Mixing
Let $f$ be the function $ymapsto (1−y^3)/y$ on $(0,infty)$. Show that $f$ is decreasing (without using...
$begingroup$
My attempt:
Let $x=1$, $y=2$.
$2>1$, $f(1)>f(2)$ so true for $x=1$, $y=2$.
Then let $x=u$, $y=v$
If $v>u$, $v=x+p$
I've then subbed them in but got $p(3u^2 -1)<3u^2$ which doesn't tell us anything.
Am I allowed to say that as $u$&$p$ are positive $1/u > 1/(u+p)$, and $1-u^3> 1-(u+p)^3$ so it's true for $1$&$2$, $u$&$v$ so true for all others?
real-analysis
edited Feb 3 at 8:02
choco_addicted
8,08461947
asked Feb 2 at 17:42
HistoryFan12HistoryFan12
255
$endgroup$
add a comment |
$begingroup$
My attempt:
Let $x=1$, $y=2$.
$2>1$, $f(1)>f(2)$ so true for $x=1$, $y=2$.
Then let $x=u$, $y=v$
If $v>u$, $v=x+p$
I've then subbed them in but got $p(3u^2 -1)<3u^2$ which doesn't tell us anything.
Am I allowed to say that as $u$&$p$ are positive $1/u > 1/(u+p)$, and $1-u^3> 1-(u+p)^3$ so it's true for $1$&$2$, $u$&$v$ so true for all others?
real-analysis
edited Feb 3 at 8:02
choco_addicted
8,08461947
asked Feb 2 at 17:42
HistoryFan12HistoryFan12
255
$endgroup$
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Feb 2 at 17:50
$begingroup$
See here
$endgroup$
– J. W. Tanner
Feb 3 at 3:13
add a comment |
$begingroup$
My attempt:
Let $x=1$, $y=2$.
$2>1$, $f(1)>f(2)$ so true for $x=1$, $y=2$.
Then let $x=u$, $y=v$
If $v>u$, $v=x+p$
I've then subbed them in but got $p(3u^2 -1)<3u^2$ which doesn't tell us anything.
Am I allowed to say that as $u$&$p$ are positive $1/u > 1/(u+p)$, and $1-u^3> 1-(u+p)^3$ so it's true for $1$&$2$, $u$&$v$ so true for all others?
real-analysis
edited Feb 3 at 8:02
choco_addicted
8,08461947
asked Feb 2 at 17:42
HistoryFan12HistoryFan12
255
$endgroup$
My attempt:
Let $x=1$, $y=2$.
$2>1$, $f(1)>f(2)$ so true for $x=1$, $y=2$.
Then let $x=u$, $y=v$
If $v>u$, $v=x+p$
I've then subbed them in but got $p(3u^2 -1)<3u^2$ which doesn't tell us anything.
Am I allowed to say that as $u$&$p$ are positive $1/u > 1/(u+p)$, and $1-u^3> 1-(u+p)^3$ so it's true for $1$&$2$, $u$&$v$ so true for all others?
real-analysis
real-analysis
edited Feb 3 at 8:02
choco_addicted
8,08461947
asked Feb 2 at 17:42
HistoryFan12HistoryFan12
255
edited Feb 3 at 8:02
choco_addicted
8,08461947
asked Feb 2 at 17:42
HistoryFan12HistoryFan12
255
edited Feb 3 at 8:02
choco_addicted
8,08461947
edited Feb 3 at 8:02
choco_addicted
8,08461947
edited Feb 3 at 8:02
choco_addicted
8,08461947
8,08461947
asked Feb 2 at 17:42
HistoryFan12HistoryFan12
255
asked Feb 2 at 17:42
HistoryFan12HistoryFan12
255
asked Feb 2 at 17:42
HistoryFan12HistoryFan12
255
255
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Feb 2 at 17:50
$begingroup$
See here
$endgroup$
– J. W. Tanner
Feb 3 at 3:13
add a comment |
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Feb 2 at 17:50
$begingroup$
See here
$endgroup$
– J. W. Tanner
Feb 3 at 3:13
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Feb 2 at 17:50
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Feb 2 at 17:50
$begingroup$
See here
$endgroup$
– J. W. Tanner
Feb 3 at 3:13
$begingroup$
See here
$endgroup$
– J. W. Tanner
Feb 3 at 3:13
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
hint
For $y>0 $
$$f(y)=frac{1-y^3}{y}=frac{1}{y}-y^2$$
What about the sum of two decreasing functions ?
