Prove or disprove that if $f$












0












$begingroup$


such that $f:[0,1] to [0,1]$ and $f$ is continuous, then exists $x_0 in [0,1]$ that satisfies $f(x_0) = x_0(4-x_0)$.



The intuition is clear that should be such an $x_0$. Since $g(x) = x(4-x)$ is strictly monotonic and continuous on $[0, 2- sqrt{3}]$, hence $f$ and $g$ have to intersect at least at one point in $[0, 2- sqrt{3}]$. But how to formalize this assertion?



Thank you.










share|cite|improve this question









$endgroup$












  • $begingroup$
    Let $h=g-f$. Then $h(0)le0$ and $h(0)>0$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:32
















0












$begingroup$


such that $f:[0,1] to [0,1]$ and $f$ is continuous, then exists $x_0 in [0,1]$ that satisfies $f(x_0) = x_0(4-x_0)$.



The intuition is clear that should be such an $x_0$. Since $g(x) = x(4-x)$ is strictly monotonic and continuous on $[0, 2- sqrt{3}]$, hence $f$ and $g$ have to intersect at least at one point in $[0, 2- sqrt{3}]$. But how to formalize this assertion?



Thank you.










share|cite|improve this question









$endgroup$












  • $begingroup$
    Let $h=g-f$. Then $h(0)le0$ and $h(0)>0$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:32














0












0








0





$begingroup$


such that $f:[0,1] to [0,1]$ and $f$ is continuous, then exists $x_0 in [0,1]$ that satisfies $f(x_0) = x_0(4-x_0)$.



The intuition is clear that should be such an $x_0$. Since $g(x) = x(4-x)$ is strictly monotonic and continuous on $[0, 2- sqrt{3}]$, hence $f$ and $g$ have to intersect at least at one point in $[0, 2- sqrt{3}]$. But how to formalize this assertion?



Thank you.










share|cite|improve this question









$endgroup$




such that $f:[0,1] to [0,1]$ and $f$ is continuous, then exists $x_0 in [0,1]$ that satisfies $f(x_0) = x_0(4-x_0)$.



The intuition is clear that should be such an $x_0$. Since $g(x) = x(4-x)$ is strictly monotonic and continuous on $[0, 2- sqrt{3}]$, hence $f$ and $g$ have to intersect at least at one point in $[0, 2- sqrt{3}]$. But how to formalize this assertion?



Thank you.







real-analysis calculus functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 20:26









J.DoeJ.Doe

8519




8519












  • $begingroup$
    Let $h=g-f$. Then $h(0)le0$ and $h(0)>0$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:32


















  • $begingroup$
    Let $h=g-f$. Then $h(0)le0$ and $h(0)>0$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:32
















$begingroup$
Let $h=g-f$. Then $h(0)le0$ and $h(0)>0$.
$endgroup$
– Mark Viola
Dec 29 '18 at 20:32




$begingroup$
Let $h=g-f$. Then $h(0)le0$ and $h(0)>0$.
$endgroup$
– Mark Viola
Dec 29 '18 at 20:32










1 Answer
1






active

oldest

votes


















2












$begingroup$

Consider $h(x) = f(x) - x(4-x)$. Then





  • $h(0) = f(0) ge 0$,


  • $h(1) = f(1) -3 le 1 - 3 < 0$.


What can you conclude?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $h(0) = 0$ then $0$ is such a point and if $h(0) > 0$ then by the intermediate point theorem exists $x_0$ in $(0,1)$ such that $h(x_0) = 0$. Is this right?
    $endgroup$
    – J.Doe
    Dec 29 '18 at 20:39










  • $begingroup$
    @J.Doe: Yes, exactly.
    $endgroup$
    – Martin R
    Dec 29 '18 at 20:39











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Consider $h(x) = f(x) - x(4-x)$. Then





  • $h(0) = f(0) ge 0$,


  • $h(1) = f(1) -3 le 1 - 3 < 0$.


What can you conclude?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $h(0) = 0$ then $0$ is such a point and if $h(0) > 0$ then by the intermediate point theorem exists $x_0$ in $(0,1)$ such that $h(x_0) = 0$. Is this right?
    $endgroup$
    – J.Doe
    Dec 29 '18 at 20:39










  • $begingroup$
    @J.Doe: Yes, exactly.
    $endgroup$
    – Martin R
    Dec 29 '18 at 20:39
















2












$begingroup$

Consider $h(x) = f(x) - x(4-x)$. Then





  • $h(0) = f(0) ge 0$,


  • $h(1) = f(1) -3 le 1 - 3 < 0$.


What can you conclude?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $h(0) = 0$ then $0$ is such a point and if $h(0) > 0$ then by the intermediate point theorem exists $x_0$ in $(0,1)$ such that $h(x_0) = 0$. Is this right?
    $endgroup$
    – J.Doe
    Dec 29 '18 at 20:39










  • $begingroup$
    @J.Doe: Yes, exactly.
    $endgroup$
    – Martin R
    Dec 29 '18 at 20:39














2












2








2





$begingroup$

Consider $h(x) = f(x) - x(4-x)$. Then





  • $h(0) = f(0) ge 0$,


  • $h(1) = f(1) -3 le 1 - 3 < 0$.


What can you conclude?






share|cite|improve this answer









$endgroup$



Consider $h(x) = f(x) - x(4-x)$. Then





  • $h(0) = f(0) ge 0$,


  • $h(1) = f(1) -3 le 1 - 3 < 0$.


What can you conclude?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 20:30









Martin RMartin R

27.5k33255




27.5k33255












  • $begingroup$
    If $h(0) = 0$ then $0$ is such a point and if $h(0) > 0$ then by the intermediate point theorem exists $x_0$ in $(0,1)$ such that $h(x_0) = 0$. Is this right?
    $endgroup$
    – J.Doe
    Dec 29 '18 at 20:39










  • $begingroup$
    @J.Doe: Yes, exactly.
    $endgroup$
    – Martin R
    Dec 29 '18 at 20:39


















  • $begingroup$
    If $h(0) = 0$ then $0$ is such a point and if $h(0) > 0$ then by the intermediate point theorem exists $x_0$ in $(0,1)$ such that $h(x_0) = 0$. Is this right?
    $endgroup$
    – J.Doe
    Dec 29 '18 at 20:39










  • $begingroup$
    @J.Doe: Yes, exactly.
    $endgroup$
    – Martin R
    Dec 29 '18 at 20:39
















$begingroup$
If $h(0) = 0$ then $0$ is such a point and if $h(0) > 0$ then by the intermediate point theorem exists $x_0$ in $(0,1)$ such that $h(x_0) = 0$. Is this right?
$endgroup$
– J.Doe
Dec 29 '18 at 20:39




$begingroup$
If $h(0) = 0$ then $0$ is such a point and if $h(0) > 0$ then by the intermediate point theorem exists $x_0$ in $(0,1)$ such that $h(x_0) = 0$. Is this right?
$endgroup$
– J.Doe
Dec 29 '18 at 20:39












$begingroup$
@J.Doe: Yes, exactly.
$endgroup$
– Martin R
Dec 29 '18 at 20:39




$begingroup$
@J.Doe: Yes, exactly.
$endgroup$
– Martin R
Dec 29 '18 at 20:39


















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