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How to “see†discontinuity of second derivative from graph of function
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Suppose we have a real function $ f: mathbbR to mathbbR$ that is two times differentiable and we draw its graph $(x,f(x)), x in mathbbR $. We know, for example, that when the first derivative is not continuous at a point we then have an "corner" in the graph. How about a discontinuity of $f''(x)$ at a point $x_0$ though? Can we spot that just by drawing the graph of $f(x)$ -not drawing the graph of the first derivative and noticing it has an "corner" at $x_0$, that's cheating. What about discontinuities of higher derivatives, which will of course be way harder to "see"?
real-analysis functions derivatives
asked 35 mins ago
Dimitris
5521617
add a comment |Â
up vote
5
down vote
favorite
Suppose we have a real function $ f: mathbbR to mathbbR$ that is two times differentiable and we draw its graph $(x,f(x)), x in mathbbR $. We know, for example, that when the first derivative is not continuous at a point we then have an "corner" in the graph. How about a discontinuity of $f''(x)$ at a point $x_0$ though? Can we spot that just by drawing the graph of $f(x)$ -not drawing the graph of the first derivative and noticing it has an "corner" at $x_0$, that's cheating. What about discontinuities of higher derivatives, which will of course be way harder to "see"?
real-analysis functions derivatives
asked 35 mins ago
Dimitris
5521617
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Suppose we have a real function $ f: mathbbR to mathbbR$ that is two times differentiable and we draw its graph $(x,f(x)), x in mathbbR $. We know, for example, that when the first derivative is not continuous at a point we then have an "corner" in the graph. How about a discontinuity of $f''(x)$ at a point $x_0$ though? Can we spot that just by drawing the graph of $f(x)$ -not drawing the graph of the first derivative and noticing it has an "corner" at $x_0$, that's cheating. What about discontinuities of higher derivatives, which will of course be way harder to "see"?
real-analysis functions derivatives
asked 35 mins ago
Dimitris
5521617
Suppose we have a real function $ f: mathbbR to mathbbR$ that is two times differentiable and we draw its graph $(x,f(x)), x in mathbbR $. We know, for example, that when the first derivative is not continuous at a point we then have an "corner" in the graph. How about a discontinuity of $f''(x)$ at a point $x_0$ though? Can we spot that just by drawing the graph of $f(x)$ -not drawing the graph of the first derivative and noticing it has an "corner" at $x_0$, that's cheating. What about discontinuities of higher derivatives, which will of course be way harder to "see"?
real-analysis functions derivatives
real-analysis functions derivatives
asked 35 mins ago
Dimitris
5521617
asked 35 mins ago
Dimitris
5521617
edited 16 mins ago
asked 35 mins ago
Dimitris
5521617
asked 35 mins ago
Dimitris
5521617
asked 35 mins ago
Dimitris
5521617
5521617
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Whether or not you can "see" this is in some sense a question about the acuity of human vision. I think the answer is "no". If you graph the function
$$
g(x) = int_0^x |t|dt
$$
you will see the usual parabola in the right half plane and its negative in the left half plane. They meet at the origin with derivative $0$. The derivative of this function is $|x|$, whose derivative is undefined at $0$. The second derivative is $-1$ on the left and $1$ on the right, undefined at the origin.
Plot that and see if you can see the second derivative. Unless you draw it really carefully and know what you are looking for the graph will look like that of $x^3$, which is quite respectable.
answered 23 mins ago
Ethan Bolker
38.4k543101
+1, I was just about to write this example.
– Ennar
20 mins ago
add a comment |Â
up vote
0
down vote
Not sure if there is a "picture rigorous" way of checking, but there are subtle clues that could indicate it's possible.
For instance, if you know the first derivative $f'$ has a "corner" at $x=a$, then the graph of $f$ has to be monotonically increasing/decreasing around $x=a$. Furthermore, this indicates a change in concavity has to be present. So places where you have
- a change in concavity AND
- don't change increasing/decreasing behavior (we allow derivative of zero here),
then it could happen at that point (in terms of limiting down the choice of possibilities). Take for instance the antiderivative of $|x|$.