answered Feb 2 at 17:46
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
So sum of 2 decreasing functions is also decreasing, but surely I then need to prove that they are both decreasing too?
$endgroup$
– HistoryFan12
Feb 2 at 17:52
$begingroup$
@HistoryFan12 What about the inverse and the opposite of an Increasing function.
$endgroup$
– hamam_Abdallah
Feb 2 at 17:54
$begingroup$
So... you can kind of say these things without proof, just stating they are the inverse and opposite? @hamam_abdallah
$endgroup$
– HistoryFan12
Feb 2 at 18:01
add a comment |
$begingroup$
For
$dfrac{1-y^3}{y}
$,
on $0, infty$,
$y$ is increasing and positive
so $y^3$ is increasing
so $1-y^3$ is decreasing
so
$dfrac{1-y^3}{y}
$
is decreasing
since a decreasing function
divided by a positive increasing function
is decreasing.
answered Feb 2 at 17:49
marty cohenmarty cohen
75.6k549130
$endgroup$
$begingroup$
I get that this is a fact, but does what you've written count as proof?
$endgroup$
– HistoryFan12
Feb 2 at 17:51
$begingroup$
If you write it out formally (e.g., $f$ increasing means $x < y implies f(x) < f(y)$) it does, at least to me.
$endgroup$
– marty cohen
Feb 2 at 18:02
add a comment |
$begingroup$
Let $0<x<y$. Then $-xy(x+y)$ is negative, hence less than $1$. This gives $(y-x)[-xy(x+y)] <y-x$ or $xy((x^{2}-y^{2}) <y-x$. This gives $x-xy^{3} <y-x^{3}y$ which says $f(y) <f(x)$.
answered Feb 3 at 12:25
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hint
For $y>0 $
$$f(y)=frac{1-y^3}{y}=frac{1}{y}-y^2$$
What about the sum of two decreasing functions ?
answered Feb 2 at 17:46
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
So sum of 2 decreasing functions is also decreasing, but surely I then need to prove that they are both decreasing too?
$endgroup$
– HistoryFan12
Feb 2 at 17:52
$begingroup$
@HistoryFan12 What about the inverse and the opposite of an Increasing function.
$endgroup$
– hamam_Abdallah
Feb 2 at 17:54
$begingroup$
So... you can kind of say these things without proof, just stating they are the inverse and opposite? @hamam_abdallah
$endgroup$
– HistoryFan12
Feb 2 at 18:01
add a comment |
$begingroup$
hint
For $y>0 $
$$f(y)=frac{1-y^3}{y}=frac{1}{y}-y^2$$
What about the sum of two decreasing functions ?
answered Feb 2 at 17:46
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
So sum of 2 decreasing functions is also decreasing, but surely I then need to prove that they are both decreasing too?
$endgroup$
– HistoryFan12
Feb 2 at 17:52
$begingroup$
@HistoryFan12 What about the inverse and the opposite of an Increasing function.
$endgroup$
– hamam_Abdallah
Feb 2 at 17:54
$begingroup$
So... you can kind of say these things without proof, just stating they are the inverse and opposite? @hamam_abdallah
$endgroup$
– HistoryFan12
Feb 2 at 18:01
add a comment |
$begingroup$
hint
For $y>0 $
$$f(y)=frac{1-y^3}{y}=frac{1}{y}-y^2$$
What about the sum of two decreasing functions ?
answered Feb 2 at 17:46
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
hint
For $y>0 $
$$f(y)=frac{1-y^3}{y}=frac{1}{y}-y^2$$
What about the sum of two decreasing functions ?
answered Feb 2 at 17:46
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Feb 2 at 17:46
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Feb 2 at 17:46
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Feb 2 at 17:46
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
$begingroup$
So sum of 2 decreasing functions is also decreasing, but surely I then need to prove that they are both decreasing too?
