Mixing
Rolle's theorem without compact domain hypothesis [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
This question already has an answer here:
$f(0)=0$ and $lim_x to inftyf(x)=0$. Show there's $x_0$ such that $f'(x_0)=0$
1 answer
Good evening everyone,
I'm asking for a proof of "Rolle's theorem generalization". The thesis is as follows:
Let be $a in mathbbR$ and $f:[a,infty) longrightarrow mathbbR$ a continuous function such that $lim_xto infty f(x) = f(a)$. Prove that if the derivative exists in $(a,infty)$ then $exists$ $x_0>a$ such that $f'(x)=0$.
I tried to prove that but I wasn't able to do it, so if you can help me I'll appreciate it a lot.
Thank you for your time.
limits rolles-theorem
edited Jul 24 at 19:19
Michael Hardy
204k23186461
asked Jul 24 at 18:42
mcat
111
marked as duplicate by Hans Lundmark, Adrian Keister, José Carlos Santos, Chris Custer, max_zorn Jul 26 at 0:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
2
down vote
favorite
This question already has an answer here:
$f(0)=0$ and $lim_x to inftyf(x)=0$. Show there's $x_0$ such that $f'(x_0)=0$
1 answer
Good evening everyone,
I'm asking for a proof of "Rolle's theorem generalization". The thesis is as follows:
Let be $a in mathbbR$ and $f:[a,infty) longrightarrow mathbbR$ a continuous function such that $lim_xto infty f(x) = f(a)$. Prove that if the derivative exists in $(a,infty)$ then $exists$ $x_0>a$ such that $f'(x)=0$.
I tried to prove that but I wasn't able to do it, so if you can help me I'll appreciate it a lot.
Thank you for your time.
limits rolles-theorem
edited Jul 24 at 19:19
Michael Hardy
204k23186461
asked Jul 24 at 18:42
mcat
111
marked as duplicate by Hans Lundmark, Adrian Keister, José Carlos Santos, Chris Custer, max_zorn Jul 26 at 0:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question already has an answer here:
$f(0)=0$ and $lim_x to inftyf(x)=0$. Show there's $x_0$ such that $f'(x_0)=0$
1 answer
Good evening everyone,
I'm asking for a proof of "Rolle's theorem generalization". The thesis is as follows:
Let be $a in mathbbR$ and $f:[a,infty) longrightarrow mathbbR$ a continuous function such that $lim_xto infty f(x) = f(a)$. Prove that if the derivative exists in $(a,infty)$ then $exists$ $x_0>a$ such that $f'(x)=0$.
I tried to prove that but I wasn't able to do it, so if you can help me I'll appreciate it a lot.
Thank you for your time.
limits rolles-theorem
edited Jul 24 at 19:19
Michael Hardy
204k23186461
asked Jul 24 at 18:42
mcat
111
This question already has an answer here:
$f(0)=0$ and $lim_x to inftyf(x)=0$. Show there's $x_0$ such that $f'(x_0)=0$
1 answer
Good evening everyone,
I'm asking for a proof of "Rolle's theorem generalization". The thesis is as follows:
Let be $a in mathbbR$ and $f:[a,infty) longrightarrow mathbbR$ a continuous function such that $lim_xto infty f(x) = f(a)$. Prove that if the derivative exists in $(a,infty)$ then $exists$ $x_0>a$ such that $f'(x)=0$.
I tried to prove that but I wasn't able to do it, so if you can help me I'll appreciate it a lot.
Thank you for your time.
This question already has an answer here:
$f(0)=0$ and $lim_x to inftyf(x)=0$. Show there's $x_0$ such that $f'(x_0)=0$
1 answer
limits rolles-theorem
edited Jul 24 at 19:19
Michael Hardy
204k23186461
asked Jul 24 at 18:42
mcat
111
edited Jul 24 at 19:19
Michael Hardy
204k23186461
edited Jul 24 at 19:19
Michael Hardy
204k23186461
edited Jul 24 at 19:19
Michael Hardy
204k23186461
204k23186461
asked Jul 24 at 18:42
mcat
111
asked Jul 24 at 18:42
mcat
111
asked Jul 24 at 18:42
mcat
111
111
marked as duplicate by Hans Lundmark, Adrian Keister, José Carlos Santos, Chris Custer, max_zorn Jul 26 at 0:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Hans Lundmark, Adrian Keister, José Carlos Santos, Chris Custer, max_zorn Jul 26 at 0:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
If there is no point $x_0$ with $f'(x_0) = 0$, then either $f'(x) > 0$ or $f'(x)<0$ for all $x$. WLOG $f'(x) >0$, so the function is strictly increasing. Thus $f(a+1) > f(a)$, and $forall x: x > a+1 implies f(x) > f(a+1)$, so $lim_x to infty f(x) ge f(a+1) > f(a)$, a contradiction.
