Proof that equation does not have real roots












0












$begingroup$



For polynom
$f(x)=ax^2+bx+c$ equation $f(x)=x$ has no real solutions.
Prove that equation $ f(f(x))=x$ also does not have does not have real solutions




Can someone explain solution to me? Why?




If equation $f(x)=x$ has no real solutions than it is either $f(x)>0$ and $a>0$ or it is $f(x)<0$ and $a<0$.
In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=0$







quadratics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your assumption about $f$ is wrong: consider $f(x)=(x+2)^2-1$.
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 19:59










  • $begingroup$
    The quoted solution is incorrect.
    $endgroup$
    – Lance
    Dec 23 '18 at 20:00










  • $begingroup$
    @John_Wick Your conclusion is wrong. Consider $f(x)=(x+2)^2-1$, then $f(f(-0.9)),f(-0.9)$.
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 20:05










  • $begingroup$
    Um. If $x = c$ is a root to $f(f(x)) $ then $x = f(c)$ is a root to $f(x)$. But $f(x)$ has no roots. That's all.
    $endgroup$
    – fleablood
    Dec 23 '18 at 20:37
















0












$begingroup$



For polynom
$f(x)=ax^2+bx+c$ equation $f(x)=x$ has no real solutions.
Prove that equation $ f(f(x))=x$ also does not have does not have real solutions




Can someone explain solution to me? Why?




If equation $f(x)=x$ has no real solutions than it is either $f(x)>0$ and $a>0$ or it is $f(x)<0$ and $a<0$.
In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=0$







quadratics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your assumption about $f$ is wrong: consider $f(x)=(x+2)^2-1$.
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 19:59










  • $begingroup$
    The quoted solution is incorrect.
    $endgroup$
    – Lance
    Dec 23 '18 at 20:00










  • $begingroup$
    @John_Wick Your conclusion is wrong. Consider $f(x)=(x+2)^2-1$, then $f(f(-0.9)),f(-0.9)$.
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 20:05










  • $begingroup$
    Um. If $x = c$ is a root to $f(f(x)) $ then $x = f(c)$ is a root to $f(x)$. But $f(x)$ has no roots. That's all.
    $endgroup$
    – fleablood
    Dec 23 '18 at 20:37














0












0








0





$begingroup$



For polynom
$f(x)=ax^2+bx+c$ equation $f(x)=x$ has no real solutions.
Prove that equation $ f(f(x))=x$ also does not have does not have real solutions




Can someone explain solution to me? Why?




If equation $f(x)=x$ has no real solutions than it is either $f(x)>0$ and $a>0$ or it is $f(x)<0$ and $a<0$.
In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=0$







quadratics







share|cite|improve this question









$endgroup$





For polynom
$f(x)=ax^2+bx+c$ equation $f(x)=x$ has no real solutions.
Prove that equation $ f(f(x))=x$ also does not have does not have real solutions




Can someone explain solution to me? Why?




If equation $f(x)=x$ has no real solutions than it is either $f(x)>0$ and $a>0$ or it is $f(x)<0$ and $a<0$.
In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=0$







quadratics




quadratics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 23 '18 at 19:47









Ivan LjiljaIvan Ljilja

205




205












  • $begingroup$
    Your assumption about $f$ is wrong: consider $f(x)=(x+2)^2-1$.
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 19:59










  • $begingroup$
    The quoted solution is incorrect.
    $endgroup$
    – Lance
    Dec 23 '18 at 20:00










  • $begingroup$
    @John_Wick Your conclusion is wrong. Consider $f(x)=(x+2)^2-1$, then $f(f(-0.9)),f(-0.9)$.
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 20:05










  • $begingroup$
    Um. If $x = c$ is a root to $f(f(x)) $ then $x = f(c)$ is a root to $f(x)$. But $f(x)$ has no roots. That's all.
    $endgroup$
    – fleablood
    Dec 23 '18 at 20:37


















  • $begingroup$
    Your assumption about $f$ is wrong: consider $f(x)=(x+2)^2-1$.
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 19:59










  • $begingroup$
    The quoted solution is incorrect.
    $endgroup$
    – Lance
    Dec 23 '18 at 20:00










  • $begingroup$
    @John_Wick Your conclusion is wrong. Consider $f(x)=(x+2)^2-1$, then $f(f(-0.9)),f(-0.9)$.
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 20:05










  • $begingroup$
    Um. If $x = c$ is a root to $f(f(x)) $ then $x = f(c)$ is a root to $f(x)$. But $f(x)$ has no roots. That's all.
    $endgroup$
    – fleablood
    Dec 23 '18 at 20:37
















$begingroup$
Your assumption about $f$ is wrong: consider $f(x)=(x+2)^2-1$.
$endgroup$
– Michael Hoppe
Dec 23 '18 at 19:59




