Are the endpoints of the domain of a function counted as critical points? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Can critical points occur at endpoints? E.g. $f(x) = frac1x$ at the interval $[1,4]$Debate on definition of Critical point and local / absolute extremasCritical points for undefined fraction on closed intervalCritical points when gradient doesn't existA silly problem on critical points?What is the definition of a Critical Point?Is it correct to say all extrema happen at critical points but not all critical points are extrema?Critical Point when not in domain of $f(x)$How do I find and classify the critical points of $x^2 sin (frac 1x)$Definition of Critical Point at endpointsFinding the derivative of function with domain empty setDetermine critical points of 2 variable functions without 2nd derivative test
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Are the endpoints of the domain of a function counted as critical points?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Can critical points occur at endpoints? E.g. $f(x) = frac1x$ at the interval $[1,4]$Debate on definition of Critical point and local / absolute extremasCritical points for undefined fraction on closed intervalCritical points when gradient doesn't existA silly problem on critical points?What is the definition of a Critical Point?Is it correct to say all extrema happen at critical points but not all critical points are extrema?Critical Point when not in domain of $f(x)$How do I find and classify the critical points of $x^2 sin (frac 1x)$Definition of Critical Point at endpointsFinding the derivative of function with domain empty setDetermine critical points of 2 variable functions without 2nd derivative test
$begingroup$
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 example:
$$ f:[0,pi] to [-1,1], f(x) = sin(x).$$
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 definition
$endgroup$
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 example:
$$ f:[0,pi] to [-1,1], f(x) = sin(x).$$
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 definition
$endgroup$
$begingroup$
Same question with different example: math.stackexchange.com/q/2880307/290189, for multivariable case: math.stackexchange.com/q/2322997/290189
$endgroup$
– GNUSupporter 8964民主女神 地下教會
3 hours ago
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 example:
$$ f:[0,pi] to [-1,1], f(x) = sin(x).$$
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 definition
$endgroup$
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 example:
$$ f:[0,pi] to [-1,1], f(x) = sin(x).$$
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 definition
calculus definition
edited 3 hours ago
GNUSupporter 8964民主女神 地下教會
14.2k82651
14.2k82651
asked 8 hours ago
rajdeep dhingrarajdeep dhingra
505
505
$begingroup$
Same question with different example: math.stackexchange.com/q/2880307/290189, for multivariable case: math.stackexchange.com/q/2322997/290189
$endgroup$
– GNUSupporter 8964民主女神 地下教會
3 hours ago
add a comment |
$begingroup$
Same question with different example: math.stackexchange.com/q/2880307/290189, for multivariable case: math.stackexchange.com/q/2322997/290189
$endgroup$
– GNUSupporter 8964民主女神 地下教會
3 hours ago
$begingroup$
Same question with different example: math.stackexchange.com/q/2880307/290189, for multivariable case: math.stackexchange.com/q/2322997/290189
$endgroup$
– GNUSupporter 8964民主女神 地下教會
3 hours ago
$begingroup$
Same question with different example: math.stackexchange.com/q/2880307/290189, for multivariable case: math.stackexchange.com/q/2322997/290189
$endgroup$
– GNUSupporter 8964民主女神 地下教會
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Edited
$$f'(x) = cos(x) = 0 iff x = fracpi2$$
The function $f$ has three critical points.
- A local maximum: $x = pi/2$ (at which $f(pi/2) = 1$.)
- The endpoints of the domain of $f$ (that is, $[0,pi]$): $x = 0$ and $x = pi$ .
Since the other answer has elaborated on OP's understanding on the definitions using the differentiability of $f$, there's no point repeating its arguments. Instead, I'll cite from a university course web page to show why we need to include the endpoints of the domain of $f$ if $f$ is defined at those points. By doing so, we learn the definitions by heart instead of by memory.
The goal of the procedure of finding critical points is to identify points in the domain at which a (global and/or local) extremum could possibly occur.