answered Jul 24 at 18:50
Ovi
11.3k935105
This is what I was looking for, I think it's correct! Thank you very much
– mcat
Jul 24 at 19:57
@mcat You're welcome
– Ovi
Jul 24 at 20:06
Ovi.A question:You proved that f' >0 everywhere leads to a contradiction.Fine. One can conclude that f' le 0 somewhere. If f'(x) =0, we are finished.If f'(x)<0 somewhere ,do we need, in addition, that f' is continuos, to conclude f'(x) somewhere in between?Thanks.
– Peter Szilas
Jul 24 at 20:07
@Ovi Peter made me think about a thing: if $f'(x)$ is not continuous, your proof should be extended somehow, because then we can not tell if $f'(x)>0$ for ALL $x$ (the same for $<0$). It could be that $f'(x)$ has a jump discontinuity so it could not assume the value 0.
– mcat
Jul 24 at 20:22
@mcat: Derivatives can't have jump discontinuities, but this a nontrivial fact (Darboux's theorem).
– Hans Lundmark
Jul 25 at 12:06
add a comment |Â
up vote
1
down vote
Suppose we say the domain is $[a,+infty],$ an interval closed at both ends, the a basic open neighborhood of $+infty$ is any interval of the form $(b,+infty],$ and we define $f(+infty)$ to be $lim_xto+infty f(x) = f(a).$ Then the extreme value theorem applies because topologically the domain $[a,+infty]$ is the same as any closed bounded interval, and $f$ is continuous on that interval. So let $M= max_x left| f(x)right|$ and we have some point $cin(a,+infty)$ for which $f(c) = M.$ Can you show $f'(c) le 0$ and $f'(c)ge 0$ by squeezing?
answered Jul 24 at 19:35
Michael Hardy
204k23186461
Not sure if I understood, but if I consider $[a,infty]$ as a closed bounded interval, could I apply Rolle's theorem? Because in that case the thesis immediately follows. If I actually can not do that, in order to complete the proof I can apply Fermat's theorem: if $f$ admits maximum or minimum in $x_0$ then if the derivative exists in $x_0$ it will be 0. Not sure if that actually works, but this is what I thought. Thank you for your time!
– mcat
Jul 24 at 20:09
@mcat : If you treat $[a,+infty]$ as a closed bounded interval you can apply the extreme value theorem for the following reason: Suppose $f(a) = f(+infty).$ Let $g(y) = a + fleft( dfrac y 1-yright)$ for $0le yle 1,$ and when $y=1$ take the value of $y/(1-y)$ to be $limlimits_y,uparrow,1 dfrac y1-y=+infty.$ Then $g$ is a continuous function on $[0,1]$ and so it assumes its extreme values. If $g$ assumes an extreme value at $cin[0,1],$ then $f$ assumes an extreme value at the number $din[a,+infty]$ for which $a + dfrac d1-d =c.$ Then$,ldots$
– Michael Hardy
Jul 24 at 23:19
$ldots,$having shown that $f$ assumes an extreme value, we have these possibilities: $f$ is constant (in which case the conclusion you want is trivial, or $f$ attains a maximum value that is$>f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint) or $f$ attains a minimum value that is$<f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint). Then work on showing that a maximum or a minimum that is not at an endpoint can occur only at a point where the derivative is $0.$ That last part is done by squeezing. $qquad$
– Michael Hardy
Jul 24 at 23:22
I.e. suppose $f(d) ge f(x)$ for all $xin[a,+infty]$ and $a<d<+infty.$ Then $$ fracf(d+h) -f(d)h left. begincases le 0 & textif h>0, \ ge 0 & textif h <0. endcases right} $$ Therefore $$f'(d) = lim_hto0 fracf(d+h) - f(d) h quad begincases le 0 \ \ textand \ \ ge 0. endcases$$
– Michael Hardy
Jul 24 at 23:32
@mcat : TYPO!!!! I should have said $g(y) = fleft( a + dfrac y1-y right). qquad$
– Michael Hardy
Jul 24 at 23:33
add a comment |Â
up vote
0
down vote
Define $gcolon [0,1/a]to mathbbR$ by
$$
g(x)=begincases
f(1/x) & x > 0\
f(a) & x = 0
endcases
$$
By your hypotheses, this function is continuous, and the derivative $g'(x)$ exists on $(0,1/a)$. Specifically,
$$
g'(x) = -fracf'(1/x)x^2,,
$$
by the chain rule.