Obviously this isn't saying this is always true (see $x^3$ for instance). Just that if you want to narrow your focus to potential points, do this trick.
Please let me know if there is something I'm missing or wrong about.
answered 18 mins ago
welshman500
358214
1
So, what exactly do you see at that plot that would tell you there is no second derivative at $0$? Why don't you plot $x^3$ next to it and tell me the qualitative difference.
– Ennar
16 mins ago
Not saying it occurs. Just that it is possible.
– welshman500
15 mins ago
1
The question is how to spot it from the graph. With $|x|$ it's quite obvious where it isn't differentiable, and similarly for any continuous function not differentiable at some point. I wouldn't say you could see the same thing about second derivative just from the graph.
– Ennar
13 mins ago
add a comment |Â
up vote
0
down vote
A vertical inflection point will have an undefined second derivative. Let $f(x):= x^1/3$, then the second derivative $f;''(x) := frac-29frac1x^5/3$ is undefined at $x = 0$. This is just one easy-to-find feature showing a failed second derivative
answered 12 mins ago
DWade64
5581413
And the first derivative of $f$?
– Ennar
11 mins ago
@Ennar It's first derivative is undefined there as well
– DWade64
7 mins ago
Thus, your test tells us when the first derivative doesn't exist. I wouldn't call it a test for second derivative.
– Ennar
6 mins ago
@Ennar I see what you are saying. The way I understood the question, we were trying to avoid corners. Also if the function is twice differentiable, then all these answers are wrong
– DWade64
5 mins ago
In a way, yes.$hphantom$
– Ennar
3 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Whether or not you can "see" this is in some sense a question about the acuity of human vision. I think the answer is "no". If you graph the function
$$
g(x) = int_0^x |t|dt
$$
you will see the usual parabola in the right half plane and its negative in the left half plane. They meet at the origin with derivative $0$. The derivative of this function is $|x|$, whose derivative is undefined at $0$. The second derivative is $-1$ on the left and $1$ on the right, undefined at the origin.
Plot that and see if you can see the second derivative. Unless you draw it really carefully and know what you are looking for the graph will look like that of $x^3$, which is quite respectable.
answered 23 mins ago
Ethan Bolker
38.4k543101
+1, I was just about to write this example.
– Ennar
20 mins ago
add a comment |Â
up vote
3
down vote
Whether or not you can "see" this is in some sense a question about the acuity of human vision. I think the answer is "no". If you graph the function
$$
g(x) = int_0^x |t|dt
$$
you will see the usual parabola in the right half plane and its negative in the left half plane. They meet at the origin with derivative $0$. The derivative of this function is $|x|$, whose derivative is undefined at $0$. The second derivative is $-1$ on the left and $1$ on the right, undefined at the origin.
Plot that and see if you can see the second derivative. Unless you draw it really carefully and know what you are looking for the graph will look like that of $x^3$, which is quite respectable.
answered 23 mins ago
Ethan Bolker
38.4k543101
+1, I was just about to write this example.
– Ennar
20 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Whether or not you can "see" this is in some sense a question about the acuity of human vision. I think the answer is "no". If you graph the function
$$
g(x) = int_0^x |t|dt
$$
you will see the usual parabola in the right half plane and its negative in the left half plane. They meet at the origin with derivative $0$. The derivative of this function is $|x|$, whose derivative is undefined at $0$. The second derivative is $-1$ on the left and $1$ on the right, undefined at the origin.
Plot that and see if you can see the second derivative. Unless you draw it really carefully and know what you are looking for the graph will look like that of $x^3$, which is quite respectable.
answered 23 mins ago
Ethan Bolker
38.4k543101
Whether or not you can "see" this is in some sense a question about the acuity of human vision. I think the answer is "no". If you graph the function
$$
g(x) = int_0^x |t|dt
$$
you will see the usual parabola in the right half plane and its negative in the left half plane. They meet at the origin with derivative $0$. The derivative of this function is $|x|$, whose derivative is undefined at $0$. The second derivative is $-1$ on the left and $1$ on the right, undefined at the origin.
