Mixing
Theorem 5.8 Baby Rudin. Some questions
$begingroup$
This is with reference to the page number 107 of Baby Rudin.
Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $xin(a,b)$ is a
local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.
I have some questions. Thanks in advance for reading and helping out.
What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.
Edits:
We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).
real-analysis
asked Jan 5 at 9:52
StammeringMathematicianStammeringMathematician
2,3441322
$endgroup$
add a comment |
$begingroup$
This is with reference to the page number 107 of Baby Rudin.
Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $xin(a,b)$ is a
local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.
I have some questions. Thanks in advance for reading and helping out.
What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.
Edits:
We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).
real-analysis
asked Jan 5 at 9:52
StammeringMathematicianStammeringMathematician
2,3441322
$endgroup$
add a comment |
$begingroup$
This is with reference to the page number 107 of Baby Rudin.
Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $xin(a,b)$ is a
local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.
I have some questions. Thanks in advance for reading and helping out.
What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.
Edits:
We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).
real-analysis
asked Jan 5 at 9:52
StammeringMathematicianStammeringMathematician
2,3441322
$endgroup$
This is with reference to the page number 107 of Baby Rudin.
Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $xin(a,b)$ is a
local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.
I have some questions. Thanks in advance for reading and helping out.
What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.
Edits:
We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).
real-analysis
real-analysis
asked Jan 5 at 9:52
StammeringMathematicianStammeringMathematician
2,3441322
asked Jan 5 at 9:52
StammeringMathematicianStammeringMathematician
2,3441322
edited Jan 5 at 9:57
StammeringMathematician
asked Jan 5 at 9:52
StammeringMathematicianStammeringMathematician
2,3441322
asked Jan 5 at 9:52
StammeringMathematicianStammeringMathematician
2,3441322
asked Jan 5 at 9:52
StammeringMathematicianStammeringMathematician
2,3441322
2,3441322
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Answer to question 1
The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.
Answer to question 2
Take $g(x)=vert x vert$, again look at zero.
answered Jan 5 at 9:59
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
add a comment |
$begingroup$
If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.
For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.
answered Jan 5 at 10:00
Gabriel RibeiroGabriel Ribeiro
1,424522
$endgroup$
add a comment |
$begingroup$
- What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
No. It could be a minimum, or it could be neither. Such examples:
$f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.
$f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.
(I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)
This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.
- The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?
Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.
answered Jan 5 at 10:03
Eevee TrainerEevee Trainer
5,7471936
$endgroup$
add a comment |
$begingroup$
Let me try:
1) $f'(x)=0; $
Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum
2) $f(x)= |x|$ , has a local minimum at $x=0.$
$f(x)= -|x|$ has a local maximum at $x=0$
(f'(0) does not exist)(Why?)
answered Jan 5 at 10:07
Peter SzilasPeter Szilas
11k2821
$endgroup$
1
$begingroup$
$f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
$endgroup$
– StammeringMathematician
Jan 5 at 10:10
$begingroup$
Stammering.Yes,exactly!!Greetings.
$endgroup$
– Peter Szilas
Jan 5 at 10:14
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Answer to question 1
The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.
Answer to question 2
Take $g(x)=vert x vert$, again look at zero.
answered Jan 5 at 9:59
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
add a comment |
$begingroup$
Answer to question 1
The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.
Answer to question 2
Take $g(x)=vert x vert$, again look at zero.
answered Jan 5 at 9:59
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
add a comment |
$begingroup$
Answer to question 1
The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.
Answer to question 2
Take $g(x)=vert x vert$, again look at zero.
answered Jan 5 at 9:59
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
Answer to question 1
The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.
Answer to question 2
Take $g(x)=vert x vert$, again look at zero.
answered Jan 5 at 9:59
mathcounterexamples.netmathcounterexamples.net
26k21955
answered Jan 5 at 9:59
mathcounterexamples.netmathcounterexamples.net
26k21955
answered Jan 5 at 9:59
mathcounterexamples.netmathcounterexamples.net
26k21955
answered Jan 5 at 9:59
mathcounterexamples.netmathcounterexamples.net
26k21955
26k21955
add a comment |
add a comment |
$begingroup$
If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.
For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.
answered Jan 5 at 10:00
Gabriel RibeiroGabriel Ribeiro
1,424522
$endgroup$
add a comment |
$begingroup$
If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.
For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.
answered Jan 5 at 10:00
Gabriel RibeiroGabriel Ribeiro
1,424522
$endgroup$
add a comment |
$begingroup$
If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.
For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.
answered Jan 5 at 10:00
Gabriel RibeiroGabriel Ribeiro
1,424522
$endgroup$
If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.