By Rolle's Theorem, we know that $g'(x)=0$ for some $xin (0,1/a)$. It follows from the formula for $g'$ that $f'(1/x)=0$.
answered Jul 24 at 23:26
John Gowers
17.6k34168
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If there is no point $x_0$ with $f'(x_0) = 0$, then either $f'(x) > 0$ or $f'(x)<0$ for all $x$. WLOG $f'(x) >0$, so the function is strictly increasing. Thus $f(a+1) > f(a)$, and $forall x: x > a+1 implies f(x) > f(a+1)$, so $lim_x to infty f(x) ge f(a+1) > f(a)$, a contradiction.
answered Jul 24 at 18:50
Ovi
11.3k935105
This is what I was looking for, I think it's correct! Thank you very much
– mcat
Jul 24 at 19:57
@mcat You're welcome
– Ovi
Jul 24 at 20:06
Ovi.A question:You proved that f' >0 everywhere leads to a contradiction.Fine. One can conclude that f' le 0 somewhere. If f'(x) =0, we are finished.If f'(x)<0 somewhere ,do we need, in addition, that f' is continuos, to conclude f'(x) somewhere in between?Thanks.
– Peter Szilas
Jul 24 at 20:07
@Ovi Peter made me think about a thing: if $f'(x)$ is not continuous, your proof should be extended somehow, because then we can not tell if $f'(x)>0$ for ALL $x$ (the same for $<0$). It could be that $f'(x)$ has a jump discontinuity so it could not assume the value 0.
– mcat
Jul 24 at 20:22
@mcat: Derivatives can't have jump discontinuities, but this a nontrivial fact (Darboux's theorem).
– Hans Lundmark
Jul 25 at 12:06
add a comment |Â
up vote
3
down vote
If there is no point $x_0$ with $f'(x_0) = 0$, then either $f'(x) > 0$ or $f'(x)<0$ for all $x$. WLOG $f'(x) >0$, so the function is strictly increasing. Thus $f(a+1) > f(a)$, and $forall x: x > a+1 implies f(x) > f(a+1)$, so $lim_x to infty f(x) ge f(a+1) > f(a)$, a contradiction.
answered Jul 24 at 18:50
Ovi
11.3k935105
This is what I was looking for, I think it's correct! Thank you very much
– mcat
Jul 24 at 19:57
@mcat You're welcome
– Ovi
Jul 24 at 20:06
Ovi.A question:You proved that f' >0 everywhere leads to a contradiction.Fine. One can conclude that f' le 0 somewhere. If f'(x) =0, we are finished.If f'(x)<0 somewhere ,do we need, in addition, that f' is continuos, to conclude f'(x) somewhere in between?Thanks.
– Peter Szilas
Jul 24 at 20:07
@Ovi Peter made me think about a thing: if $f'(x)$ is not continuous, your proof should be extended somehow, because then we can not tell if $f'(x)>0$ for ALL $x$ (the same for $<0$). It could be that $f'(x)$ has a jump discontinuity so it could not assume the value 0.
– mcat
Jul 24 at 20:22
@mcat: Derivatives can't have jump discontinuities, but this a nontrivial fact (Darboux's theorem).
– Hans Lundmark
Jul 25 at 12:06
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If there is no point $x_0$ with $f'(x_0) = 0$, then either $f'(x) > 0$ or $f'(x)<0$ for all $x$. WLOG $f'(x) >0$, so the function is strictly increasing. Thus $f(a+1) > f(a)$, and $forall x: x > a+1 implies f(x) > f(a+1)$, so $lim_x to infty f(x) ge f(a+1) > f(a)$, a contradiction.
answered Jul 24 at 18:50
Ovi
11.3k935105
If there is no point $x_0$ with $f'(x_0) = 0$, then either $f'(x) > 0$ or $f'(x)<0$ for all $x$. WLOG $f'(x) >0$, so the function is strictly increasing. Thus $f(a+1) > f(a)$, and $forall x: x > a+1 implies f(x) > f(a+1)$, so $lim_x to infty f(x) ge f(a+1) > f(a)$, a contradiction.
answered Jul 24 at 18:50
Ovi
11.3k935105
answered Jul 24 at 18:50
Ovi
11.3k935105
answered Jul 24 at 18:50
Ovi
11.3k935105
answered Jul 24 at 18:50
Ovi
11.3k935105
11.3k935105
This is what I was looking for, I think it's correct! Thank you very much
– mcat
Jul 24 at 19:57
@mcat You're welcome
– Ovi
Jul 24 at 20:06
Ovi.A question:You proved that f' >0 everywhere leads to a contradiction.Fine. One can conclude that f' le 0 somewhere. If f'(x) =0, we are finished.If f'(x)<0 somewhere ,do we need, in addition, that f' is continuos, to conclude f'(x) somewhere in between?Thanks.