- vanishing derivatives:
- endpoints of interval: (image source: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png)
- derivative undefined: , including points of discontinuity
Source: © CalculusQuest™
$endgroup$
1
$begingroup$
This is much much better. Thanks
$endgroup$
– rajdeep dhingra
3 hours ago
$begingroup$
@Carmeister Thanks for suggestion. Edit done.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
2 hours ago
add a comment |
$begingroup$
Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sinx, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sinx, xin[0,pi]$ 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 $f(x)=sinx, xin[0,pi]$ at $x=0$? Well, it should be:
$$
lim_xto0fracsinx-sin0x-0=lim_xto0fracsinxx
$$
Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):
$$
lim_xto0^-fracsinxx, lim_xto0^+fracsinxx
$$
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 $f(x)=sinx, xin[0,pi]$. 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.
$endgroup$
$begingroup$
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
7 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
7 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
7 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
7 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
7 hours ago
|
show 5 more comments
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2 Answers
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$begingroup$
Edited
$$f'(x) = cos(x) = 0 iff x = fracpi2$$
The function $f$ has three critical points.
- A local maximum: $x = pi/2$ (at which $f(pi/2) = 1$.)
- The endpoints of the domain of $f$ (that is, $[0,pi]$): $x = 0$ and $x = pi$ .
Since the other answer has elaborated on OP's understanding on the definitions using the differentiability of $f$, there's no point repeating its arguments. Instead, I'll cite from a university course web page to show why we need to include the endpoints of the domain of $f$ if $f$ is defined at those points. By doing so, we learn the definitions by heart instead of by memory.
The goal of the procedure of finding critical points is to identify points in the domain at which a (global and/or local) extremum could possibly occur.
- vanishing derivatives:
- endpoints of interval: (image source: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png)
- derivative undefined: , including points of discontinuity
Source: © CalculusQuest™
$endgroup$
1
$begingroup$
This is much much better. Thanks
$endgroup$
– rajdeep dhingra
3 hours ago
$begingroup$
@Carmeister Thanks for suggestion. Edit done.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
2 hours ago
add a comment |
$begingroup$
Edited
$$f'(x) = cos(x) = 0 iff x = fracpi2$$
The function $f$ has three critical points.
- A local maximum: $x = pi/2$ (at which $f(pi/2) = 1$.)
- The endpoints of the domain of $f$ (that is, $[0,pi]$): $x = 0$ and $x = pi$ .
Since the other answer has elaborated on OP's understanding on the definitions using the differentiability of $f$, there's no point repeating its arguments. Instead, I'll cite from a university course web page to show why we need to include the endpoints of the domain of $f$ if $f$ is defined at those points. By doing so, we learn the definitions by heart instead of by memory.
The goal of the procedure of finding critical points is to identify points in the domain at which a (global and/or local) extremum could possibly occur.
- vanishing derivatives:
- endpoints of interval: (image source: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png)
- derivative undefined: , including points of discontinuity
Source: © CalculusQuest™
$endgroup$
1
$begingroup$
This is much much better. Thanks
$endgroup$
– rajdeep dhingra
3 hours ago
$begingroup$
@Carmeister Thanks for suggestion. Edit done.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
2 hours ago
add a comment |
$begingroup$
Edited
$$f'(x) = cos(x) = 0 iff x = fracpi2$$
The function $f$ has three critical points.
- A local maximum: $x = pi/2$ (at which $f(pi/2) = 1$.)
- The endpoints of the domain of $f$ (that is, $[0,pi]$): $x = 0$ and $x = pi$ .
Since the other answer has elaborated on OP's understanding on the definitions using the differentiability of $f$, there's no point repeating its arguments. Instead, I'll cite from a university course web page to show why we need to include the endpoints of the domain of $f$ if $f$ is defined at those points. By doing so, we learn the definitions by heart instead of by memory.
The goal of the procedure of finding critical points is to identify points in the domain at which a (global and/or local) extremum could possibly occur.
- vanishing derivatives:
- endpoints of interval: (image source: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png)
- derivative undefined: , including points of discontinuity
Source: © CalculusQuest™
$endgroup$
Edited
$$f'(x) = cos(x) = 0 iff x = fracpi2$$
The function $f$ has three critical points.
- A local maximum: $x = pi/2$ (at which $f(pi/2) = 1$.)
- The endpoints of the domain of $f$ (that is, $[0,pi]$): $x = 0$ and $x = pi$ .
Since the other answer has elaborated on OP's understanding on the definitions using the differentiability of $f$, there's no point repeating its arguments. Instead, I'll cite from a university course web page to show why we need to include the endpoints of the domain of $f$ if $f$ is defined at those points. By doing so, we learn the definitions by heart instead of by memory.
