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Are the endpoints of the domain of a function counted as critical points?

1

$begingroup$

Do the end points of a domain come under critical points ?
I know we say critical point is a point where the derivative is zero or the derivative doesn't exist.

For ex:-
$$ f:[0,pi] rightarrow [-1,1] , f(x) = sin(x)$$
Does this have 1 critical point or 3 critical points(0 and $pi$ included) ?

NOTE:- This question is limited to only Single Variable Functions.Although I really would love an insight to this for Multivariable as well.

calculus

share|cite|improve this question

edited 4 hours ago

Michael Rybkin

4,504522

asked 4 hours ago

rajdeep dhingrarajdeep dhingra

405

$endgroup$

add a comment | 

1

$begingroup$

Do the end points of a domain come under critical points ?
I know we say critical point is a point where the derivative is zero or the derivative doesn't exist.

For ex:-
$$ f:[0,pi] rightarrow [-1,1] , f(x) = sin(x)$$
Does this have 1 critical point or 3 critical points(0 and $pi$ included) ?

NOTE:- This question is limited to only Single Variable Functions.Although I really would love an insight to this for Multivariable as well.

calculus

share|cite|improve this question

edited 4 hours ago

Michael Rybkin

4,504522

asked 4 hours ago

rajdeep dhingrarajdeep dhingra

405

$endgroup$

add a comment | 

$begingroup$

Do the end points of a domain come under critical points ?
I know we say critical point is a point where the derivative is zero or the derivative doesn't exist.

For ex:-
$$ f:[0,pi] rightarrow [-1,1] , f(x) = sin(x)$$
Does this have 1 critical point or 3 critical points(0 and $pi$ included) ?

NOTE:- This question is limited to only Single Variable Functions.Although I really would love an insight to this for Multivariable as well.

calculus

share|cite|improve this question

edited 4 hours ago

Michael Rybkin

4,504522

asked 4 hours ago

rajdeep dhingrarajdeep dhingra

405

$endgroup$

share|cite|improve this question

edited 4 hours ago

Michael Rybkin

4,504522

asked 4 hours ago

rajdeep dhingrarajdeep dhingra

405

share|cite|improve this question

edited 4 hours ago

Michael Rybkin

4,504522

asked 4 hours ago

rajdeep dhingrarajdeep dhingra

405

edited 4 hours ago

Michael Rybkin

4,504522

edited 4 hours ago

Michael Rybkin

4,504522

asked 4 hours ago

rajdeep dhingrarajdeep dhingra

405

asked 4 hours ago

rajdeep dhingrarajdeep dhingra

405

2 Answers
2

active

oldest

votes

2

$begingroup$

Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sin{x}, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sin{x}, xin[0,pi]$ is non-differentiable at those points.

Do you remember what it means for a function to be differentiable at a point? The function has to have a derivative at that point. What is the derivative of the function $f(x)=sin{x}, xin[0,pi]$ at $x=0$? Well, it should be:

$$
lim_{xto0}frac{sin{x}-sin{0}}{x-0}=lim_{xto0}frac{sin{x}}{x}
$$

Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):

$$
lim_{xto0^-}frac{sin{x}}{x}, lim_{xto0^+}frac{sin{x}}{x}
$$
But the first of those two limits for all intents and purposes is nonexistent because all x values that lie to the left of $0$ are not in the domain of the function $f(x)=sin{x}, xin[0,pi]$. For a limit to exist, you need two one-sided limits. But you've got only one! Thus, the derivative at $x=0$ does not exist which makes it a critical number. The exact same idea applies to the other endpoint.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,504522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get your logic. But don't we have a definition for derivative at end points. We only take one sided derivatives at the end points. Do you have any book/source to back up your answer ? Thank you for your help
  $endgroup$
  – rajdeep dhingra
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you can look it up on Wikipedia: en.wikipedia.org/wiki/Differentiable_function Wikipedia is more or less a reliable and authoritative source of information when it comes to mathematics.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They have mentioned on that Wikipedia page that , the they are taking in an open interval U. Whereas what I am asking here is a closed interval case.
  $endgroup$
  – rajdeep dhingra
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It says quite clearly at the beginning of the second paragraph: More generally, if $x_0$ is a point in the domain of a function $f$, then $f$ is said to be differentiable at $x_0$ if the derivative $f '(x_0)$ exists. It follows that if the point $x_0$ is in the domain of the function and the derivative $f '(x_0)$ does not exist, then the function is non-differentiable at that point.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To put it simply, a derivative is basically a formula for the slope of the tangent line to a curve. If there is no curve to speak of to the left of a point, there cannot be a tangle line at that point.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  

 | 
show 4 more comments

2

$begingroup$

$$f'(x) = cos(x) = 0 iff x = frac{pi}{2}$$
The function $f$ has only one critical point, since the derivative ($f'(x) = cos(x)$) exists at the endpoints. ($0$ and $pi$.)

share|cite|improve this answer

answered 4 hours ago

GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

14.1k82651

$endgroup$

add a comment | 

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2

$begingroup$

Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sin{x}, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sin{x}, xin[0,pi]$ is non-differentiable at those points.