– Peter Szilas
Jul 24 at 20:07
@Ovi Peter made me think about a thing: if $f'(x)$ is not continuous, your proof should be extended somehow, because then we can not tell if $f'(x)>0$ for ALL $x$ (the same for $<0$). It could be that $f'(x)$ has a jump discontinuity so it could not assume the value 0.
– mcat
Jul 24 at 20:22
@mcat: Derivatives can't have jump discontinuities, but this a nontrivial fact (Darboux's theorem).
– Hans Lundmark
Jul 25 at 12:06
add a comment |Â
This is what I was looking for, I think it's correct! Thank you very much
– mcat
Jul 24 at 19:57
@mcat You're welcome
– Ovi
Jul 24 at 20:06
Ovi.A question:You proved that f' >0 everywhere leads to a contradiction.Fine. One can conclude that f' le 0 somewhere. If f'(x) =0, we are finished.If f'(x)<0 somewhere ,do we need, in addition, that f' is continuos, to conclude f'(x) somewhere in between?Thanks.
– Peter Szilas
Jul 24 at 20:07
@Ovi Peter made me think about a thing: if $f'(x)$ is not continuous, your proof should be extended somehow, because then we can not tell if $f'(x)>0$ for ALL $x$ (the same for $<0$). It could be that $f'(x)$ has a jump discontinuity so it could not assume the value 0.
– mcat
Jul 24 at 20:22
@mcat: Derivatives can't have jump discontinuities, but this a nontrivial fact (Darboux's theorem).
– Hans Lundmark
Jul 25 at 12:06
This is what I was looking for, I think it's correct! Thank you very much
– mcat
Jul 24 at 19:57
This is what I was looking for, I think it's correct! Thank you very much
– mcat
Jul 24 at 19:57
@mcat You're welcome
– Ovi
Jul 24 at 20:06
@mcat You're welcome
– Ovi
Jul 24 at 20:06
Ovi.A question:You proved that f' >0 everywhere leads to a contradiction.Fine. One can conclude that f' le 0 somewhere. If f'(x) =0, we are finished.If f'(x)<0 somewhere ,do we need, in addition, that f' is continuos, to conclude f'(x) somewhere in between?Thanks.
– Peter Szilas
Jul 24 at 20:07
Ovi.A question:You proved that f' >0 everywhere leads to a contradiction.Fine. One can conclude that f' le 0 somewhere. If f'(x) =0, we are finished.If f'(x)<0 somewhere ,do we need, in addition, that f' is continuos, to conclude f'(x) somewhere in between?Thanks.
– Peter Szilas
Jul 24 at 20:07
@Ovi Peter made me think about a thing: if $f'(x)$ is not continuous, your proof should be extended somehow, because then we can not tell if $f'(x)>0$ for ALL $x$ (the same for $<0$). It could be that $f'(x)$ has a jump discontinuity so it could not assume the value 0.
– mcat
Jul 24 at 20:22
@Ovi Peter made me think about a thing: if $f'(x)$ is not continuous, your proof should be extended somehow, because then we can not tell if $f'(x)>0$ for ALL $x$ (the same for $<0$). It could be that $f'(x)$ has a jump discontinuity so it could not assume the value 0.
– mcat
Jul 24 at 20:22
@mcat: Derivatives can't have jump discontinuities, but this a nontrivial fact (Darboux's theorem).
– Hans Lundmark
Jul 25 at 12:06
@mcat: Derivatives can't have jump discontinuities, but this a nontrivial fact (Darboux's theorem).
– Hans Lundmark
Jul 25 at 12:06
add a comment |Â
up vote
1
down vote
Suppose we say the domain is $[a,+infty],$ an interval closed at both ends, the a basic open neighborhood of $+infty$ is any interval of the form $(b,+infty],$ and we define $f(+infty)$ to be $lim_xto+infty f(x) = f(a).$ Then the extreme value theorem applies because topologically the domain $[a,+infty]$ is the same as any closed bounded interval, and $f$ is continuous on that interval. So let $M= max_x left| f(x)right|$ and we have some point $cin(a,+infty)$ for which $f(c) = M.$ Can you show $f'(c) le 0$ and $f'(c)ge 0$ by squeezing?
answered Jul 24 at 19:35
Michael Hardy
204k23186461
Not sure if I understood, but if I consider $[a,infty]$ as a closed bounded interval, could I apply Rolle's theorem? Because in that case the thesis immediately follows. If I actually can not do that, in order to complete the proof I can apply Fermat's theorem: if $f$ admits maximum or minimum in $x_0$ then if the derivative exists in $x_0$ it will be 0. Not sure if that actually works, but this is what I thought. Thank you for your time!
