For any real number $c$, the polynomial $x^3 + x + c$ has exactly one real root












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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







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    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Nov 23 '12 at 13:01









      gumtigumti

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          5 Answers
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          $begingroup$

          You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            strictly increasing even.
            $endgroup$
            – lhf
            Nov 23 '12 at 13:14



















          3












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              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






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                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



















              0












              $begingroup$

              $ 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} $.



              graph of y=x(x^2 + 1)



              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 $y_0$.






              share|cite|improve this answer









              $endgroup$













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                5 Answers
                5






                active

                oldest

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                5 Answers
                5






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                8












                $begingroup$

                You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  strictly increasing even.
                  $endgroup$
                  – lhf
                  Nov 23 '12 at 13:14
















                8












                $begingroup$

                You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  strictly increasing even.
                  $endgroup$
                  – lhf
                  Nov 23 '12 at 13:14














                8












                8








                8





                $begingroup$

                You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.






                share|cite|improve this answer









                $endgroup$



                You can show (for example using the derivative) that the function $x mapsto x^3 + x + c$ is increasing for any real $c$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 '12 at 13:06









                xenxen

                3,0661624




                3,0661624












                • $begingroup$
                  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




                $begingroup$
                strictly increasing even.
                $endgroup$
                – lhf
                Nov 23 '12 at 13:14











                3












                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$


















                  3












                  $begingroup$

                  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.






                  share|cite|improve this answer











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    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.






                    share|cite|improve this answer











                    $endgroup$



                    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.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 7 '18 at 22:26

























                    answered Nov 23 '12 at 13:12









                    P..P..

                    13.4k22348




                    13.4k22348























                        2












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            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.






                            share|cite|improve this answer









                            $endgroup$



                            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.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 23 '12 at 15:06









                            LubinLubin

                            44.2k44585




                            44.2k44585























                                0












                                $begingroup$

                                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






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  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
















                                0












                                $begingroup$

                                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






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  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














                                0












                                0








                                0





                                $begingroup$

                                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






                                share|cite|improve this answer











                                $endgroup$



                                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







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Nov 23 '12 at 13:22

























                                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


















                                • $begingroup$
                                  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











                                0












                                $begingroup$

                                $ 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} $.



                                graph of y=x(x^2 + 1)



                                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 $y_0$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  $ 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} $.



                                  graph of y=x(x^2 + 1)



                                  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 $y_0$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    $ 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} $.



                                    graph of y=x(x^2 + 1)



                                    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 $y_0$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    $ 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} $.



                                    graph of y=x(x^2 + 1)



                                    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 $y_0$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 23 '12 at 13:23









                                    hkBattousaihkBattousai

                                    2,43042950




                                    2,43042950






























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