)~~forall x>a$ or just prouve that $lim_{xto+infty} (f(x)-g(x))>0$. In your case $f(x)-g(x)=x^2+x-1to+infty$ when $xto+infty$ so you can deduce that the set is bounded from above.
edited Feb 1 at 15:56
NetherGranite
1478
answered Feb 1 at 15:32
MahranMahran
1235
$endgroup$
add a comment |
$begingroup$
You just need to prouve that there exists an $a>0$ such that $f(x)geq g(x)~~forall x>a$ or just prouve that $lim_{xto+infty} (f(x)-g(x))>0$. In your case $f(x)-g(x)=x^2+x-1to+infty$ when $xto+infty$ so you can deduce that the set is bounded from above.
edited Feb 1 at 15:56
NetherGranite
1478
answered Feb 1 at 15:32
MahranMahran
1235
$endgroup$
add a comment |
$begingroup$
You just need to prouve that there exists an $a>0$ such that $f(x)geq g(x)~~forall x>a$ or just prouve that $lim_{xto+infty} (f(x)-g(x))>0$. In your case $f(x)-g(x)=x^2+x-1to+infty$ when $xto+infty$ so you can deduce that the set is bounded from above.
edited Feb 1 at 15:56
NetherGranite
1478
answered Feb 1 at 15:32
MahranMahran
1235
$endgroup$
You just need to prouve that there exists an $a>0$ such that $f(x)geq g(x)~~forall x>a$ or just prouve that $lim_{xto+infty} (f(x)-g(x))>0$. In your case $f(x)-g(x)=x^2+x-1to+infty$ when $xto+infty$ so you can deduce that the set is bounded from above.
edited Feb 1 at 15:56
NetherGranite
1478
answered Feb 1 at 15:32
MahranMahran
1235
edited Feb 1 at 15:56
NetherGranite
1478
edited Feb 1 at 15:56
NetherGranite
1478
edited Feb 1 at 15:56
NetherGranite
1478
1478
answered Feb 1 at 15:32
MahranMahran
1235
answered Feb 1 at 15:32
MahranMahran
1235
answered Feb 1 at 15:32
MahranMahran
1235
1235
add a comment |
add a comment |
$begingroup$
The first thing I would do, and what Harnack suggested, is solve the given inequality. We are given that $x^2< 1- x$. That is the same as $x^2+ x- 1= 0$. Completing the square, $x^2+ x+ 1/4- 1/4- 1= (x- 1/2)^2- 5/4= 0$. $(x- 1/2)^2= 5/4$ so $x- 1/2= pmfrac{sqrt{5}}{2}$ so that $x= frac{1}{2}pmfrac{sqrt{5}}{2}$. $y= x^2+ x- 1$ is a parabola opening upward, crossing the x-axis at $x= frac{1- sqrt{5}}{2}$ and $x= frac{1+ sqrt{5}}{2}$. The parabola is below the x-axis, so the original inequality is true, between those two numbers.
answered Feb 1 at 15:40
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
The first thing I would do, and what Harnack suggested, is solve the given inequality. We are given that $x^2< 1- x$. That is the same as $x^2+ x- 1= 0$. Completing the square, $x^2+ x+ 1/4- 1/4- 1= (x- 1/2)^2- 5/4= 0$. $(x- 1/2)^2= 5/4$ so $x- 1/2= pmfrac{sqrt{5}}{2}$ so that $x= frac{1}{2}pmfrac{sqrt{5}}{2}$. $y= x^2+ x- 1$ is a parabola opening upward, crossing the x-axis at $x= frac{1- sqrt{5}}{2}$ and $x= frac{1+ sqrt{5}}{2}$. The parabola is below the x-axis, so the original inequality is true, between those two numbers.
answered Feb 1 at 15:40
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
The first thing I would do, and what Harnack suggested, is solve the given inequality. We are given that $x^2< 1- x$. That is the same as $x^2+ x- 1= 0$. Completing the square, $x^2+ x+ 1/4- 1/4- 1= (x- 1/2)^2- 5/4= 0$. $(x- 1/2)^2= 5/4$ so $x- 1/2= pmfrac{sqrt{5}}{2}$ so that $x= frac{1}{2}pmfrac{sqrt{5}}{2}$. $y= x^2+ x- 1$ is a parabola opening upward, crossing the x-axis at $x= frac{1- sqrt{5}}{2}$ and $x= frac{1+ sqrt{5}}{2}$. The parabola is below the x-axis, so the original inequality is true, between those two numbers.
answered Feb 1 at 15:40
user247327user247327
11.6k1516
$endgroup$
The first thing I would do, and what Harnack suggested, is solve the given inequality. We are given that $x^2< 1- x$. That is the same as $x^2+ x- 1= 0$. Completing the square, $x^2+ x+ 1/4- 1/4- 1= (x- 1/2)^2- 5/4= 0$. $(x- 1/2)^2= 5/4$ so $x- 1/2= pmfrac{sqrt{5}}{2}$ so that $x= frac{1}{2}pmfrac{sqrt{5}}{2}$. $y= x^2+ x- 1$ is a parabola opening upward, crossing the x-axis at $x= frac{1- sqrt{5}}{2}$ and $x= frac{1+ sqrt{5}}{2}$. The parabola is below the x-axis, so the original inequality is true, between those two numbers.
answered Feb 1 at 15:40
user247327user247327
11.6k1516
answered Feb 1 at 15:40
user247327user247327
11.6k1516
answered Feb 1 at 15:40
user247327user247327
11.6k1516
answered Feb 1 at 15:40
user247327user247327
11.6k1516
11.6k1516
add a comment |
add a comment |
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