– mcat
Jul 24 at 20:09
@mcat : If you treat $[a,+infty]$ as a closed bounded interval you can apply the extreme value theorem for the following reason: Suppose $f(a) = f(+infty).$ Let $g(y) = a + fleft( dfrac y 1-yright)$ for $0le yle 1,$ and when $y=1$ take the value of $y/(1-y)$ to be $limlimits_y,uparrow,1 dfrac y1-y=+infty.$ Then $g$ is a continuous function on $[0,1]$ and so it assumes its extreme values. If $g$ assumes an extreme value at $cin[0,1],$ then $f$ assumes an extreme value at the number $din[a,+infty]$ for which $a + dfrac d1-d =c.$ Then$,ldots$
– Michael Hardy
Jul 24 at 23:19
$ldots,$having shown that $f$ assumes an extreme value, we have these possibilities: $f$ is constant (in which case the conclusion you want is trivial, or $f$ attains a maximum value that is$>f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint) or $f$ attains a minimum value that is$<f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint). Then work on showing that a maximum or a minimum that is not at an endpoint can occur only at a point where the derivative is $0.$ That last part is done by squeezing. $qquad$
– Michael Hardy
Jul 24 at 23:22
I.e. suppose $f(d) ge f(x)$ for all $xin[a,+infty]$ and $a<d<+infty.$ Then $$ fracf(d+h) -f(d)h left. begincases le 0 & textif h>0, \ ge 0 & textif h <0. endcases right} $$ Therefore $$f'(d) = lim_hto0 fracf(d+h) - f(d) h quad begincases le 0 \ \ textand \ \ ge 0. endcases$$
– Michael Hardy
Jul 24 at 23:32
@mcat : TYPO!!!! I should have said $g(y) = fleft( a + dfrac y1-y right). qquad$
– Michael Hardy
Jul 24 at 23:33
add a comment |Â
up vote
1
down vote
Suppose we say the domain is $[a,+infty],$ an interval closed at both ends, the a basic open neighborhood of $+infty$ is any interval of the form $(b,+infty],$ and we define $f(+infty)$ to be $lim_xto+infty f(x) = f(a).$ Then the extreme value theorem applies because topologically the domain $[a,+infty]$ is the same as any closed bounded interval, and $f$ is continuous on that interval. So let $M= max_x left| f(x)right|$ and we have some point $cin(a,+infty)$ for which $f(c) = M.$ Can you show $f'(c) le 0$ and $f'(c)ge 0$ by squeezing?
answered Jul 24 at 19:35
Michael Hardy
204k23186461
Not sure if I understood, but if I consider $[a,infty]$ as a closed bounded interval, could I apply Rolle's theorem? Because in that case the thesis immediately follows. If I actually can not do that, in order to complete the proof I can apply Fermat's theorem: if $f$ admits maximum or minimum in $x_0$ then if the derivative exists in $x_0$ it will be 0. Not sure if that actually works, but this is what I thought. Thank you for your time!
– mcat
Jul 24 at 20:09
@mcat : If you treat $[a,+infty]$ as a closed bounded interval you can apply the extreme value theorem for the following reason: Suppose $f(a) = f(+infty).$ Let $g(y) = a + fleft( dfrac y 1-yright)$ for $0le yle 1,$ and when $y=1$ take the value of $y/(1-y)$ to be $limlimits_y,uparrow,1 dfrac y1-y=+infty.$ Then $g$ is a continuous function on $[0,1]$ and so it assumes its extreme values. If $g$ assumes an extreme value at $cin[0,1],$ then $f$ assumes an extreme value at the number $din[a,+infty]$ for which $a + dfrac d1-d =c.$ Then$,ldots$
– Michael Hardy
Jul 24 at 23:19
$ldots,$having shown that $f$ assumes an extreme value, we have these possibilities: $f$ is constant (in which case the conclusion you want is trivial, or $f$ attains a maximum value that is$>f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint) or $f$ attains a minimum value that is$<f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint). Then work on showing that a maximum or a minimum that is not at an endpoint can occur only at a point where the derivative is $0.$ That last part is done by squeezing. $qquad$
– Michael Hardy
Jul 24 at 23:22
I.e. suppose $f(d) ge f(x)$ for all $xin[a,+infty]$ and $a<d<+infty.$ Then $$ fracf(d+h) -f(d)h left. begincases le 0 & textif h>0, \ ge 0 & textif h <0. endcases right} $$ Therefore $$f'(d) = lim_hto0 fracf(d+h) - f(d) h quad begincases le 0 \ \ textand \ \ ge 0. endcases$$
– Michael Hardy
Jul 24 at 23:32
@mcat : TYPO!!!! I should have said $g(y) = fleft( a + dfrac y1-y right). qquad$
– Michael Hardy
Jul 24 at 23:33
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose we say the domain is $[a,+infty],$ an interval closed at both ends, the a basic open neighborhood of $+infty$ is any interval of the form $(b,+infty],$ and we define $f(+infty)$ to be $lim_xto+infty f(x) = f(a).$ Then the extreme value theorem applies because topologically the domain $[a,+infty]$ is the same as any closed bounded interval, and $f$ is continuous on that interval. So let $M= max_x left| f(x)right|$ and we have some point $cin(a,+infty)$ for which $f(c) = M.$ Can you show $f'(c) le 0$ and $f'(c)ge 0$ by squeezing?
