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Relative Extrema

1

$begingroup$

I did a question to find relative extrema for the following function:
$f(x)=x^2$ on $[−2,2].$

The answer said that there is no relative maxima for this function because relative extrema cannot occur at the end points of a domain.
Why is this so ?
Thank you.

functions

share|cite|improve this question

edited Mar 23 at 13:51

Allawonder

2,281616

asked Mar 23 at 11:23

Ashok SharmaAshok Sharma

111

$endgroup$

add a comment | 

1

$begingroup$

I did a question to find relative extrema for the following function:
$f(x)=x^2$ on $[−2,2].$

The answer said that there is no relative maxima for this function because relative extrema cannot occur at the end points of a domain.
Why is this so ?
Thank you.

functions

share|cite|improve this question

edited Mar 23 at 13:51

Allawonder

2,281616

asked Mar 23 at 11:23

Ashok SharmaAshok Sharma

111

$endgroup$

add a comment | 

$begingroup$

I did a question to find relative extrema for the following function:
$f(x)=x^2$ on $[−2,2].$

The answer said that there is no relative maxima for this function because relative extrema cannot occur at the end points of a domain.
Why is this so ?
Thank you.

functions

share|cite|improve this question

edited Mar 23 at 13:51

Allawonder

2,281616

asked Mar 23 at 11:23

Ashok SharmaAshok Sharma

111

$endgroup$

share|cite|improve this question

edited Mar 23 at 13:51

Allawonder

2,281616

asked Mar 23 at 11:23

Ashok SharmaAshok Sharma

111

share|cite|improve this question

edited Mar 23 at 13:51

Allawonder

2,281616

asked Mar 23 at 11:23

Ashok SharmaAshok Sharma

111

edited Mar 23 at 13:51

Allawonder

2,281616

edited Mar 23 at 13:51

Allawonder

2,281616

asked Mar 23 at 11:23

Ashok SharmaAshok Sharma

111

asked Mar 23 at 11:23

Ashok SharmaAshok Sharma

111

2 Answers
2

active

oldest

votes

0

$begingroup$

There is a relative minimum at $0$. But there are also relative maxima at $2$ and at $-2$.

The definition for a relative maximum at a point $x_0$ is that $f(x)leq f(x_0)$ for all $x$ in a sufficiently small neighborhood of $x_0$ (intersected with the function's domain).

For $x_0=2$, for example, it is true that $f(x)leq f(2)$ for all $x$ in the sufficiently small intervall $(1.5,2.5)cap[-2,2]=(1.5,2]$ around the point $x_0=2$. So, it is a relative maximum.

There is no reason to believe that relative extrema cannot occur at the boundaries of intervals.

share|cite|improve this answer

edited Mar 23 at 12:29

answered Mar 23 at 11:28

st.mathst.math

1,233115

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AshokSharma I have edited my answer to make it more clear. :)
  $endgroup$
  – st.math
  Mar 23 at 11:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Allawonder The common definition is that there is a maximum at $x_0$ if and only if $$exists_{delta>0}forall_{xin U_delta(x_0)cap D},f(x_0)geq f(x)$$ if $D$ is the domain of the function. I don't see a mistake there.
  $endgroup$
  – st.math
  Mar 23 at 12:25
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Allawonder In two calculus scripts of my university, this is the exact definition of a relative (!) maximum. (Neighborhoods of a point intersected with the function's domain.) The fact that there are different definitions of concepts is not a reason to blame another person's answer as false or confusing.
  $endgroup$
  – st.math
  Mar 23 at 12:34
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Because the definition does not force it to be a neighborhood, but a non-empty set which results from a neighborhood intersected with the domain (in this case: $(1.5,2]$).
  $endgroup$
  – st.math
  Mar 23 at 12:39
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, it does not. See my comment to your answer.
  $endgroup$
  – st.math
  Mar 23 at 14:13
  

  
  
  
  

 | 
show 7 more comments

-2

$begingroup$

The answer is correct. Why it is so follows immediately from the definition.

