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
3,0341624
$endgroup$
$begingroup$
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
44.1k44585
$endgroup$
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
8,54843163
$endgroup$
$begingroup$
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
2,42042950
$endgroup$
add a comment |
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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
3,0341624
$endgroup$
$begingroup$
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
3,0341624
$endgroup$
$begingroup$
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
3,0341624
$endgroup$
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
3,0341624
answered Nov 23 '12 at 13:06
xenxen
3,0341624
answered Nov 23 '12 at 13:06
xenxen
3,0341624
answered Nov 23 '12 at 13:06
xenxen
3,0341624
3,0341624
$begingroup$
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
44.1k44585
$endgroup$
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
44.1k44585
$endgroup$
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
44.1k44585
$endgroup$
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
44.1k44585
answered Nov 23 '12 at 15:06
LubinLubin
44.1k44585
answered Nov 23 '12 at 15:06
LubinLubin
44.1k44585
answered Nov 23 '12 at 15:06
LubinLubin
44.1k44585
44.1k44585
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
8,54843163
$endgroup$
$begingroup$
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
8,54843163
$endgroup$
$begingroup$
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
8,54843163
$endgroup$
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
8,54843163
edited Nov 23 '12 at 13:22
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,54843163
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,54843163
answered Nov 23 '12 at 13:12
MathOverviewMathOverview
8,54843163
8,54843163
$begingroup$
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
2,42042950
$endgroup$
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
2,42042950
$endgroup$
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
2,42042950
$endgroup$
$ 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
2,42042950
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,42042950
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,42042950
answered Nov 23 '12 at 13:23
hkBattousaihkBattousai
2,42042950
2,42042950
add a comment |
add a comment |
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