answered Jul 24 at 19:35
Michael Hardy
204k23186461
Suppose we say the domain is $[a,+infty],$ an interval closed at both ends, the a basic open neighborhood of $+infty$ is any interval of the form $(b,+infty],$ and we define $f(+infty)$ to be $lim_xto+infty f(x) = f(a).$ Then the extreme value theorem applies because topologically the domain $[a,+infty]$ is the same as any closed bounded interval, and $f$ is continuous on that interval. So let $M= max_x left| f(x)right|$ and we have some point $cin(a,+infty)$ for which $f(c) = M.$ Can you show $f'(c) le 0$ and $f'(c)ge 0$ by squeezing?
answered Jul 24 at 19:35
Michael Hardy
204k23186461
edited Jul 24 at 19:53
answered Jul 24 at 19:35
Michael Hardy
204k23186461
answered Jul 24 at 19:35
Michael Hardy
204k23186461
answered Jul 24 at 19:35
Michael Hardy
204k23186461
204k23186461
Not sure if I understood, but if I consider $[a,infty]$ as a closed bounded interval, could I apply Rolle's theorem? Because in that case the thesis immediately follows. If I actually can not do that, in order to complete the proof I can apply Fermat's theorem: if $f$ admits maximum or minimum in $x_0$ then if the derivative exists in $x_0$ it will be 0. Not sure if that actually works, but this is what I thought. Thank you for your time!
– mcat
Jul 24 at 20:09
@mcat : If you treat $[a,+infty]$ as a closed bounded interval you can apply the extreme value theorem for the following reason: Suppose $f(a) = f(+infty).$ Let $g(y) = a + fleft( dfrac y 1-yright)$ for $0le yle 1,$ and when $y=1$ take the value of $y/(1-y)$ to be $limlimits_y,uparrow,1 dfrac y1-y=+infty.$ Then $g$ is a continuous function on $[0,1]$ and so it assumes its extreme values. If $g$ assumes an extreme value at $cin[0,1],$ then $f$ assumes an extreme value at the number $din[a,+infty]$ for which $a + dfrac d1-d =c.$ Then$,ldots$
– Michael Hardy
Jul 24 at 23:19
$ldots,$having shown that $f$ assumes an extreme value, we have these possibilities: $f$ is constant (in which case the conclusion you want is trivial, or $f$ attains a maximum value that is$>f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint) or $f$ attains a minimum value that is$<f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint). Then work on showing that a maximum or a minimum that is not at an endpoint can occur only at a point where the derivative is $0.$ That last part is done by squeezing. $qquad$
– Michael Hardy
Jul 24 at 23:22
I.e. suppose $f(d) ge f(x)$ for all $xin[a,+infty]$ and $a<d<+infty.$ Then $$ fracf(d+h) -f(d)h left. begincases le 0 & textif h>0, \ ge 0 & textif h <0. endcases right} $$ Therefore $$f'(d) = lim_hto0 fracf(d+h) - f(d) h quad begincases le 0 \ \ textand \ \ ge 0. endcases$$
– Michael Hardy
Jul 24 at 23:32
@mcat : TYPO!!!! I should have said $g(y) = fleft( a + dfrac y1-y right). qquad$
– Michael Hardy
Jul 24 at 23:33
add a comment |Â
Not sure if I understood, but if I consider $[a,infty]$ as a closed bounded interval, could I apply Rolle's theorem? Because in that case the thesis immediately follows. If I actually can not do that, in order to complete the proof I can apply Fermat's theorem: if $f$ admits maximum or minimum in $x_0$ then if the derivative exists in $x_0$ it will be 0. Not sure if that actually works, but this is what I thought. Thank you for your time!