$begingroup$
Your assumption about $f$ is wrong: consider $f(x)=(x+2)^2-1$.
$endgroup$
– Michael Hoppe
Dec 23 '18 at 19:59












$begingroup$
The quoted solution is incorrect.
$endgroup$
– Lance
Dec 23 '18 at 20:00




$begingroup$
The quoted solution is incorrect.
$endgroup$
– Lance
Dec 23 '18 at 20:00












$begingroup$
@John_Wick Your conclusion is wrong. Consider $f(x)=(x+2)^2-1$, then $f(f(-0.9)),f(-0.9)$.
$endgroup$
– Michael Hoppe
Dec 23 '18 at 20:05




$begingroup$
@John_Wick Your conclusion is wrong. Consider $f(x)=(x+2)^2-1$, then $f(f(-0.9)),f(-0.9)$.
$endgroup$
– Michael Hoppe
Dec 23 '18 at 20:05












$begingroup$
Um. If $x = c$ is a root to $f(f(x)) $ then $x = f(c)$ is a root to $f(x)$. But $f(x)$ has no roots. That's all.
$endgroup$
– fleablood
Dec 23 '18 at 20:37




$begingroup$
Um. If $x = c$ is a root to $f(f(x)) $ then $x = f(c)$ is a root to $f(x)$. But $f(x)$ has no roots. That's all.
$endgroup$
– fleablood
Dec 23 '18 at 20:37










3 Answers
3






active

oldest

votes


















2












$begingroup$

The quoted solution should read




If equation $f(x)=x$ has no real solutions than it is either $f(x)>color{red}x$ and $a>0$ or it is $f(x)<color{red}x$ and $a<0$.
In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=color{red}x$.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Doesn't that hold for any continuous function $f$ defined on an interval?
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 20:33



















3












$begingroup$

If $a>0$ then $f(x)=ax^2+bx+c>x$ for sufficiently large $x$. And as $f(x)=x$ has no real root then $f(x)>x$ for all $x$ (otherwise there will be a real root). Hence $f(f(x))>f(x)>x$ for all $x$. So, $f(f(x))=x$ has no real root. Similarly for $a<0.$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Let



    $$g(x)=f(x)-x=ax^2+(b-1)x+c$$



    no real roots implies that



    $$(b-1)^2-4ac<0$$
    and $$(forall xin Bbb R);; g(x) text{ has a constant sign} :$$
    $(+ text{ if } a>0 text{ and } - text{ if } a<0)$.



    thus
    $$f(f(x))-x=$$
    $$f(f(x))-f(x)+f(x)-x=$$
    $$g(f(x))+g(x)$$
    will have a constant sign and no root.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice one, definitely +1.
      $endgroup$
      – Michael Hoppe
      Dec 23 '18 at 20:11












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The quoted solution should read




    If equation $f(x)=x$ has no real solutions than it is either $f(x)>color{red}x$ and $a>0$ or it is $f(x)<color{red}x$ and $a<0$.
    In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=color{red}x$.







    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Doesn't that hold for any continuous function $f$ defined on an interval?
      $endgroup$
      – Michael Hoppe
      Dec 23 '18 at 20:33
















    2












    $begingroup$

    The quoted solution should read




    If equation $f(x)=x$ has no real solutions than it is either $f(x)>color{red}x$ and $a>0$ or it is $f(x)<color{red}x$ and $a<0$.
    In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=color{red}x$.







    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Doesn't that hold for any continuous function $f$ defined on an interval?
      $endgroup$
      – Michael Hoppe
      Dec 23 '18 at 20:33














    2












    2








    2





    $begingroup$

    The quoted solution should read




    If equation $f(x)=x$ has no real solutions than it is either $f(x)>color{red}x$ and $a>0$ or it is $f(x)<color{red}x$ and $a<0$.
    In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=color{red}x$.







    share|cite|improve this answer









    $endgroup$



    The quoted solution should read




    If equation $f(x)=x$ has no real solutions than it is either $f(x)>color{red}x$ and $a>0$ or it is $f(x)<color{red}x$ and $a<0$.
    In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=color{red}x$.








