Prove or disprove that if $f$
$begingroup$
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.
real-analysis calculus functions
asked Dec 29 '18 at 20:26
J.DoeJ.Doe
8519
$endgroup$
add a comment |
$begingroup$
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.
real-analysis calculus functions
asked Dec 29 '18 at 20:26
J.DoeJ.Doe
8519
$endgroup$
$begingroup$
Let $h=g-f$. Then $h(0)le0$ and $h(0)>0$.
$endgroup$
– Mark Viola
Dec 29 '18 at 20:32
add a comment |
$begingroup$
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.
real-analysis calculus functions
asked Dec 29 '18 at 20:26
J.DoeJ.Doe
8519
$endgroup$
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
real-analysis calculus functions
asked Dec 29 '18 at 20:26
J.DoeJ.Doe
8519
asked Dec 29 '18 at 20:26
J.DoeJ.Doe
8519
asked Dec 29 '18 at 20:26
J.DoeJ.Doe
8519
asked Dec 29 '18 at 20:26
J.DoeJ.Doe
8519
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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?
answered Dec 29 '18 at 20:30
Martin RMartin R
27.5k33255
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
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active
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active
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$begingroup$
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?
answered Dec 29 '18 at 20:30
Martin RMartin R
27.5k33255
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
answered Dec 29 '18 at 20:30
Martin RMartin R
27.5k33255
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
answered Dec 29 '18 at 20:30
Martin RMartin R
27.5k33255
$endgroup$
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?
answered Dec 29 '18 at 20:30
Martin RMartin R
27.5k33255
answered Dec 29 '18 at 20:30
Martin RMartin R
27.5k33255
answered Dec 29 '18 at 20:30
Martin RMartin R
27.5k33255
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Let $h=g-f$. Then $h(0)le0$ and $h(0)>0$.
$endgroup$
– Mark Viola
Dec 29 '18 at 20:32