– mcat
Jul 24 at 20:09
@mcat : If you treat $[a,+infty]$ as a closed bounded interval you can apply the extreme value theorem for the following reason: Suppose $f(a) = f(+infty).$ Let $g(y) = a + fleft( dfrac y 1-yright)$ for $0le yle 1,$ and when $y=1$ take the value of $y/(1-y)$ to be $limlimits_y,uparrow,1 dfrac y1-y=+infty.$ Then $g$ is a continuous function on $[0,1]$ and so it assumes its extreme values. If $g$ assumes an extreme value at $cin[0,1],$ then $f$ assumes an extreme value at the number $din[a,+infty]$ for which $a + dfrac d1-d =c.$ Then$,ldots$
– Michael Hardy
Jul 24 at 23:19
$ldots,$having shown that $f$ assumes an extreme value, we have these possibilities: $f$ is constant (in which case the conclusion you want is trivial, or $f$ attains a maximum value that is$>f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint) or $f$ attains a minimum value that is$<f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint). Then work on showing that a maximum or a minimum that is not at an endpoint can occur only at a point where the derivative is $0.$ That last part is done by squeezing. $qquad$
– Michael Hardy
Jul 24 at 23:22
I.e. suppose $f(d) ge f(x)$ for all $xin[a,+infty]$ and $a<d<+infty.$ Then $$ fracf(d+h) -f(d)h left. begincases le 0 & textif h>0, \ ge 0 & textif h <0. endcases right} $$ Therefore $$f'(d) = lim_hto0 fracf(d+h) - f(d) h quad begincases le 0 \ \ textand \ \ ge 0. endcases$$
– Michael Hardy
Jul 24 at 23:32
@mcat : TYPO!!!! I should have said $g(y) = fleft( a + dfrac y1-y right). qquad$
– Michael Hardy
Jul 24 at 23:33
Not sure if I understood, but if I consider $[a,infty]$ as a closed bounded interval, could I apply Rolle's theorem? Because in that case the thesis immediately follows. If I actually can not do that, in order to complete the proof I can apply Fermat's theorem: if $f$ admits maximum or minimum in $x_0$ then if the derivative exists in $x_0$ it will be 0. Not sure if that actually works, but this is what I thought. Thank you for your time!
– mcat
Jul 24 at 20:09
Not sure if I understood, but if I consider $[a,infty]$ as a closed bounded interval, could I apply Rolle's theorem? Because in that case the thesis immediately follows. If I actually can not do that, in order to complete the proof I can apply Fermat's theorem: if $f$ admits maximum or minimum in $x_0$ then if the derivative exists in $x_0$ it will be 0. Not sure if that actually works, but this is what I thought. Thank you for your time!
– mcat
Jul 24 at 20:09
@mcat : If you treat $[a,+infty]$ as a closed bounded interval you can apply the extreme value theorem for the following reason: Suppose $f(a) = f(+infty).$ Let $g(y) = a + fleft( dfrac y 1-yright)$ for $0le yle 1,$ and when $y=1$ take the value of $y/(1-y)$ to be $limlimits_y,uparrow,1 dfrac y1-y=+infty.$ Then $g$ is a continuous function on $[0,1]$ and so it assumes its extreme values. If $g$ assumes an extreme value at $cin[0,1],$ then $f$ assumes an extreme value at the number $din[a,+infty]$ for which $a + dfrac d1-d =c.$ Then$,ldots$
– Michael Hardy
Jul 24 at 23:19
@mcat : If you treat $[a,+infty]$ as a closed bounded interval you can apply the extreme value theorem for the following reason: Suppose $f(a) = f(+infty).$ Let $g(y) = a + fleft( dfrac y 1-yright)$ for $0le yle 1,$ and when $y=1$ take the value of $y/(1-y)$ to be $limlimits_y,uparrow,1 dfrac y1-y=+infty.$ Then $g$ is a continuous function on $[0,1]$ and so it assumes its extreme values. If $g$ assumes an extreme value at $cin[0,1],$ then $f$ assumes an extreme value at the number $din[a,+infty]$ for which $a + dfrac d1-d =c.$ Then$,ldots$
– Michael Hardy
Jul 24 at 23:19