$endgroup$
– HistoryFan12
Feb 2 at 17:52
$begingroup$
@HistoryFan12 What about the inverse and the opposite of an Increasing function.
$endgroup$
– hamam_Abdallah
Feb 2 at 17:54
$begingroup$
So... you can kind of say these things without proof, just stating they are the inverse and opposite? @hamam_abdallah
$endgroup$
– HistoryFan12
Feb 2 at 18:01
add a comment |
$begingroup$
So sum of 2 decreasing functions is also decreasing, but surely I then need to prove that they are both decreasing too?
$endgroup$
– HistoryFan12
Feb 2 at 17:52
$begingroup$
@HistoryFan12 What about the inverse and the opposite of an Increasing function.
$endgroup$
– hamam_Abdallah
Feb 2 at 17:54
$begingroup$
So... you can kind of say these things without proof, just stating they are the inverse and opposite? @hamam_abdallah
$endgroup$
– HistoryFan12
Feb 2 at 18:01
$begingroup$
So sum of 2 decreasing functions is also decreasing, but surely I then need to prove that they are both decreasing too?
$endgroup$
– HistoryFan12
Feb 2 at 17:52
$begingroup$
So sum of 2 decreasing functions is also decreasing, but surely I then need to prove that they are both decreasing too?
$endgroup$
– HistoryFan12
Feb 2 at 17:52
$begingroup$
@HistoryFan12 What about the inverse and the opposite of an Increasing function.
$endgroup$
– hamam_Abdallah
Feb 2 at 17:54
$begingroup$
@HistoryFan12 What about the inverse and the opposite of an Increasing function.
$endgroup$
– hamam_Abdallah
Feb 2 at 17:54
$begingroup$
So... you can kind of say these things without proof, just stating they are the inverse and opposite? @hamam_abdallah
$endgroup$
– HistoryFan12
Feb 2 at 18:01
$begingroup$
So... you can kind of say these things without proof, just stating they are the inverse and opposite? @hamam_abdallah
$endgroup$
– HistoryFan12
Feb 2 at 18:01
add a comment |
$begingroup$
For
$dfrac{1-y^3}{y}
$,
on $0, infty$,
$y$ is increasing and positive
so $y^3$ is increasing
so $1-y^3$ is decreasing
so
$dfrac{1-y^3}{y}
$
is decreasing
since a decreasing function
divided by a positive increasing function
is decreasing.
answered Feb 2 at 17:49
marty cohenmarty cohen
75.6k549130
$endgroup$
$begingroup$
I get that this is a fact, but does what you've written count as proof?
$endgroup$
– HistoryFan12
Feb 2 at 17:51
$begingroup$
If you write it out formally (e.g., $f$ increasing means $x < y implies f(x) < f(y)$) it does, at least to me.
$endgroup$
– marty cohen
Feb 2 at 18:02
add a comment |
$begingroup$
For
$dfrac{1-y^3}{y}
$,
on $0, infty$,
$y$ is increasing and positive
so $y^3$ is increasing
so $1-y^3$ is decreasing
so
$dfrac{1-y^3}{y}
$
is decreasing
since a decreasing function
divided by a positive increasing function
is decreasing.
answered Feb 2 at 17:49
marty cohenmarty cohen
75.6k549130
$endgroup$
$begingroup$
I get that this is a fact, but does what you've written count as proof?
$endgroup$
– HistoryFan12
Feb 2 at 17:51
$begingroup$
If you write it out formally (e.g., $f$ increasing means $x < y implies f(x) < f(y)$) it does, at least to me.
