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Proving $f'$ can't have a jump discontinuity point [duplicate]
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Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
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I was given a hw question that I just cant seem to solve. It goes:
Let $f$ be defined and differentiable in $(a,b)$. Prove that $f'$ can't have a jump discontinuity point.
I think that by using Darboux's theorem you could get to that conclusion but we were instructed to use Lagrange's theorem.
What I did do:
Suppose $x_0in(a,b)$ is a jump discontinuity point. We can look at $(x_0,x_0+delta)subset(a,b)$ all the conditions of Lagrange's theorem are met so we have a point $cin(x_0,x_0+delta)$ such that $f'(c)=fracf(x_0+delta)-f(x_0)delta$
And this is where I'm stuck. I understand that the right side reminds the derivative definition of $f(x_0)$ but I don't understand how to connect it all.
calculus
edited Jul 24 at 17:56
Mark Viola
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asked Jul 24 at 16:29
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marked as duplicate by Mark Viola calculus
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Jul 24 at 18:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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favorite
This question already has an answer here:
Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
2 answers
I was given a hw question that I just cant seem to solve. It goes:
Let $f$ be defined and differentiable in $(a,b)$. Prove that $f'$ can't have a jump discontinuity point.
I think that by using Darboux's theorem you could get to that conclusion but we were instructed to use Lagrange's theorem.
What I did do:
Suppose $x_0in(a,b)$ is a jump discontinuity point. We can look at $(x_0,x_0+delta)subset(a,b)$ all the conditions of Lagrange's theorem are met so we have a point $cin(x_0,x_0+delta)$ such that $f'(c)=fracf(x_0+delta)-f(x_0)delta$
And this is where I'm stuck. I understand that the right side reminds the derivative definition of $f(x_0)$ but I don't understand how to connect it all.
calculus
edited Jul 24 at 17:56
Mark Viola
126k1172167
asked Jul 24 at 16:29
override
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marked as duplicate by Mark Viola calculus
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Jul 24 at 18:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
this can be easily proven by using the definition of limit and derivative at $x_0$
– Vasya
Jul 24 at 16:41
add a comment |Â
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up vote
2
down vote
favorite
This question already has an answer here:
Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
2 answers
I was given a hw question that I just cant seem to solve. It goes:
Let $f$ be defined and differentiable in $(a,b)$. Prove that $f'$ can't have a jump discontinuity point.
I think that by using Darboux's theorem you could get to that conclusion but we were instructed to use Lagrange's theorem.
What I did do:
Suppose $x_0in(a,b)$ is a jump discontinuity point. We can look at $(x_0,x_0+delta)subset(a,b)$ all the conditions of Lagrange's theorem are met so we have a point $cin(x_0,x_0+delta)$ such that $f'(c)=fracf(x_0+delta)-f(x_0)delta$
And this is where I'm stuck. I understand that the right side reminds the derivative definition of $f(x_0)$ but I don't understand how to connect it all.
calculus
edited Jul 24 at 17:56
Mark Viola
126k1172167
asked Jul 24 at 16:29
override
142
This question already has an answer here:
Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
2 answers
I was given a hw question that I just cant seem to solve. It goes:
Let $f$ be defined and differentiable in $(a,b)$. Prove that $f'$ can't have a jump discontinuity point.
I think that by using Darboux's theorem you could get to that conclusion but we were instructed to use Lagrange's theorem.
What I did do:
Suppose $x_0in(a,b)$ is a jump discontinuity point. We can look at $(x_0,x_0+delta)subset(a,b)$ all the conditions of Lagrange's theorem are met so we have a point $cin(x_0,x_0+delta)$ such that $f'(c)=fracf(x_0+delta)-f(x_0)delta$
And this is where I'm stuck. I understand that the right side reminds the derivative definition of $f(x_0)$ but I don't understand how to connect it all.
