Geometrical interpretation of differentiability
$begingroup$
We check the differentiability of $f(x) = xleft| x right| $ at $x=0$.
Evaluating derivatives from first principles, we find that the left hand and right hand derivatives both equate to $0$, which in principle proves that the function is differentiable at that point.
Now, let us try to interpret this geometrically.
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. This means that the graph of $f$ has a non-vertical tangent line at the point $(x_{0}, f(x_{0}))$.
Source: Wikipedia
If so, where is the so-called unique tangent at $x=0$?
Source: WolframAlpha
Please find flaws in my interpretation, if any.
Thank you.
calculus derivatives
asked Jun 2 '18 at 10:55
Swapnil DasSwapnil Das
614517
$endgroup$
add a comment |
$begingroup$
We check the differentiability of $f(x) = xleft| x right| $ at $x=0$.
Evaluating derivatives from first principles, we find that the left hand and right hand derivatives both equate to $0$, which in principle proves that the function is differentiable at that point.
Now, let us try to interpret this geometrically.
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. This means that the graph of $f$ has a non-vertical tangent line at the point $(x_{0}, f(x_{0}))$.
Source: Wikipedia
If so, where is the so-called unique tangent at $x=0$?
Source: WolframAlpha
Please find flaws in my interpretation, if any.
Thank you.
calculus derivatives
asked Jun 2 '18 at 10:55
Swapnil DasSwapnil Das
614517
$endgroup$
add a comment |
$begingroup$
We check the differentiability of $f(x) = xleft| x right| $ at $x=0$.
Evaluating derivatives from first principles, we find that the left hand and right hand derivatives both equate to $0$, which in principle proves that the function is differentiable at that point.
Now, let us try to interpret this geometrically.
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. This means that the graph of $f$ has a non-vertical tangent line at the point $(x_{0}, f(x_{0}))$.
Source: Wikipedia
If so, where is the so-called unique tangent at $x=0$?
Source: WolframAlpha
Please find flaws in my interpretation, if any.
Thank you.
calculus derivatives
asked Jun 2 '18 at 10:55
Swapnil DasSwapnil Das
614517
$endgroup$
We check the differentiability of $f(x) = xleft| x right| $ at $x=0$.
Evaluating derivatives from first principles, we find that the left hand and right hand derivatives both equate to $0$, which in principle proves that the function is differentiable at that point.
Now, let us try to interpret this geometrically.
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. This means that the graph of $f$ has a non-vertical tangent line at the point $(x_{0}, f(x_{0}))$.
Source: Wikipedia
If so, where is the so-called unique tangent at $x=0$?
Source: WolframAlpha
Please find flaws in my interpretation, if any.
Thank you.
calculus derivatives
calculus derivatives
asked Jun 2 '18 at 10:55
Swapnil DasSwapnil Das
614517
asked Jun 2 '18 at 10:55
Swapnil DasSwapnil Das
614517
asked Jun 2 '18 at 10:55
Swapnil DasSwapnil Das
614517
asked Jun 2 '18 at 10:55
Swapnil DasSwapnil Das
614517
asked Jun 2 '18 at 10:55
Swapnil DasSwapnil Das
614517
614517
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Geometrically the derivative at a point can be interpreted for strictly convex or strictly concave graphs easily: the derivative at a point $(c,f(c))$ of a strictly convex (or strictly concave) graph exists if there is a unique straight line $g(x):= ax+b$ that intersect the graph of $f$ at the point $(c,f(c))$ such that $f(x)ge g(x)$ for all $x$ if $f$ is convex, or $f(x)le g(x)$ for all $x$ when $f$ is concave. Then we says that $a$ is the derivative of $f$ at $c$, that is, that $f'(c)=a$.
For other functions, not necessarily strictly concave or strictly convex, the geometric interpretation is not so clear because there are infinite straight lines that can cut the graph of a function uniquely at a point as in the function of your picture.
But the geometric definition of the derivative at points of strictly convex or strictly concave graphs define analitically the derivative through a limit, and we can extend this definition to other cases not necessarily strictly concave or strictly convex.
Note that there are strictly convex and strictly concave graphs of functions that doesn't have derivative at some points, because such straight line doesn't exists or because there are more than one straight line that intersect the graph of the function uniquely in some point.
answered Jun 2 '18 at 11:12
MasacrosoMasacroso
13k41746
$endgroup$
$begingroup$
Nice answer. Can you provide us some examples of those 'patological' cases, please?