There's a difference between an extreme value and a relative extreme value of a function at a point. While your function has a maximum value at $pm2,$ it does not have a relative maximum at all. The reason is that relative extreme values by definition can only be defined at a point $c$ first, when the function is defined over an interval of the form $[c-delta,c+delta],$ where $delta>0.$ Another example is $sqrt x,$ which has no relative minimum at $x=0,$ though it has a minimum there.

In mathematics, the words are very important and never superfluous. Pay attention to such innocuous qualifiers as relative.

Good luck in your studies.

share|cite|improve this answer

answered Mar 23 at 12:25

AllawonderAllawonder

2,281616

$endgroup$

add a comment | 

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0

$begingroup$

There is a relative minimum at $0$. But there are also relative maxima at $2$ and at $-2$.

The definition for a relative maximum at a point $x_0$ is that $f(x)leq f(x_0)$ for all $x$ in a sufficiently small neighborhood of $x_0$ (intersected with the function's domain).

For $x_0=2$, for example, it is true that $f(x)leq f(2)$ for all $x$ in the sufficiently small intervall $(1.5,2.5)cap[-2,2]=(1.5,2]$ around the point $x_0=2$. So, it is a relative maximum.

There is no reason to believe that relative extrema cannot occur at the boundaries of intervals.

share|cite|improve this answer

edited Mar 23 at 12:29

answered Mar 23 at 11:28

st.mathst.math

1,233115

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AshokSharma I have edited my answer to make it more clear. :)
  $endgroup$
  – st.math
  Mar 23 at 11:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Allawonder The common definition is that there is a maximum at $x_0$ if and only if $$exists_{delta>0}forall_{xin U_delta(x_0)cap D},f(x_0)geq f(x)$$ if $D$ is the domain of the function. I don't see a mistake there.
  $endgroup$
  – st.math
  Mar 23 at 12:25
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Allawonder In two calculus scripts of my university, this is the exact definition of a relative (!) maximum. (Neighborhoods of a point intersected with the function's domain.) The fact that there are different definitions of concepts is not a reason to blame another person's answer as false or confusing.
  $endgroup$
  – st.math
  Mar 23 at 12:34
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Because the definition does not force it to be a neighborhood, but a non-empty set which results from a neighborhood intersected with the domain (in this case: $(1.5,2]$).
  $endgroup$
  – st.math
  Mar 23 at 12:39
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, it does not. See my comment to your answer.
  $endgroup$
  – st.math
  Mar 23 at 14:13
  

  
  
  
  

 | 
show 7 more comments

0

$begingroup$

There is a relative minimum at $0$. But there are also relative maxima at $2$ and at $-2$.

The definition for a relative maximum at a point $x_0$ is that $f(x)leq f(x_0)$ for all $x$ in a sufficiently small neighborhood of $x_0$ (intersected with the function's domain).

For $x_0=2$, for example, it is true that $f(x)leq f(2)$ for all $x$ in the sufficiently small intervall $(1.5,2.5)cap[-2,2]=(1.5,2]$ around the point $x_0=2$. So, it is a relative maximum.

There is no reason to believe that relative extrema cannot occur at the boundaries of intervals.

share|cite|improve this answer

edited Mar 23 at 12:29

answered Mar 23 at 11:28

st.mathst.math

1,233115

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AshokSharma I have edited my answer to make it more clear. :)
  $endgroup$
  – st.math
  Mar 23 at 11:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Allawonder The common definition is that there is a maximum at $x_0$ if and only if $$exists_{delta>0}forall_{xin U_delta(x_0)cap D},f(x_0)geq f(x)$$ if $D$ is the domain of the function. I don't see a mistake there.
  $endgroup$
  – st.math
  Mar 23 at 12:25
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Allawonder In two calculus scripts of my university, this is the exact definition of a relative (!) maximum. (Neighborhoods of a point intersected with the function's domain.) The fact that there are different definitions of concepts is not a reason to blame another person's answer as false or confusing.
  $endgroup$
  – st.math
  Mar 23 at 12:34
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Because the definition does not force it to be a neighborhood, but a non-empty set which results from a neighborhood intersected with the domain (in this case: $(1.5,2]$).
  $endgroup$
  – st.math
  Mar 23 at 12:39
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, it does not. See my comment to your answer.
  $endgroup$
  – st.math
  Mar 23 at 14:13
  

  
  
  
  

 | 
show 7 more comments

$begingroup$

There is a relative minimum at $0$. But there are also relative maxima at $2$ and at $-2$.