$ldots,$having shown that $f$ assumes an extreme value, we have these possibilities: $f$ is constant (in which case the conclusion you want is trivial, or $f$ attains a maximum value that is$>f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint) or $f$ attains a minimum value that is$<f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint). Then work on showing that a maximum or a minimum that is not at an endpoint can occur only at a point where the derivative is $0.$ That last part is done by squeezing. $qquad$
– Michael Hardy
Jul 24 at 23:22
$ldots,$having shown that $f$ assumes an extreme value, we have these possibilities: $f$ is constant (in which case the conclusion you want is trivial, or $f$ attains a maximum value that is$>f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint) or $f$ attains a minimum value that is$<f(a)$ at some point $din(a,+infty)$ (thus not at an endpoint). Then work on showing that a maximum or a minimum that is not at an endpoint can occur only at a point where the derivative is $0.$ That last part is done by squeezing. $qquad$
– Michael Hardy
Jul 24 at 23:22
I.e. suppose $f(d) ge f(x)$ for all $xin[a,+infty]$ and $a<d<+infty.$ Then $$ fracf(d+h) -f(d)h left. begincases le 0 & textif h>0, \ ge 0 & textif h <0. endcases right} $$ Therefore $$f'(d) = lim_hto0 fracf(d+h) - f(d) h quad begincases le 0 \ \ textand \ \ ge 0. endcases$$
– Michael Hardy
Jul 24 at 23:32
I.e. suppose $f(d) ge f(x)$ for all $xin[a,+infty]$ and $a<d<+infty.$ Then $$ fracf(d+h) -f(d)h left. begincases le 0 & textif h>0, \ ge 0 & textif h <0. endcases right} $$ Therefore $$f'(d) = lim_hto0 fracf(d+h) - f(d) h quad begincases le 0 \ \ textand \ \ ge 0. endcases$$
– Michael Hardy
Jul 24 at 23:32
@mcat : TYPO!!!! I should have said $g(y) = fleft( a + dfrac y1-y right). qquad$
– Michael Hardy
Jul 24 at 23:33
@mcat : TYPO!!!! I should have said $g(y) = fleft( a + dfrac y1-y right). qquad$
– Michael Hardy
Jul 24 at 23:33
add a comment |Â
up vote
0
down vote
Define $gcolon [0,1/a]to mathbbR$ by
$$
g(x)=begincases
f(1/x) & x > 0\
f(a) & x = 0
endcases
$$
By your hypotheses, this function is continuous, and the derivative $g'(x)$ exists on $(0,1/a)$. Specifically,
$$
g'(x) = -fracf'(1/x)x^2,,
$$
by the chain rule.
By Rolle's Theorem, we know that $g'(x)=0$ for some $xin (0,1/a)$. It follows from the formula for $g'$ that $f'(1/x)=0$.
answered Jul 24 at 23:26
John Gowers
17.6k34168
add a comment |Â
up vote
0
down vote
Define $gcolon [0,1/a]to mathbbR$ by
$$
g(x)=begincases
f(1/x) & x > 0\
f(a) & x = 0
endcases
$$
By your hypotheses, this function is continuous, and the derivative $g'(x)$ exists on $(0,1/a)$. Specifically,
$$
g'(x) = -fracf'(1/x)x^2,,
$$
by the chain rule.
By Rolle's Theorem, we know that $g'(x)=0$ for some $xin (0,1/a)$. It follows from the formula for $g'$ that $f'(1/x)=0$.
answered Jul 24 at 23:26
John Gowers
17.6k34168
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Define $gcolon [0,1/a]to mathbbR$ by
$$
g(x)=begincases
f(1/x) & x > 0\
f(a) & x = 0
endcases
$$
By your hypotheses, this function is continuous, and the derivative $g'(x)$ exists on $(0,1/a)$. Specifically,
$$
g'(x) = -fracf'(1/x)x^2,,
$$
by the chain rule.
By Rolle's Theorem, we know that $g'(x)=0$ for some $xin (0,1/a)$. It follows from the formula for $g'$ that $f'(1/x)=0$.
answered Jul 24 at 23:26
John Gowers
17.6k34168
Define $gcolon [0,1/a]to mathbbR$ by
$$
g(x)=begincases
f(1/x) & x > 0\
f(a) & x = 0
endcases
$$
By your hypotheses, this function is continuous, and the derivative $g'(x)$ exists on $(0,1/a)$. Specifically,
$$
g'(x) = -fracf'(1/x)x^2,,
$$
by the chain rule.
By Rolle's Theorem, we know that $g'(x)=0$ for some $xin (0,1/a)$. It follows from the formula for $g'$ that $f'(1/x)=0$.
answered Jul 24 at 23:26
John Gowers
17.6k34168
answered Jul 24 at 23:26
John Gowers
17.6k34168
answered Jul 24 at 23:26
John Gowers
17.6k34168
answered Jul 24 at 23:26
John Gowers
17.6k34168
17.6k34168
add a comment |Â
add a comment |Â