Plot that and see if you can see the second derivative. Unless you draw it really carefully and know what you are looking for the graph will look like that of $x^3$, which is quite respectable.
answered 23 mins ago
Ethan Bolker
38.4k543101
answered 23 mins ago
Ethan Bolker
38.4k543101
answered 23 mins ago
Ethan Bolker
38.4k543101
answered 23 mins ago
Ethan Bolker
38.4k543101
38.4k543101
+1, I was just about to write this example.
– Ennar
20 mins ago
add a comment |Â
+1, I was just about to write this example.
– Ennar
20 mins ago
+1, I was just about to write this example.
– Ennar
20 mins ago
+1, I was just about to write this example.
– Ennar
20 mins ago
add a comment |Â
up vote
0
down vote
Not sure if there is a "picture rigorous" way of checking, but there are subtle clues that could indicate it's possible.
For instance, if you know the first derivative $f'$ has a "corner" at $x=a$, then the graph of $f$ has to be monotonically increasing/decreasing around $x=a$. Furthermore, this indicates a change in concavity has to be present. So places where you have
- a change in concavity AND
- don't change increasing/decreasing behavior (we allow derivative of zero here),
then it could happen at that point (in terms of limiting down the choice of possibilities). Take for instance the antiderivative of $|x|$.
Obviously this isn't saying this is always true (see $x^3$ for instance). Just that if you want to narrow your focus to potential points, do this trick.
Please let me know if there is something I'm missing or wrong about.
answered 18 mins ago
welshman500
358214
1
So, what exactly do you see at that plot that would tell you there is no second derivative at $0$? Why don't you plot $x^3$ next to it and tell me the qualitative difference.
– Ennar
16 mins ago
Not saying it occurs. Just that it is possible.
– welshman500
15 mins ago
1
The question is how to spot it from the graph. With $|x|$ it's quite obvious where it isn't differentiable, and similarly for any continuous function not differentiable at some point. I wouldn't say you could see the same thing about second derivative just from the graph.
– Ennar
13 mins ago
add a comment |Â
up vote
0
down vote
Not sure if there is a "picture rigorous" way of checking, but there are subtle clues that could indicate it's possible.
For instance, if you know the first derivative $f'$ has a "corner" at $x=a$, then the graph of $f$ has to be monotonically increasing/decreasing around $x=a$. Furthermore, this indicates a change in concavity has to be present. So places where you have
- a change in concavity AND
- don't change increasing/decreasing behavior (we allow derivative of zero here),
then it could happen at that point (in terms of limiting down the choice of possibilities). Take for instance the antiderivative of $|x|$.
Obviously this isn't saying this is always true (see $x^3$ for instance). Just that if you want to narrow your focus to potential points, do this trick.
Please let me know if there is something I'm missing or wrong about.
answered 18 mins ago
welshman500
358214
1
So, what exactly do you see at that plot that would tell you there is no second derivative at $0$? Why don't you plot $x^3$ next to it and tell me the qualitative difference.
– Ennar
16 mins ago
Not saying it occurs. Just that it is possible.
– welshman500
15 mins ago
1
The question is how to spot it from the graph. With $|x|$ it's quite obvious where it isn't differentiable, and similarly for any continuous function not differentiable at some point. I wouldn't say you could see the same thing about second derivative just from the graph.
– Ennar
13 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Not sure if there is a "picture rigorous" way of checking, but there are subtle clues that could indicate it's possible.
For instance, if you know the first derivative $f'$ has a "corner" at $x=a$, then the graph of $f$ has to be monotonically increasing/decreasing around $x=a$. Furthermore, this indicates a change in concavity has to be present. So places where you have
- a change in concavity AND
- don't change increasing/decreasing behavior (we allow derivative of zero here),
then it could happen at that point (in terms of limiting down the choice of possibilities). Take for instance the antiderivative of $|x|$.