Do you remember what it means for a function to be differentiable at a point? The function has to have a derivative at that point. What is the derivative of the function $f(x)=sin{x}, xin[0,pi]$ at $x=0$? Well, it should be:

$$
lim_{xto0}frac{sin{x}-sin{0}}{x-0}=lim_{xto0}frac{sin{x}}{x}
$$

Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):

$$
lim_{xto0^-}frac{sin{x}}{x}, lim_{xto0^+}frac{sin{x}}{x}
$$
But the first of those two limits for all intents and purposes is nonexistent because all x values that lie to the left of $0$ are not in the domain of the function $f(x)=sin{x}, xin[0,pi]$. For a limit to exist, you need two one-sided limits. But you've got only one! Thus, the derivative at $x=0$ does not exist which makes it a critical number. The exact same idea applies to the other endpoint.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,504522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get your logic. But don't we have a definition for derivative at end points. We only take one sided derivatives at the end points. Do you have any book/source to back up your answer ? Thank you for your help
  $endgroup$
  – rajdeep dhingra
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you can look it up on Wikipedia: en.wikipedia.org/wiki/Differentiable_function Wikipedia is more or less a reliable and authoritative source of information when it comes to mathematics.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They have mentioned on that Wikipedia page that , the they are taking in an open interval U. Whereas what I am asking here is a closed interval case.
  $endgroup$
  – rajdeep dhingra
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It says quite clearly at the beginning of the second paragraph: More generally, if $x_0$ is a point in the domain of a function $f$, then $f$ is said to be differentiable at $x_0$ if the derivative $f '(x_0)$ exists. It follows that if the point $x_0$ is in the domain of the function and the derivative $f '(x_0)$ does not exist, then the function is non-differentiable at that point.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To put it simply, a derivative is basically a formula for the slope of the tangent line to a curve. If there is no curve to speak of to the left of a point, there cannot be a tangle line at that point.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  

 | 
show 4 more comments

2

$begingroup$

Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sin{x}, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sin{x}, xin[0,pi]$ is non-differentiable at those points.

Do you remember what it means for a function to be differentiable at a point? The function has to have a derivative at that point. What is the derivative of the function $f(x)=sin{x}, xin[0,pi]$ at $x=0$? Well, it should be:

$$
lim_{xto0}frac{sin{x}-sin{0}}{x-0}=lim_{xto0}frac{sin{x}}{x}
$$

Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):

$$
lim_{xto0^-}frac{sin{x}}{x}, lim_{xto0^+}frac{sin{x}}{x}
$$
But the first of those two limits for all intents and purposes is nonexistent because all x values that lie to the left of $0$ are not in the domain of the function $f(x)=sin{x}, xin[0,pi]$. For a limit to exist, you need two one-sided limits. But you've got only one! Thus, the derivative at $x=0$ does not exist which makes it a critical number. The exact same idea applies to the other endpoint.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,504522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get your logic. But don't we have a definition for derivative at end points. We only take one sided derivatives at the end points. Do you have any book/source to back up your answer ? Thank you for your help
  $endgroup$
  – rajdeep dhingra
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you can look it up on Wikipedia: en.wikipedia.org/wiki/Differentiable_function Wikipedia is more or less a reliable and authoritative source of information when it comes to mathematics.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They have mentioned on that Wikipedia page that , the they are taking in an open interval U. Whereas what I am asking here is a closed interval case.
  $endgroup$
  – rajdeep dhingra
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It says quite clearly at the beginning of the second paragraph: More generally, if $x_0$ is a point in the domain of a function $f$, then $f$ is said to be differentiable at $x_0$ if the derivative $f '(x_0)$ exists. It follows that if the point $x_0$ is in the domain of the function and the derivative $f '(x_0)$ does not exist, then the function is non-differentiable at that point.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To put it simply, a derivative is basically a formula for the slope of the tangent line to a curve. If there is no curve to speak of to the left of a point, there cannot be a tangle line at that point.
  $endgroup$
  – Michael Rybkin
  4 hours ago
  
  
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sin{x}, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sin{x}, xin[0,pi]$ is non-differentiable at those points.

Do you remember what it means for a function to be differentiable at a point? The function has to have a derivative at that point. What is the derivative of the function $f(x)=sin{x}, xin[0,pi]$ at $x=0$? Well, it should be:

$$
lim_{xto0}frac{sin{x}-sin{0}}{x-0}=lim_{xto0}frac{sin{x}}{x}
$$

Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):

$$
lim_{xto0^-}frac{sin{x}}{x}, lim_{xto0^+}frac{sin{x}}{x}
$$
But the first of those two limits for all intents and purposes is nonexistent because all x values that lie to the left of $0$ are not in the domain of the function $f(x)=sin{x}, xin[0,pi]$. For a limit to exist, you need two one-sided limits. But you've got only one! Thus, the derivative at $x=0$ does not exist which makes it a critical number. The exact same idea applies to the other endpoint.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,504522

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,504522

answered 4 hours ago

Michael RybkinMichael Rybkin

4,504522

answered 4 hours ago

Michael RybkinMichael Rybkin

4,504522

2

$begingroup$

$$f'(x) = cos(x) = 0 iff x = frac{pi}{2}$$
The function $f$ has only one critical point, since the derivative ($f'(x) = cos(x)$) exists at the endpoints. ($0$ and $pi$.)

share|cite|improve this answer

answered 4 hours ago

GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

14.1k82651

$endgroup$

add a comment | 

2

$begingroup$

$$f'(x) = cos(x) = 0 iff x = frac{pi}{2}$$
The function $f$ has only one critical point, since the derivative ($f'(x) = cos(x)$) exists at the endpoints. ($0$ and $pi$.)

share|cite|improve this answer

answered 4 hours ago

GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

14.1k82651

$endgroup$

add a comment | 

$begingroup$

$$f'(x) = cos(x) = 0 iff x = frac{pi}{2}$$
The function $f$ has only one critical point, since the derivative ($f'(x) = cos(x)$) exists at the endpoints. ($0$ and $pi$.)

share|cite|improve this answer

answered 4 hours ago

GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

14.1k82651

$endgroup$

share|cite|improve this answer

answered 4 hours ago

GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

14.1k82651

answered 4 hours ago

GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

14.1k82651

answered 4 hours ago

GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

14.1k82651