$endgroup$
– marty cohen
Feb 2 at 18:02
add a comment |
$begingroup$
For
$dfrac{1-y^3}{y}
$,
on $0, infty$,
$y$ is increasing and positive
so $y^3$ is increasing
so $1-y^3$ is decreasing
so
$dfrac{1-y^3}{y}
$
is decreasing
since a decreasing function
divided by a positive increasing function
is decreasing.
answered Feb 2 at 17:49
marty cohenmarty cohen
75.6k549130
$endgroup$
For
$dfrac{1-y^3}{y}
$,
on $0, infty$,
$y$ is increasing and positive
so $y^3$ is increasing
so $1-y^3$ is decreasing
so
$dfrac{1-y^3}{y}
$
is decreasing
since a decreasing function
divided by a positive increasing function
is decreasing.
answered Feb 2 at 17:49
marty cohenmarty cohen
75.6k549130
answered Feb 2 at 17:49
marty cohenmarty cohen
75.6k549130
answered Feb 2 at 17:49
marty cohenmarty cohen
75.6k549130
answered Feb 2 at 17:49
marty cohenmarty cohen
75.6k549130
75.6k549130
$begingroup$
I get that this is a fact, but does what you've written count as proof?
$endgroup$
– HistoryFan12
Feb 2 at 17:51
$begingroup$
If you write it out formally (e.g., $f$ increasing means $x < y implies f(x) < f(y)$) it does, at least to me.
$endgroup$
– marty cohen
Feb 2 at 18:02
add a comment |
$begingroup$
I get that this is a fact, but does what you've written count as proof?
$endgroup$
– HistoryFan12
Feb 2 at 17:51
$begingroup$
If you write it out formally (e.g., $f$ increasing means $x < y implies f(x) < f(y)$) it does, at least to me.
$endgroup$
– marty cohen
Feb 2 at 18:02
$begingroup$
I get that this is a fact, but does what you've written count as proof?
$endgroup$
– HistoryFan12
Feb 2 at 17:51
$begingroup$
I get that this is a fact, but does what you've written count as proof?
$endgroup$
– HistoryFan12
Feb 2 at 17:51
$begingroup$
If you write it out formally (e.g., $f$ increasing means $x < y implies f(x) < f(y)$) it does, at least to me.
$endgroup$
– marty cohen
Feb 2 at 18:02
$begingroup$
If you write it out formally (e.g., $f$ increasing means $x < y implies f(x) < f(y)$) it does, at least to me.
$endgroup$
– marty cohen
Feb 2 at 18:02
add a comment |
$begingroup$
Let $0<x<y$. Then $-xy(x+y)$ is negative, hence less than $1$. This gives $(y-x)[-xy(x+y)] <y-x$ or $xy((x^{2}-y^{2}) <y-x$. This gives $x-xy^{3} <y-x^{3}y$ which says $f(y) <f(x)$.
answered Feb 3 at 12:25
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
Let $0<x<y$. Then $-xy(x+y)$ is negative, hence less than $1$. This gives $(y-x)[-xy(x+y)] <y-x$ or $xy((x^{2}-y^{2}) <y-x$. This gives $x-xy^{3} <y-x^{3}y$ which says $f(y) <f(x)$.
answered Feb 3 at 12:25
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
Let $0<x<y$. Then $-xy(x+y)$ is negative, hence less than $1$. This gives $(y-x)[-xy(x+y)] <y-x$ or $xy((x^{2}-y^{2}) <y-x$. This gives $x-xy^{3} <y-x^{3}y$ which says $f(y) <f(x)$.
answered Feb 3 at 12:25
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
Let $0<x<y$. Then $-xy(x+y)$ is negative, hence less than $1$. This gives $(y-x)[-xy(x+y)] <y-x$ or $xy((x^{2}-y^{2}) <y-x$. This gives $x-xy^{3} <y-x^{3}y$ which says $f(y) <f(x)$.
answered Feb 3 at 12:25
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Feb 3 at 12:25
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Feb 3 at 12:25
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Feb 3 at 12:25
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
add a comment |
add a comment |
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$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Feb 2 at 17:50
$begingroup$
See here
$endgroup$
– J. W. Tanner
Feb 3 at 3:13