Upper Bound of $x(1-x)$ for $xin [0,1]$?
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Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?
real-analysis
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Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?
real-analysis
2
The maximum value is $frac 1 4$
– Kavi Rama Murthy
Nov 14 at 8:05
2
For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
– Yadati Kiran
Nov 14 at 8:05
3
By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
– Robert Z
Nov 14 at 8:06
You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 15 at 13:54
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?
real-analysis
Is there a good upper bound for $x(1-x)$ if $xin [0,1]$?
real-analysis
real-analysis
edited Nov 14 at 8:05
asked Nov 14 at 8:03
Mog
2007
2007
2
The maximum value is $frac 1 4$
– Kavi Rama Murthy
Nov 14 at 8:05
2
For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
– Yadati Kiran
Nov 14 at 8:05
3
By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
– Robert Z
Nov 14 at 8:06
You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 15 at 13:54
add a comment |
2
The maximum value is $frac 1 4$
– Kavi Rama Murthy
Nov 14 at 8:05
2
For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
– Yadati Kiran
Nov 14 at 8:05
3
By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
– Robert Z
Nov 14 at 8:06
You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 15 at 13:54
2
2
The maximum value is $frac 1 4$
– Kavi Rama Murthy
Nov 14 at 8:05
The maximum value is $frac 1 4$
– Kavi Rama Murthy
Nov 14 at 8:05
2
2
For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
– Yadati Kiran
Nov 14 at 8:05
For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
– Yadati Kiran
Nov 14 at 8:05
3
3
By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
– Robert Z
Nov 14 at 8:06
By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
– Robert Z
Nov 14 at 8:06
You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 15 at 13:54
You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 15 at 13:54
add a comment |
4 Answers
4
active
oldest
votes
up vote
5
down vote
accepted
Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.
You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as
$$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$
add a comment |
up vote
5
down vote
An upper bound is given by $frac{1}{4}$.
Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.
New contributor
add a comment |
up vote
3
down vote
We have $x(1-x) ge 0$ and by AM-GM
$$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$
with equality for
- $x=1-x implies x=frac12 $
therefore as $x in[0,1]$
$$0le x(1-x)le frac14$$
add a comment |
up vote
2
down vote
Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.
You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as
$$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$
add a comment |
up vote
5
down vote
accepted
Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.
You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as
$$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.
You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as
$$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$
Depends on what you mean by "good". A lot of times, $1$ is a perfectly useful upper bound for the expression. If you need a stricter upper bound, the maximum of the expression (and therefore the smallest upper bound possible) is $frac14$.
You can easily see this is the upper bound by calculating the derivative and setting it to $0$, or by rewriting the expression (completing the square) as
$$x(1-x) = x-x^2 = - left(x-frac12right)^2 + frac14$$
answered Nov 14 at 8:06
5xum
88.3k392160
88.3k392160
add a comment |
add a comment |
up vote
5
down vote
An upper bound is given by $frac{1}{4}$.
Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.
New contributor
add a comment |
up vote
5
down vote
An upper bound is given by $frac{1}{4}$.
Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.
New contributor
add a comment |
up vote
5
down vote
up vote
5
down vote
An upper bound is given by $frac{1}{4}$.
Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.
New contributor
An upper bound is given by $frac{1}{4}$.
Define $f(x) = x(1 - x) = x - x^{2}$. Then, $f'(x) = 1 - 2x$, and we have a critical point at $x = frac{1}{2}$. Plugging this value back into our original equation yields $frac{1}{4}$.
New contributor
New contributor
answered Nov 14 at 8:06
Ekesh
3414
3414
New contributor
New contributor
add a comment |
add a comment |
up vote
3
down vote
We have $x(1-x) ge 0$ and by AM-GM
$$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$
with equality for
- $x=1-x implies x=frac12 $
therefore as $x in[0,1]$
$$0le x(1-x)le frac14$$
add a comment |
up vote
3
down vote
We have $x(1-x) ge 0$ and by AM-GM
$$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$
with equality for
- $x=1-x implies x=frac12 $
therefore as $x in[0,1]$
$$0le x(1-x)le frac14$$
add a comment |
up vote
3
down vote
up vote
3
down vote
We have $x(1-x) ge 0$ and by AM-GM
$$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$
with equality for
- $x=1-x implies x=frac12 $
therefore as $x in[0,1]$
$$0le x(1-x)le frac14$$
We have $x(1-x) ge 0$ and by AM-GM
$$x(1-x)le left(frac{x+1-x}{2}right)^2=frac14$$
with equality for
- $x=1-x implies x=frac12 $
therefore as $x in[0,1]$
$$0le x(1-x)le frac14$$
answered Nov 14 at 8:18
gimusi
85.5k74294
85.5k74294
add a comment |
add a comment |
up vote
2
down vote
Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.
add a comment |
up vote
2
down vote
Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.
add a comment |
up vote
2
down vote
up vote
2
down vote
Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.
Let $f(x)=x(1-x)$. $f'(x)=(1-x)+x=0$, we get $x=dfrac{1}{2}$ as the critical point. To be sure of the nature of the function (whether concave up or down) use second derivative test to confirm $f''(x)<0$. This says that the function $f$ at $x=dfrac{1}{2}$ attains a local maximum thereby giving us the upper bound $fBig(dfrac{1}{2}Big)$.
answered Nov 14 at 8:12
Yadati Kiran
358111
358111
add a comment |
add a comment |
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2
The maximum value is $frac 1 4$
– Kavi Rama Murthy
Nov 14 at 8:05
2
For this calculate the derivative of $x(1-x)$ and check the fucntion value at critical points.
– Yadati Kiran
Nov 14 at 8:05
3
By AM-GM inequality $x(1-x)leq [(x+(1-x))/2]^2=1/4$.
– Robert Z
Nov 14 at 8:06
You recieved 4 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 15 at 13:54