$endgroup$
– Dog_69
Jun 2 '18 at 16:34
2
$begingroup$
@Dog_69 sorry, my previous comment was wrong :P. The function $$f(x):=begin{cases}x^2,&x<0\(x+1)^2-1,&xge 0end{cases}$$ is strictly convex but is not differentiable at zero because both straight lines $g_1(x):=0$ and $g_2(x):=2x$ cut the graph uniquely at the point $(0,0)$. Indeed any straight line of the form $g(x):=kx$ for $kin[0,2]$ does.
$endgroup$
– Masacroso
Jun 2 '18 at 17:05
$begingroup$
don't worry. Thanks for the example.
$endgroup$
– Dog_69
Jun 2 '18 at 18:15
add a comment |
$begingroup$
If you want to see the tangent at x=0, go to a graph plotter app (DESMOS), plot your graph (eg. x^3) and zoom it at x=0. Zoom upto that extent until the line becomes horizontal. This acts as the tangent at that point.
Isn't it amazing that a curve becomes parallel if zoomed to a large extent. Think it in a sense that Earth appears to be flat (because we are so small that Earth appears to be zoomed), but actually its not. So the flat surface acts as a tangent at any point where you are standing.
Hope you are clear with the geometrical approach.
answered Jun 2 '18 at 11:20
Yash MittalYash Mittal
1
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Geometrically the derivative at a point can be interpreted for strictly convex or strictly concave graphs easily: the derivative at a point $(c,f(c))$ of a strictly convex (or strictly concave) graph exists if there is a unique straight line $g(x):= ax+b$ that intersect the graph of $f$ at the point $(c,f(c))$ such that $f(x)ge g(x)$ for all $x$ if $f$ is convex, or $f(x)le g(x)$ for all $x$ when $f$ is concave. Then we says that $a$ is the derivative of $f$ at $c$, that is, that $f'(c)=a$.
For other functions, not necessarily strictly concave or strictly convex, the geometric interpretation is not so clear because there are infinite straight lines that can cut the graph of a function uniquely at a point as in the function of your picture.
But the geometric definition of the derivative at points of strictly convex or strictly concave graphs define analitically the derivative through a limit, and we can extend this definition to other cases not necessarily strictly concave or strictly convex.
Note that there are strictly convex and strictly concave graphs of functions that doesn't have derivative at some points, because such straight line doesn't exists or because there are more than one straight line that intersect the graph of the function uniquely in some point.
answered Jun 2 '18 at 11:12
MasacrosoMasacroso
13k41746
$endgroup$
$begingroup$
Nice answer. Can you provide us some examples of those 'patological' cases, please?
$endgroup$
– Dog_69
Jun 2 '18 at 16:34
2
$begingroup$
@Dog_69 sorry, my previous comment was wrong :P. The function $$f(x):=begin{cases}x^2,&x<0\(x+1)^2-1,&xge 0end{cases}$$ is strictly convex but is not differentiable at zero because both straight lines $g_1(x):=0$ and $g_2(x):=2x$ cut the graph uniquely at the point $(0,0)$. Indeed any straight line of the form $g(x):=kx$ for $kin[0,2]$ does.
$endgroup$
– Masacroso
Jun 2 '18 at 17:05
$begingroup$
don't worry. Thanks for the example.
$endgroup$
– Dog_69
Jun 2 '18 at 18:15
add a comment |
$begingroup$
Geometrically the derivative at a point can be interpreted for strictly convex or strictly concave graphs easily: the derivative at a point $(c,f(c))$ of a strictly convex (or strictly concave) graph exists if there is a unique straight line $g(x):= ax+b$ that intersect the graph of $f$ at the point $(c,f(c))$ such that $f(x)ge g(x)$ for all $x$ if $f$ is convex, or $f(x)le g(x)$ for all $x$ when $f$ is concave. Then we says that $a$ is the derivative of $f$ at $c$, that is, that $f'(c)=a$.
For other functions, not necessarily strictly concave or strictly convex, the geometric interpretation is not so clear because there are infinite straight lines that can cut the graph of a function uniquely at a point as in the function of your picture.
But the geometric definition of the derivative at points of strictly convex or strictly concave graphs define analitically the derivative through a limit, and we can extend this definition to other cases not necessarily strictly concave or strictly convex.
Note that there are strictly convex and strictly concave graphs of functions that doesn't have derivative at some points, because such straight line doesn't exists or because there are more than one straight line that intersect the graph of the function uniquely in some point.
answered Jun 2 '18 at 11:12
MasacrosoMasacroso
13k41746
$endgroup$
$begingroup$
Nice answer. Can you provide us some examples of those 'patological' cases, please?
