For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root
$begingroup$
For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root.
how can i show that above statement is true or false.can anyone help me please
real-analysis
asked Nov 23 '12 at 13:01
gumtigumti
92
$endgroup$
add a comment |
$begingroup$
For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root.
how can i show that above statement is true or false.can anyone help me please
real-analysis
asked Nov 23 '12 at 13:01
gumtigumti
92
$endgroup$
add a comment |
$begingroup$
For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root.
how can i show that above statement is true or false.can anyone help me please
real-analysis
asked Nov 23 '12 at 13:01
gumtigumti
92
$endgroup$
For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root.
how can i show that above statement is true or false.can anyone help me please
real-analysis
real-analysis
asked Nov 23 '12 at 13:01
gumtigumti
92
asked Nov 23 '12 at 13:01
gumtigumti
92
asked Nov 23 '12 at 13:01
gumtigumti
92
asked Nov 23 '12 at 13:01
gumtigumti
92
asked Nov 23 '12 at 13:01
gumtigumti
92
92
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.
answered Nov 23 '12 at 13:06
xenxen
3,0661624
$endgroup$
$begingroup$
strictly increasing even.
$endgroup$
– lhf
Nov 23 '12 at 13:14
add a comment |
$begingroup$
There are several ways to do it:
- Show that $x mapsto x^3+x+c$ is increasing
- Use Rolle's theorem.
- Use Descartes' rule of signs: If $c>0$ it has a negative root while if $cleq0$ it has a nonnegative root.
answered Nov 23 '12 at 13:12
P..P..
13.4k22348
$endgroup$
add a comment |
$begingroup$
Here’s yet another argument, not using calculus. As @xen remarks, a strictly increasing real-valued function will have at most one root. Letting $f$ be our function, we find that $f(x+varepsilon)-f(x)=varepsilonlbrack 3x^2+3xvarepsilon+varepsilon^2+1rbrack$, and the quantity in brackets is a sum of squares, namely $3x^2/4+(3x/2+varepsilon)^2+1$. So the computed difference has the same sign as $varepsilon$, the function is strictly increasing.
answered Nov 23 '12 at 15:06
LubinLubin
44.2k44585
$endgroup$
add a comment |
$begingroup$
See wikipedia for a algebric condition in wikipedia section "Roots of a cubic function".
Let
$$
(1) quad ax^3+bx^2+cx+d=0.
$$
Every cubic equation $(1)$ with real coefficients has at least one solution ''x'' among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminante $Delta$ of $(1)$,
$$
Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2.
$$
The following cases need to be considered:
If $Delta > 0$, then the equation has three distinct real roots.
If $Delta = 0$, then the equation has multiple root and all its roots are real.
If $Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.
For reference see the book
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,66343163
$endgroup$
$begingroup$
For a complet dedution of $Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped.
$endgroup$
– MathOverview
Nov 23 '12 at 13:34
add a comment |
$begingroup$
$ x^3 + x + c = 0 $
$ x^3 + x = y = -c $
$ x(x^2 + 1) = y $
$ (x^2 + 1) $ is always positive and greater than zero, and $ x $ takes both negative and positive values. So it is possible to sweep the entire real numbers, including $ d $, by giving $ x $ values from $ -infty $ to $ +infty $. In other words, $ y=x(x^2 + 1) $ is a smooth function from $ mathbb{R} $ to $ mathbb{R} $.
So, $ x^3 + x + c = 0 $ must have at least real root, since $ y = x(x^2 + 1) $ always has a solution $x_0$ for any given $y_0$.
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,43042950
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f243152%2ffor-any-real-number-c-the-polynomial-x3-x-c-has-exactly-one-real-root%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.
answered Nov 23 '12 at 13:06
xenxen
3,0661624
$endgroup$
$begingroup$
strictly increasing even.
$endgroup$
– lhf
Nov 23 '12 at 13:14
add a comment |
$begingroup$
You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.
answered Nov 23 '12 at 13:06
xenxen
3,0661624
$endgroup$
$begingroup$
strictly increasing even.
$endgroup$
– lhf
Nov 23 '12 at 13:14
add a comment |
$begingroup$
You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.
answered Nov 23 '12 at 13:06
xenxen
3,0661624
$endgroup$
You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.
answered Nov 23 '12 at 13:06
xenxen
3,0661624
answered Nov 23 '12 at 13:06
xenxen
3,0661624
answered Nov 23 '12 at 13:06
xenxen
3,0661624
answered Nov 23 '12 at 13:06
xenxen
3,0661624
3,0661624
$begingroup$
strictly increasing even.
$endgroup$
– lhf
Nov 23 '12 at 13:14
add a comment |
$begingroup$
strictly increasing even.
$endgroup$
– lhf
Nov 23 '12 at 13:14
$begingroup$
strictly increasing even.
$endgroup$
– lhf
Nov 23 '12 at 13:14
$begingroup$
strictly increasing even.
$endgroup$
– lhf
Nov 23 '12 at 13:14
add a comment |
$begingroup$
There are several ways to do it:
- Show that $x mapsto x^3+x+c$ is increasing
- Use Rolle's theorem.