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 23 '18 at 20:17









    NyfikenNyfiken

    805612




    805612












    • $begingroup$
      Doesn't that hold for any continuous function $f$ defined on an interval?
      $endgroup$
      – Michael Hoppe
      Dec 23 '18 at 20:33


















    • $begingroup$
      Doesn't that hold for any continuous function $f$ defined on an interval?
      $endgroup$
      – Michael Hoppe
      Dec 23 '18 at 20:33
















    $begingroup$
    Doesn't that hold for any continuous function $f$ defined on an interval?
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 20:33




    $begingroup$
    Doesn't that hold for any continuous function $f$ defined on an interval?
    $endgroup$
    – Michael Hoppe
    Dec 23 '18 at 20:33











    3












    $begingroup$

    If $a>0$ then $f(x)=ax^2+bx+c>x$ for sufficiently large $x$. And as $f(x)=x$ has no real root then $f(x)>x$ for all $x$ (otherwise there will be a real root). Hence $f(f(x))>f(x)>x$ for all $x$. So, $f(f(x))=x$ has no real root. Similarly for $a<0.$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      If $a>0$ then $f(x)=ax^2+bx+c>x$ for sufficiently large $x$. And as $f(x)=x$ has no real root then $f(x)>x$ for all $x$ (otherwise there will be a real root). Hence $f(f(x))>f(x)>x$ for all $x$. So, $f(f(x))=x$ has no real root. Similarly for $a<0.$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        If $a>0$ then $f(x)=ax^2+bx+c>x$ for sufficiently large $x$. And as $f(x)=x$ has no real root then $f(x)>x$ for all $x$ (otherwise there will be a real root). Hence $f(f(x))>f(x)>x$ for all $x$. So, $f(f(x))=x$ has no real root. Similarly for $a<0.$






        share|cite|improve this answer









        $endgroup$



        If $a>0$ then $f(x)=ax^2+bx+c>x$ for sufficiently large $x$. And as $f(x)=x$ has no real root then $f(x)>x$ for all $x$ (otherwise there will be a real root). Hence $f(f(x))>f(x)>x$ for all $x$. So, $f(f(x))=x$ has no real root. Similarly for $a<0.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 20:03









        John_WickJohn_Wick

        1,626111




        1,626111























            2












            $begingroup$

            Let



            $$g(x)=f(x)-x=ax^2+(b-1)x+c$$



            no real roots implies that



            $$(b-1)^2-4ac<0$$
            and $$(forall xin Bbb R);; g(x) text{ has a constant sign} :$$
            $(+ text{ if } a>0 text{ and } - text{ if } a<0)$.



            thus
            $$f(f(x))-x=$$
            $$f(f(x))-f(x)+f(x)-x=$$
            $$g(f(x))+g(x)$$
            will have a constant sign and no root.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice one, definitely +1.
              $endgroup$
              – Michael Hoppe
              Dec 23 '18 at 20:11
















            2












            $begingroup$

            Let



            $$g(x)=f(x)-x=ax^2+(b-1)x+c$$



            no real roots implies that



            $$(b-1)^2-4ac<0$$
            and $$(forall xin Bbb R);; g(x) text{ has a constant sign} :$$
            $(+ text{ if } a>0 text{ and } - text{ if } a<0)$.



            thus
            $$f(f(x))-x=$$
            $$f(f(x))-f(x)+f(x)-x=$$
            $$g(f(x))+g(x)$$
            will have a constant sign and no root.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice one, definitely +1.
              $endgroup$
              – Michael Hoppe
              Dec 23 '18 at 20:11














            2












            2








            2





            $begingroup$

            Let



            $$g(x)=f(x)-x=ax^2+(b-1)x+c$$



            no real roots implies that



            $$(b-1)^2-4ac<0$$
            and $$(forall xin Bbb R);; g(x) text{ has a constant sign} :$$
            $(+ text{ if } a>0 text{ and } - text{ if } a<0)$.



            thus
            $$f(f(x))-x=$$
            $$f(f(x))-f(x)+f(x)-x=$$
            $$g(f(x))+g(x)$$
            will have a constant sign and no root.






            share|cite|improve this answer











            $endgroup$



            Let



            $$g(x)=f(x)-x=ax^2+(b-1)x+c$$



            no real roots implies that



            $$(b-1)^2-4ac<0$$
            and $$(forall xin Bbb R);; g(x) text{ has a constant sign} :$$
            $(+ text{ if } a>0 text{ and } - text{ if } a<0)$.



            thus
            $$f(f(x))-x=$$
            $$f(f(x))-f(x)+f(x)-x=$$
            $$g(f(x))+g(x)$$
            will have a constant sign and no root.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 23 '18 at 20:14

























            answered Dec 23 '18 at 20:07









            hamam_Abdallahhamam_Abdallah

            38.1k21634




            38.1k21634












            • $begingroup$
              Nice one, definitely +1.
              $endgroup$
              – Michael Hoppe
              Dec 23 '18 at 20:11


















            • $begingroup$
              Nice one, definitely +1.
              $endgroup$
              – Michael Hoppe
              Dec 23 '18 at 20:11
















            $begingroup$
            Nice one, definitely +1.
            $endgroup$
            – Michael Hoppe
            Dec 23 '18 at 20:11




            $begingroup$
            Nice one, definitely +1.
            $endgroup$
            – Michael Hoppe
            Dec 23 '18 at 20:11


















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