$endgroup$
– Dog_69
Jun 2 '18 at 16:34
2
$begingroup$
@Dog_69 sorry, my previous comment was wrong :P. The function $$f(x):=begin{cases}x^2,&x<0\(x+1)^2-1,&xge 0end{cases}$$ is strictly convex but is not differentiable at zero because both straight lines $g_1(x):=0$ and $g_2(x):=2x$ cut the graph uniquely at the point $(0,0)$. Indeed any straight line of the form $g(x):=kx$ for $kin[0,2]$ does.
$endgroup$
– Masacroso
Jun 2 '18 at 17:05
$begingroup$
don't worry. Thanks for the example.
$endgroup$
– Dog_69
Jun 2 '18 at 18:15
add a comment |
$begingroup$
Geometrically the derivative at a point can be interpreted for strictly convex or strictly concave graphs easily: the derivative at a point $(c,f(c))$ of a strictly convex (or strictly concave) graph exists if there is a unique straight line $g(x):= ax+b$ that intersect the graph of $f$ at the point $(c,f(c))$ such that $f(x)ge g(x)$ for all $x$ if $f$ is convex, or $f(x)le g(x)$ for all $x$ when $f$ is concave. Then we says that $a$ is the derivative of $f$ at $c$, that is, that $f'(c)=a$.
For other functions, not necessarily strictly concave or strictly convex, the geometric interpretation is not so clear because there are infinite straight lines that can cut the graph of a function uniquely at a point as in the function of your picture.
But the geometric definition of the derivative at points of strictly convex or strictly concave graphs define analitically the derivative through a limit, and we can extend this definition to other cases not necessarily strictly concave or strictly convex.
Note that there are strictly convex and strictly concave graphs of functions that doesn't have derivative at some points, because such straight line doesn't exists or because there are more than one straight line that intersect the graph of the function uniquely in some point.
answered Jun 2 '18 at 11:12
MasacrosoMasacroso
13k41746
$endgroup$
Geometrically the derivative at a point can be interpreted for strictly convex or strictly concave graphs easily: the derivative at a point $(c,f(c))$ of a strictly convex (or strictly concave) graph exists if there is a unique straight line $g(x):= ax+b$ that intersect the graph of $f$ at the point $(c,f(c))$ such that $f(x)ge g(x)$ for all $x$ if $f$ is convex, or $f(x)le g(x)$ for all $x$ when $f$ is concave. Then we says that $a$ is the derivative of $f$ at $c$, that is, that $f'(c)=a$.
For other functions, not necessarily strictly concave or strictly convex, the geometric interpretation is not so clear because there are infinite straight lines that can cut the graph of a function uniquely at a point as in the function of your picture.
But the geometric definition of the derivative at points of strictly convex or strictly concave graphs define analitically the derivative through a limit, and we can extend this definition to other cases not necessarily strictly concave or strictly convex.
Note that there are strictly convex and strictly concave graphs of functions that doesn't have derivative at some points, because such straight line doesn't exists or because there are more than one straight line that intersect the graph of the function uniquely in some point.
answered Jun 2 '18 at 11:12
MasacrosoMasacroso
13k41746
edited Dec 6 '18 at 2:27
answered Jun 2 '18 at 11:12
MasacrosoMasacroso
13k41746
answered Jun 2 '18 at 11:12
MasacrosoMasacroso
13k41746
answered Jun 2 '18 at 11:12
MasacrosoMasacroso
13k41746
13k41746
$begingroup$
Nice answer. Can you provide us some examples of those 'patological' cases, please?
$endgroup$
– Dog_69
Jun 2 '18 at 16:34
2
$begingroup$
@Dog_69 sorry, my previous comment was wrong :P. The function $$f(x):=begin{cases}x^2,&x<0\(x+1)^2-1,&xge 0end{cases}$$ is strictly convex but is not differentiable at zero because both straight lines $g_1(x):=0$ and $g_2(x):=2x$ cut the graph uniquely at the point $(0,0)$. Indeed any straight line of the form $g(x):=kx$ for $kin[0,2]$ does.
$endgroup$
– Masacroso
Jun 2 '18 at 17:05
$begingroup$
don't worry. Thanks for the example.
$endgroup$
– Dog_69
Jun 2 '18 at 18:15
add a comment |
$begingroup$
Nice answer. Can you provide us some examples of those 'patological' cases, please?
$endgroup$
– Dog_69
Jun 2 '18 at 16:34
2
$begingroup$
@Dog_69 sorry, my previous comment was wrong :P. The function $$f(x):=begin{cases}x^2,&x<0\(x+1)^2-1,&xge 0end{cases}$$ is strictly convex but is not differentiable at zero because both straight lines $g_1(x):=0$ and $g_2(x):=2x$ cut the graph uniquely at the point $(0,0)$. Indeed any straight line of the form $g(x):=kx$ for $kin[0,2]$ does.