- Use Descartes' rule of signs: If $c>0$ it has a negative root while if $cleq0$ it has a nonnegative root.
answered Nov 23 '12 at 13:12
P..P..
13.4k22348
$endgroup$
add a comment |
$begingroup$
There are several ways to do it:
- Show that $x mapsto x^3+x+c$ is increasing
- Use Rolle's theorem.
- Use Descartes' rule of signs: If $c>0$ it has a negative root while if $cleq0$ it has a nonnegative root.
answered Nov 23 '12 at 13:12
P..P..
13.4k22348
$endgroup$
add a comment |
$begingroup$
There are several ways to do it:
- Show that $x mapsto x^3+x+c$ is increasing
- Use Rolle's theorem.
- Use Descartes' rule of signs: If $c>0$ it has a negative root while if $cleq0$ it has a nonnegative root.
answered Nov 23 '12 at 13:12
P..P..
13.4k22348
$endgroup$
There are several ways to do it:
- Show that $x mapsto x^3+x+c$ is increasing
- Use Rolle's theorem.
- Use Descartes' rule of signs: If $c>0$ it has a negative root while if $cleq0$ it has a nonnegative root.
answered Nov 23 '12 at 13:12
P..P..
13.4k22348
edited Dec 7 '18 at 22:26
answered Nov 23 '12 at 13:12
P..P..
13.4k22348
answered Nov 23 '12 at 13:12
P..P..
13.4k22348
answered Nov 23 '12 at 13:12
P..P..
13.4k22348
13.4k22348
add a comment |
add a comment |
$begingroup$
Here’s yet another argument, not using calculus. As @xen remarks, a strictly increasing real-valued function will have at most one root. Letting $f$ be our function, we find that $f(x+varepsilon)-f(x)=varepsilonlbrack 3x^2+3xvarepsilon+varepsilon^2+1rbrack$, and the quantity in brackets is a sum of squares, namely $3x^2/4+(3x/2+varepsilon)^2+1$. So the computed difference has the same sign as $varepsilon$, the function is strictly increasing.
answered Nov 23 '12 at 15:06
LubinLubin
44.2k44585
$endgroup$
add a comment |
$begingroup$
Here’s yet another argument, not using calculus. As @xen remarks, a strictly increasing real-valued function will have at most one root. Letting $f$ be our function, we find that $f(x+varepsilon)-f(x)=varepsilonlbrack 3x^2+3xvarepsilon+varepsilon^2+1rbrack$, and the quantity in brackets is a sum of squares, namely $3x^2/4+(3x/2+varepsilon)^2+1$. So the computed difference has the same sign as $varepsilon$, the function is strictly increasing.
answered Nov 23 '12 at 15:06
LubinLubin
44.2k44585
$endgroup$
add a comment |
$begingroup$
Here’s yet another argument, not using calculus. As @xen remarks, a strictly increasing real-valued function will have at most one root. Letting $f$ be our function, we find that $f(x+varepsilon)-f(x)=varepsilonlbrack 3x^2+3xvarepsilon+varepsilon^2+1rbrack$, and the quantity in brackets is a sum of squares, namely $3x^2/4+(3x/2+varepsilon)^2+1$. So the computed difference has the same sign as $varepsilon$, the function is strictly increasing.
answered Nov 23 '12 at 15:06
LubinLubin
44.2k44585
$endgroup$
Here’s yet another argument, not using calculus. As @xen remarks, a strictly increasing real-valued function will have at most one root. Letting $f$ be our function, we find that $f(x+varepsilon)-f(x)=varepsilonlbrack 3x^2+3xvarepsilon+varepsilon^2+1rbrack$, and the quantity in brackets is a sum of squares, namely $3x^2/4+(3x/2+varepsilon)^2+1$. So the computed difference has the same sign as $varepsilon$, the function is strictly increasing.
answered Nov 23 '12 at 15:06
LubinLubin
44.2k44585
answered Nov 23 '12 at 15:06
LubinLubin
44.2k44585
answered Nov 23 '12 at 15:06
LubinLubin
44.2k44585
answered Nov 23 '12 at 15:06
LubinLubin
44.2k44585
44.2k44585
add a comment |
add a comment |
$begingroup$
See wikipedia for a algebric condition in wikipedia section "Roots of a cubic function".
Let
$$
(1) quad ax^3+bx^2+cx+d=0.
$$
Every cubic equation $(1)$ with real coefficients has at least one solution ''x'' among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminante $Delta$ of $(1)$,
$$
Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2.
$$
The following cases need to be considered:
If $Delta > 0$, then the equation has three distinct real roots.
If $Delta = 0$, then the equation has multiple root and all its roots are real.
If $Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.
For reference see the book
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,66343163
$endgroup$
$begingroup$
For a complet dedution of $Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped.
$endgroup$
– MathOverview
Nov 23 '12 at 13:34
add a comment |
$begingroup$
See wikipedia for a algebric condition in wikipedia section "Roots of a cubic function".
Let
$$
(1) quad ax^3+bx^2+cx+d=0.
$$
Every cubic equation $(1)$ with real coefficients has at least one solution ''x'' among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminante $Delta$ of $(1)$,
$$
Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2.
$$
The following cases need to be considered:
If $Delta > 0$, then the equation has three distinct real roots.