Obviously this isn't saying this is always true (see $x^3$ for instance). Just that if you want to narrow your focus to potential points, do this trick.
Please let me know if there is something I'm missing or wrong about.
answered 18 mins ago
welshman500
358214
Not sure if there is a "picture rigorous" way of checking, but there are subtle clues that could indicate it's possible.
For instance, if you know the first derivative $f'$ has a "corner" at $x=a$, then the graph of $f$ has to be monotonically increasing/decreasing around $x=a$. Furthermore, this indicates a change in concavity has to be present. So places where you have
- a change in concavity AND
- don't change increasing/decreasing behavior (we allow derivative of zero here),
then it could happen at that point (in terms of limiting down the choice of possibilities). Take for instance the antiderivative of $|x|$.
Obviously this isn't saying this is always true (see $x^3$ for instance). Just that if you want to narrow your focus to potential points, do this trick.
Please let me know if there is something I'm missing or wrong about.
answered 18 mins ago
welshman500
358214
edited 15 mins ago
answered 18 mins ago
welshman500
358214
answered 18 mins ago
welshman500
358214
answered 18 mins ago
welshman500
358214
358214
1
So, what exactly do you see at that plot that would tell you there is no second derivative at $0$? Why don't you plot $x^3$ next to it and tell me the qualitative difference.
– Ennar
16 mins ago
Not saying it occurs. Just that it is possible.
– welshman500
15 mins ago
1
The question is how to spot it from the graph. With $|x|$ it's quite obvious where it isn't differentiable, and similarly for any continuous function not differentiable at some point. I wouldn't say you could see the same thing about second derivative just from the graph.
– Ennar
13 mins ago
add a comment |Â
1
So, what exactly do you see at that plot that would tell you there is no second derivative at $0$? Why don't you plot $x^3$ next to it and tell me the qualitative difference.
– Ennar
16 mins ago
Not saying it occurs. Just that it is possible.
– welshman500
15 mins ago
1
The question is how to spot it from the graph. With $|x|$ it's quite obvious where it isn't differentiable, and similarly for any continuous function not differentiable at some point. I wouldn't say you could see the same thing about second derivative just from the graph.
– Ennar
13 mins ago
1
1
So, what exactly do you see at that plot that would tell you there is no second derivative at $0$? Why don't you plot $x^3$ next to it and tell me the qualitative difference.
– Ennar
16 mins ago
So, what exactly do you see at that plot that would tell you there is no second derivative at $0$? Why don't you plot $x^3$ next to it and tell me the qualitative difference.
– Ennar
16 mins ago
Not saying it occurs. Just that it is possible.
– welshman500
15 mins ago
Not saying it occurs. Just that it is possible.
– welshman500
15 mins ago
1
1
The question is how to spot it from the graph. With $|x|$ it's quite obvious where it isn't differentiable, and similarly for any continuous function not differentiable at some point. I wouldn't say you could see the same thing about second derivative just from the graph.
– Ennar
13 mins ago
The question is how to spot it from the graph. With $|x|$ it's quite obvious where it isn't differentiable, and similarly for any continuous function not differentiable at some point. I wouldn't say you could see the same thing about second derivative just from the graph.
– Ennar
13 mins ago
add a comment |Â
up vote
0
down vote
A vertical inflection point will have an undefined second derivative. Let $f(x):= x^1/3$, then the second derivative $f;''(x) := frac-29frac1x^5/3$ is undefined at $x = 0$. This is just one easy-to-find feature showing a failed second derivative
answered 12 mins ago
DWade64
5581413
And the first derivative of $f$?
– Ennar
11 mins ago
@Ennar It's first derivative is undefined there as well
– DWade64
7 mins ago
Thus, your test tells us when the first derivative doesn't exist. I wouldn't call it a test for second derivative.