$endgroup$
– Masacroso
Jun 2 '18 at 17:05
$begingroup$
don't worry. Thanks for the example.
$endgroup$
– Dog_69
Jun 2 '18 at 18:15
$begingroup$
Nice answer. Can you provide us some examples of those 'patological' cases, please?
$endgroup$
– Dog_69
Jun 2 '18 at 16:34
$begingroup$
Nice answer. Can you provide us some examples of those 'patological' cases, please?
$endgroup$
– Dog_69
Jun 2 '18 at 16:34
2
2
$begingroup$
@Dog_69 sorry, my previous comment was wrong :P. The function $$f(x):=begin{cases}x^2,&x<0\(x+1)^2-1,&xge 0end{cases}$$ is strictly convex but is not differentiable at zero because both straight lines $g_1(x):=0$ and $g_2(x):=2x$ cut the graph uniquely at the point $(0,0)$. Indeed any straight line of the form $g(x):=kx$ for $kin[0,2]$ does.
$endgroup$
– Masacroso
Jun 2 '18 at 17:05
$begingroup$
@Dog_69 sorry, my previous comment was wrong :P. The function $$f(x):=begin{cases}x^2,&x<0\(x+1)^2-1,&xge 0end{cases}$$ is strictly convex but is not differentiable at zero because both straight lines $g_1(x):=0$ and $g_2(x):=2x$ cut the graph uniquely at the point $(0,0)$. Indeed any straight line of the form $g(x):=kx$ for $kin[0,2]$ does.
$endgroup$
– Masacroso
Jun 2 '18 at 17:05
$begingroup$
don't worry. Thanks for the example.
$endgroup$
– Dog_69
Jun 2 '18 at 18:15
$begingroup$
don't worry. Thanks for the example.
$endgroup$
– Dog_69
Jun 2 '18 at 18:15
add a comment |
$begingroup$
If you want to see the tangent at x=0, go to a graph plotter app (DESMOS), plot your graph (eg. x^3) and zoom it at x=0. Zoom upto that extent until the line becomes horizontal. This acts as the tangent at that point.
Isn't it amazing that a curve becomes parallel if zoomed to a large extent. Think it in a sense that Earth appears to be flat (because we are so small that Earth appears to be zoomed), but actually its not. So the flat surface acts as a tangent at any point where you are standing.
Hope you are clear with the geometrical approach.
answered Jun 2 '18 at 11:20
Yash MittalYash Mittal
1
$endgroup$
add a comment |
$begingroup$
If you want to see the tangent at x=0, go to a graph plotter app (DESMOS), plot your graph (eg. x^3) and zoom it at x=0. Zoom upto that extent until the line becomes horizontal. This acts as the tangent at that point.
Isn't it amazing that a curve becomes parallel if zoomed to a large extent. Think it in a sense that Earth appears to be flat (because we are so small that Earth appears to be zoomed), but actually its not. So the flat surface acts as a tangent at any point where you are standing.
Hope you are clear with the geometrical approach.
answered Jun 2 '18 at 11:20
Yash MittalYash Mittal
1
$endgroup$
add a comment |
$begingroup$
If you want to see the tangent at x=0, go to a graph plotter app (DESMOS), plot your graph (eg. x^3) and zoom it at x=0. Zoom upto that extent until the line becomes horizontal. This acts as the tangent at that point.
Isn't it amazing that a curve becomes parallel if zoomed to a large extent. Think it in a sense that Earth appears to be flat (because we are so small that Earth appears to be zoomed), but actually its not. So the flat surface acts as a tangent at any point where you are standing.
Hope you are clear with the geometrical approach.
answered Jun 2 '18 at 11:20
Yash MittalYash Mittal
1
$endgroup$
If you want to see the tangent at x=0, go to a graph plotter app (DESMOS), plot your graph (eg. x^3) and zoom it at x=0. Zoom upto that extent until the line becomes horizontal. This acts as the tangent at that point.
Isn't it amazing that a curve becomes parallel if zoomed to a large extent. Think it in a sense that Earth appears to be flat (because we are so small that Earth appears to be zoomed), but actually its not. So the flat surface acts as a tangent at any point where you are standing.
Hope you are clear with the geometrical approach.
answered Jun 2 '18 at 11:20
Yash MittalYash Mittal
1
answered Jun 2 '18 at 11:20
Yash MittalYash Mittal
1
answered Jun 2 '18 at 11:20
Yash MittalYash Mittal
1
answered Jun 2 '18 at 11:20
Yash MittalYash Mittal
1
1
add a comment |
add a comment |
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