If $Delta = 0$, then the equation has multiple root and all its roots are real.
If $Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.
For reference see the book
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,66343163
$endgroup$
$begingroup$
For a complet dedution of $Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped.
$endgroup$
– MathOverview
Nov 23 '12 at 13:34
add a comment |
$begingroup$
See wikipedia for a algebric condition in wikipedia section "Roots of a cubic function".
Let
$$
(1) quad ax^3+bx^2+cx+d=0.
$$
Every cubic equation $(1)$ with real coefficients has at least one solution ''x'' among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminante $Delta$ of $(1)$,
$$
Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2.
$$
The following cases need to be considered:
If $Delta > 0$, then the equation has three distinct real roots.
If $Delta = 0$, then the equation has multiple root and all its roots are real.
If $Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.
For reference see the book
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,66343163
$endgroup$
See wikipedia for a algebric condition in wikipedia section "Roots of a cubic function".
Let
$$
(1) quad ax^3+bx^2+cx+d=0.
$$
Every cubic equation $(1)$ with real coefficients has at least one solution ''x'' among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminante $Delta$ of $(1)$,
$$
Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2.
$$
The following cases need to be considered:
If $Delta > 0$, then the equation has three distinct real roots.
If $Delta = 0$, then the equation has multiple root and all its roots are real.
If $Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.
For reference see the book
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,66343163
edited Nov 23 '12 at 13:22
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,66343163
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,66343163
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,66343163
8,66343163
$begingroup$
For a complet dedution of $Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped.
$endgroup$
– MathOverview
Nov 23 '12 at 13:34
add a comment |
$begingroup$
For a complet dedution of $Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped.
$endgroup$
– MathOverview
Nov 23 '12 at 13:34
$begingroup$
For a complet dedution of $Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped.
$endgroup$
– MathOverview
Nov 23 '12 at 13:34
$begingroup$
For a complet dedution of $Delta$ see mathworld.wolfram.com/PolynomialDiscriminant.html and mathworld.wolfram.com/CubicFormula.html. I hope helped.
$endgroup$
– MathOverview
Nov 23 '12 at 13:34
add a comment |
$begingroup$
$ x^3 + x + c = 0 $
$ x^3 + x = y = -c $
$ x(x^2 + 1) = y $
$ (x^2 + 1) $ is always positive and greater than zero, and $ x $ takes both negative and positive values. So it is possible to sweep the entire real numbers, including $ d $, by giving $ x $ values from $ -infty $ to $ +infty $. In other words, $ y=x(x^2 + 1) $ is a smooth function from $ mathbb{R} $ to $ mathbb{R} $.
So, $ x^3 + x + c = 0 $ must have at least real root, since $ y = x(x^2 + 1) $ always has a solution $x_0$ for any given $y_0$.
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,43042950
$endgroup$
add a comment |
$begingroup$
$ x^3 + x + c = 0 $
$ x^3 + x = y = -c $
$ x(x^2 + 1) = y $
$ (x^2 + 1) $ is always positive and greater than zero, and $ x $ takes both negative and positive values. So it is possible to sweep the entire real numbers, including $ d $, by giving $ x $ values from $ -infty $ to $ +infty $. In other words, $ y=x(x^2 + 1) $ is a smooth function from $ mathbb{R} $ to $ mathbb{R} $.
So, $ x^3 + x + c = 0 $ must have at least real root, since $ y = x(x^2 + 1) $ always has a solution $x_0$ for any given $y_0$.
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,43042950
$endgroup$
add a comment |
$begingroup$
$ x^3 + x + c = 0 $
$ x^3 + x = y = -c $
$ x(x^2 + 1) = y $
$ (x^2 + 1) $ is always positive and greater than zero, and $ x $ takes both negative and positive values. So it is possible to sweep the entire real numbers, including $ d $, by giving $ x $ values from $ -infty $ to $ +infty $. In other words, $ y=x(x^2 + 1) $ is a smooth function from $ mathbb{R} $ to $ mathbb{R} $.
So, $ x^3 + x + c = 0 $ must have at least real root, since $ y = x(x^2 + 1) $ always has a solution $x_0$ for any given $y_0$.
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,43042950
$endgroup$
$ x^3 + x + c = 0 $
$ x^3 + x = y = -c $
$ x(x^2 + 1) = y $
$ (x^2 + 1) $ is always positive and greater than zero, and $ x $ takes both negative and positive values. So it is possible to sweep the entire real numbers, including $ d $, by giving $ x $ values from $ -infty $ to $ +infty $. In other words, $ y=x(x^2 + 1) $ is a smooth function from $ mathbb{R} $ to $ mathbb{R} $.
So, $ x^3 + x + c = 0 $ must have at least real root, since $ y = x(x^2 + 1) $ always has a solution $x_0$ for any given $y_0$.
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,43042950
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,43042950
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,43042950
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,43042950
2,43042950
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f243152%2ffor-any-real-number-c-the-polynomial-x3-x-c-has-exactly-one-real-root%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