– Ennar
6 mins ago
@Ennar I see what you are saying. The way I understood the question, we were trying to avoid corners. Also if the function is twice differentiable, then all these answers are wrong
– DWade64
5 mins ago
In a way, yes.$hphantom$
– Ennar
3 mins ago
add a comment |Â
up vote
0
down vote
A vertical inflection point will have an undefined second derivative. Let $f(x):= x^1/3$, then the second derivative $f;''(x) := frac-29frac1x^5/3$ is undefined at $x = 0$. This is just one easy-to-find feature showing a failed second derivative
answered 12 mins ago
DWade64
5581413
And the first derivative of $f$?
– Ennar
11 mins ago
@Ennar It's first derivative is undefined there as well
– DWade64
7 mins ago
Thus, your test tells us when the first derivative doesn't exist. I wouldn't call it a test for second derivative.
– Ennar
6 mins ago
@Ennar I see what you are saying. The way I understood the question, we were trying to avoid corners. Also if the function is twice differentiable, then all these answers are wrong
– DWade64
5 mins ago
In a way, yes.$hphantom$
– Ennar
3 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A vertical inflection point will have an undefined second derivative. Let $f(x):= x^1/3$, then the second derivative $f;''(x) := frac-29frac1x^5/3$ is undefined at $x = 0$. This is just one easy-to-find feature showing a failed second derivative
answered 12 mins ago
DWade64
5581413
A vertical inflection point will have an undefined second derivative. Let $f(x):= x^1/3$, then the second derivative $f;''(x) := frac-29frac1x^5/3$ is undefined at $x = 0$. This is just one easy-to-find feature showing a failed second derivative
answered 12 mins ago
DWade64
5581413
answered 12 mins ago
DWade64
5581413
answered 12 mins ago
DWade64
5581413
answered 12 mins ago
DWade64
5581413
5581413
And the first derivative of $f$?
– Ennar
11 mins ago
@Ennar It's first derivative is undefined there as well
– DWade64
7 mins ago
Thus, your test tells us when the first derivative doesn't exist. I wouldn't call it a test for second derivative.
– Ennar
6 mins ago
@Ennar I see what you are saying. The way I understood the question, we were trying to avoid corners. Also if the function is twice differentiable, then all these answers are wrong
– DWade64
5 mins ago
In a way, yes.$hphantom$
– Ennar
3 mins ago
add a comment |Â
And the first derivative of $f$?
– Ennar
11 mins ago
@Ennar It's first derivative is undefined there as well
– DWade64
7 mins ago
Thus, your test tells us when the first derivative doesn't exist. I wouldn't call it a test for second derivative.
– Ennar
6 mins ago
@Ennar I see what you are saying. The way I understood the question, we were trying to avoid corners. Also if the function is twice differentiable, then all these answers are wrong
– DWade64
5 mins ago
In a way, yes.$hphantom$
– Ennar
3 mins ago
And the first derivative of $f$?
– Ennar
11 mins ago
And the first derivative of $f$?
– Ennar
11 mins ago
@Ennar It's first derivative is undefined there as well
– DWade64
7 mins ago
@Ennar It's first derivative is undefined there as well
– DWade64
7 mins ago
Thus, your test tells us when the first derivative doesn't exist. I wouldn't call it a test for second derivative.
– Ennar
6 mins ago
Thus, your test tells us when the first derivative doesn't exist. I wouldn't call it a test for second derivative.
– Ennar
6 mins ago
@Ennar I see what you are saying. The way I understood the question, we were trying to avoid corners. Also if the function is twice differentiable, then all these answers are wrong
– DWade64
5 mins ago
@Ennar I see what you are saying. The way I understood the question, we were trying to avoid corners. Also if the function is twice differentiable, then all these answers are wrong
– DWade64
5 mins ago
In a way, yes.$hphantom$
– Ennar
3 mins ago
In a way, yes.$hphantom$
– Ennar
3 mins ago
add a comment |Â
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