The goal of the procedure of finding critical points is to identify points in the domain at which a (global and/or local) extremum could possibly occur.
- vanishing derivatives:
- endpoints of interval: (image source: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues_Files/image002.png)
- derivative undefined: , including points of discontinuity
Source: © CalculusQuest™
edited 2 hours ago
answered 8 hours ago
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14.2k82651
14.2k82651
1
$begingroup$
This is much much better. Thanks
$endgroup$
– rajdeep dhingra
3 hours ago
$begingroup$
@Carmeister Thanks for suggestion. Edit done.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
2 hours ago
add a comment |
1
$begingroup$
This is much much better. Thanks
$endgroup$
– rajdeep dhingra
3 hours ago
$begingroup$
@Carmeister Thanks for suggestion. Edit done.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
2 hours ago
1
1
$begingroup$
This is much much better. Thanks
$endgroup$
– rajdeep dhingra
3 hours ago
$begingroup$
This is much much better. Thanks
$endgroup$
– rajdeep dhingra
3 hours ago
$begingroup$
@Carmeister Thanks for suggestion. Edit done.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
2 hours ago
$begingroup$
@Carmeister Thanks for suggestion. Edit done.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
2 hours ago
add a comment |
$begingroup$
Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sinx, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sinx, xin[0,pi]$ 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 $f(x)=sinx, xin[0,pi]$ at $x=0$? Well, it should be:
$$
lim_xto0fracsinx-sin0x-0=lim_xto0fracsinxx
$$
Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):
$$
lim_xto0^-fracsinxx, lim_xto0^+fracsinxx
$$
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 $f(x)=sinx, xin[0,pi]$. 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.
$endgroup$
$begingroup$
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
7 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
7 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
7 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
7 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
7 hours ago
|
show 5 more comments
$begingroup$
Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sinx, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sinx, xin[0,pi]$ 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 $f(x)=sinx, xin[0,pi]$ at $x=0$? Well, it should be:
$$
lim_xto0fracsinx-sin0x-0=lim_xto0fracsinxx
$$
Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):
$$
lim_xto0^-fracsinxx, lim_xto0^+fracsinxx
$$
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 $f(x)=sinx, xin[0,pi]$. 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.
$endgroup$
$begingroup$
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
7 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
7 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
7 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
7 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
7 hours ago
|
show 5 more comments
$begingroup$
Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sinx, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sinx, xin[0,pi]$ 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 $f(x)=sinx, xin[0,pi]$ at $x=0$? Well, it should be:
$$
lim_xto0fracsinx-sin0x-0=lim_xto0fracsinxx
$$
Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):
$$
lim_xto0^-fracsinxx, lim_xto0^+fracsinxx
$$
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 $f(x)=sinx, xin[0,pi]$. 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.
$endgroup$
Yes, the function has 3 critical numbers. One is where the derivative of the function $f(x)=sinx, xin[0,pi]$ is zero and the other two happen to be the endpoints $x=0$ and $x=pi$ because the function $f(x)=sinx, xin[0,pi]$ 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 $f(x)=sinx, xin[0,pi]$ at $x=0$? Well, it should be:
$$
lim_xto0fracsinx-sin0x-0=lim_xto0fracsinxx
$$
Which is nothing more than two one-sided limits (if those two limits exist and are equal to each other, the limit itself exists):
$$
lim_xto0^-fracsinxx, lim_xto0^+fracsinxx
$$
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 $f(x)=sinx, xin[0,pi]$. 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.
edited 3 hours ago
GNUSupporter 8964民主女神 地下教會
14.2k82651
14.2k82651
answered 7 hours ago
Michael RybkinMichael Rybkin
4,524522
4,524522
$begingroup$
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
7 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
7 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
7 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
7 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.
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– Michael Rybkin
7 hours ago
|
show 5 more comments
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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
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– rajdeep dhingra
7 hours ago
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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.
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– Michael Rybkin
7 hours ago
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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
7 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
7 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
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
7 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
7 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
7 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
7 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
7 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
7 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
7 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
7 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
7 hours ago
|
show 5 more comments
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Same question with different example: math.stackexchange.com/q/2880307/290189, for multivariable case: math.stackexchange.com/q/2322997/290189
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– GNUSupporter 8964民主女神 地下教會
3 hours ago