The definition for a relative maximum at a point $x_0$ is that $f(x)leq f(x_0)$ for all $x$ in a sufficiently small neighborhood of $x_0$ (intersected with the function's domain).

For $x_0=2$, for example, it is true that $f(x)leq f(2)$ for all $x$ in the sufficiently small intervall $(1.5,2.5)cap[-2,2]=(1.5,2]$ around the point $x_0=2$. So, it is a relative maximum.

There is no reason to believe that relative extrema cannot occur at the boundaries of intervals.

share|cite|improve this answer

edited Mar 23 at 12:29

answered Mar 23 at 11:28

st.mathst.math

1,233115

$endgroup$

share|cite|improve this answer

edited Mar 23 at 12:29

answered Mar 23 at 11:28

st.mathst.math

1,233115

answered Mar 23 at 11:28

st.mathst.math

1,233115

answered Mar 23 at 11:28

st.mathst.math

1,233115

-2

$begingroup$

The answer is correct. Why it is so follows immediately from the definition.

There's a difference between an extreme value and a relative extreme value of a function at a point. While your function has a maximum value at $pm2,$ it does not have a relative maximum at all. The reason is that relative extreme values by definition can only be defined at a point $c$ first, when the function is defined over an interval of the form $[c-delta,c+delta],$ where $delta>0.$ Another example is $sqrt x,$ which has no relative minimum at $x=0,$ though it has a minimum there.

In mathematics, the words are very important and never superfluous. Pay attention to such innocuous qualifiers as relative.

Good luck in your studies.

share|cite|improve this answer

answered Mar 23 at 12:25

AllawonderAllawonder

2,281616

$endgroup$

add a comment | 

-2

$begingroup$

The answer is correct. Why it is so follows immediately from the definition.

There's a difference between an extreme value and a relative extreme value of a function at a point. While your function has a maximum value at $pm2,$ it does not have a relative maximum at all. The reason is that relative extreme values by definition can only be defined at a point $c$ first, when the function is defined over an interval of the form $[c-delta,c+delta],$ where $delta>0.$ Another example is $sqrt x,$ which has no relative minimum at $x=0,$ though it has a minimum there.

In mathematics, the words are very important and never superfluous. Pay attention to such innocuous qualifiers as relative.

Good luck in your studies.

share|cite|improve this answer

answered Mar 23 at 12:25

AllawonderAllawonder

2,281616

$endgroup$

add a comment | 

$begingroup$

The answer is correct. Why it is so follows immediately from the definition.

There's a difference between an extreme value and a relative extreme value of a function at a point. While your function has a maximum value at $pm2,$ it does not have a relative maximum at all. The reason is that relative extreme values by definition can only be defined at a point $c$ first, when the function is defined over an interval of the form $[c-delta,c+delta],$ where $delta>0.$ Another example is $sqrt x,$ which has no relative minimum at $x=0,$ though it has a minimum there.

In mathematics, the words are very important and never superfluous. Pay attention to such innocuous qualifiers as relative.

Good luck in your studies.

share|cite|improve this answer

answered Mar 23 at 12:25

AllawonderAllawonder

2,281616

$endgroup$

share|cite|improve this answer

answered Mar 23 at 12:25

AllawonderAllawonder

2,281616

answered Mar 23 at 12:25

AllawonderAllawonder

2,281616

answered Mar 23 at 12:25

AllawonderAllawonder

2,281616