For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.
answered Jan 5 at 10:00
Gabriel RibeiroGabriel Ribeiro
1,424522
answered Jan 5 at 10:00
Gabriel RibeiroGabriel Ribeiro
1,424522
answered Jan 5 at 10:00
Gabriel RibeiroGabriel Ribeiro
1,424522
answered Jan 5 at 10:00
Gabriel RibeiroGabriel Ribeiro
1,424522
1,424522
add a comment |
add a comment |
$begingroup$
- What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
No. It could be a minimum, or it could be neither. Such examples:
$f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.
$f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.
(I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)
This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.
- The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?
Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.
answered Jan 5 at 10:03
Eevee TrainerEevee Trainer
5,7471936
$endgroup$
add a comment |
$begingroup$
- What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
No. It could be a minimum, or it could be neither. Such examples:
$f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.
$f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.
(I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)
This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.
- The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?
Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.
answered Jan 5 at 10:03
Eevee TrainerEevee Trainer
5,7471936
$endgroup$
add a comment |
$begingroup$
- What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
No. It could be a minimum, or it could be neither. Such examples:
$f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.
$f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.
(I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)
This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.
- The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?
Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.
answered Jan 5 at 10:03
Eevee TrainerEevee Trainer
5,7471936
$endgroup$
- What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?
No. It could be a minimum, or it could be neither. Such examples:
$f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.
$f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.
(I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)
This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.
- The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?
Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.
answered Jan 5 at 10:03
Eevee TrainerEevee Trainer
5,7471936
answered Jan 5 at 10:03
Eevee TrainerEevee Trainer
5,7471936
answered Jan 5 at 10:03
Eevee TrainerEevee Trainer
5,7471936
answered Jan 5 at 10:03
Eevee TrainerEevee Trainer
5,7471936
5,7471936
add a comment |
add a comment |
$begingroup$
Let me try:
1) $f'(x)=0; $
Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum
2) $f(x)= |x|$ , has a local minimum at $x=0.$
$f(x)= -|x|$ has a local maximum at $x=0$
(f'(0) does not exist)(Why?)
answered Jan 5 at 10:07
Peter SzilasPeter Szilas
11k2821
$endgroup$
1
$begingroup$
$f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
$endgroup$
– StammeringMathematician
Jan 5 at 10:10
$begingroup$
Stammering.Yes,exactly!!Greetings.
$endgroup$
– Peter Szilas
Jan 5 at 10:14
add a comment |
$begingroup$
Let me try:
1) $f'(x)=0; $
Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum
2) $f(x)= |x|$ , has a local minimum at $x=0.$
$f(x)= -|x|$ has a local maximum at $x=0$
(f'(0) does not exist)(Why?)
answered Jan 5 at 10:07
Peter SzilasPeter Szilas
11k2821
$endgroup$
1
$begingroup$
$f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
$endgroup$
– StammeringMathematician
Jan 5 at 10:10
$begingroup$
Stammering.Yes,exactly!!Greetings.
$endgroup$
– Peter Szilas
Jan 5 at 10:14
add a comment |
$begingroup$
Let me try:
1) $f'(x)=0; $
Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum
2) $f(x)= |x|$ , has a local minimum at $x=0.$
$f(x)= -|x|$ has a local maximum at $x=0$
(f'(0) does not exist)(Why?)
answered Jan 5 at 10:07
Peter SzilasPeter Szilas
11k2821
$endgroup$
Let me try:
1) $f'(x)=0; $
Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum
2) $f(x)= |x|$ , has a local minimum at $x=0.$
$f(x)= -|x|$ has a local maximum at $x=0$
(f'(0) does not exist)(Why?)
answered Jan 5 at 10:07
Peter SzilasPeter Szilas
11k2821
answered Jan 5 at 10:07
Peter SzilasPeter Szilas
11k2821
answered Jan 5 at 10:07
Peter SzilasPeter Szilas
11k2821
answered Jan 5 at 10:07
Peter SzilasPeter Szilas
11k2821
11k2821
1
$begingroup$
$f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
$endgroup$
– StammeringMathematician
Jan 5 at 10:10
$begingroup$
Stammering.Yes,exactly!!Greetings.
$endgroup$
– Peter Szilas
Jan 5 at 10:14
add a comment |
1
$begingroup$
$f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
$endgroup$
– StammeringMathematician
Jan 5 at 10:10
$begingroup$
Stammering.Yes,exactly!!Greetings.
$endgroup$
– Peter Szilas
Jan 5 at 10:14
1
1
$begingroup$
$f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
$endgroup$
– StammeringMathematician
Jan 5 at 10:10
$begingroup$
$f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
$endgroup$
– StammeringMathematician
Jan 5 at 10:10
$begingroup$
Stammering.Yes,exactly!!Greetings.
$endgroup$
– Peter Szilas
Jan 5 at 10:14
$begingroup$
Stammering.Yes,exactly!!Greetings.
$endgroup$
– Peter Szilas
Jan 5 at 10:14
add a comment |
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