This question already has an answer here:
Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
2 answers
calculus
edited Jul 24 at 17:56
Mark Viola
126k1172167
asked Jul 24 at 16:29
override
142
edited Jul 24 at 17:56
Mark Viola
126k1172167
edited Jul 24 at 17:56
Mark Viola
126k1172167
edited Jul 24 at 17:56
Mark Viola
126k1172167
126k1172167
asked Jul 24 at 16:29
override
142
asked Jul 24 at 16:29
override
142
asked Jul 24 at 16:29
override
142
142
marked as duplicate by Mark Viola calculus
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Jul 24 at 18:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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this can be easily proven by using the definition of limit and derivative at $x_0$
– Vasya
Jul 24 at 16:41
add a comment |Â
this can be easily proven by using the definition of limit and derivative at $x_0$
– Vasya
Jul 24 at 16:41
this can be easily proven by using the definition of limit and derivative at $x_0$
– Vasya
Jul 24 at 16:41
this can be easily proven by using the definition of limit and derivative at $x_0$
– Vasya
Jul 24 at 16:41
add a comment |Â
1 Answer
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I will prove that $lim_xto x_0f'(x)$, if it exists, is equal to $f'(x_0)$ (this is the meaning of $f'$ not having a jump discontinuity at $x_0$). Take $varepsilon>0$. Since the limit $lim_xto x_0f'(x)$ exists, there is a $delta>0$ such that$$|x-x_0|<deltaimpliesleft|f'(x)-lim_xto x_0f'(x)right|<varepsilon.$$So, if $|x-x_0|<delta$ and $xneq x_0$, $fracf(x)-f(x_0)x-x_0=f'(c)$ for some $c$ between $x_0$ and $x$ and therefore $|c-x_0|<delta$ too. Therefore$$left|fracf(x)-f(x_0)x-x_0-lim_xto x_0f'(x)right|=left|f'(c)-lim_xto x_0f'(x)right|<varepsilon.$$
answered Jul 24 at 16:48
José Carlos Santos
113k1697176
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1 Answer
1
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1 Answer
1
active
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active
oldest
votes
active
oldest
votes
up vote
0
down vote
I will prove that $lim_xto x_0f'(x)$, if it exists, is equal to $f'(x_0)$ (this is the meaning of $f'$ not having a jump discontinuity at $x_0$). Take $varepsilon>0$. Since the limit $lim_xto x_0f'(x)$ exists, there is a $delta>0$ such that$$|x-x_0|<deltaimpliesleft|f'(x)-lim_xto x_0f'(x)right|<varepsilon.$$So, if $|x-x_0|<delta$ and $xneq x_0$, $fracf(x)-f(x_0)x-x_0=f'(c)$ for some $c$ between $x_0$ and $x$ and therefore $|c-x_0|<delta$ too. Therefore$$left|fracf(x)-f(x_0)x-x_0-lim_xto x_0f'(x)right|=left|f'(c)-lim_xto x_0f'(x)right|<varepsilon.$$
answered Jul 24 at 16:48
José Carlos Santos
113k1697176
add a comment |Â
up vote
0
down vote
I will prove that $lim_xto x_0f'(x)$, if it exists, is equal to $f'(x_0)$ (this is the meaning of $f'$ not having a jump discontinuity at $x_0$). Take $varepsilon>0$. Since the limit $lim_xto x_0f'(x)$ exists, there is a $delta>0$ such that$$|x-x_0|<deltaimpliesleft|f'(x)-lim_xto x_0f'(x)right|<varepsilon.$$So, if $|x-x_0|<delta$ and $xneq x_0$, $fracf(x)-f(x_0)x-x_0=f'(c)$ for some $c$ between $x_0$ and $x$ and therefore $|c-x_0|<delta$ too. Therefore$$left|fracf(x)-f(x_0)x-x_0-lim_xto x_0f'(x)right|=left|f'(c)-lim_xto x_0f'(x)right|<varepsilon.$$
answered Jul 24 at 16:48
José Carlos Santos
113k1697176
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I will prove that $lim_xto x_0f'(x)$, if it exists, is equal to $f'(x_0)$ (this is the meaning of $f'$ not having a jump discontinuity at $x_0$). Take $varepsilon>0$. Since the limit $lim_xto x_0f'(x)$ exists, there is a $delta>0$ such that$$|x-x_0|<deltaimpliesleft|f'(x)-lim_xto x_0f'(x)right|<varepsilon.$$So, if $|x-x_0|<delta$ and $xneq x_0$, $fracf(x)-f(x_0)x-x_0=f'(c)$ for some $c$ between $x_0$ and $x$ and therefore $|c-x_0|<delta$ too. Therefore$$left|fracf(x)-f(x_0)x-x_0-lim_xto x_0f'(x)right|=left|f'(c)-lim_xto x_0f'(x)right|<varepsilon.$$
answered Jul 24 at 16:48
José Carlos Santos
113k1697176
I will prove that $lim_xto x_0f'(x)$, if it exists, is equal to $f'(x_0)$ (this is the meaning of $f'$ not having a jump discontinuity at $x_0$). Take $varepsilon>0$. Since the limit $lim_xto x_0f'(x)$ exists, there is a $delta>0$ such that$$|x-x_0|<deltaimpliesleft|f'(x)-lim_xto x_0f'(x)right|<varepsilon.$$So, if $|x-x_0|<delta$ and $xneq x_0$, $fracf(x)-f(x_0)x-x_0=f'(c)$ for some $c$ between $x_0$ and $x$ and therefore $|c-x_0|<delta$ too. Therefore$$left|fracf(x)-f(x_0)x-x_0-lim_xto x_0f'(x)right|=left|f'(c)-lim_xto x_0f'(x)right|<varepsilon.$$
answered Jul 24 at 16:48
José Carlos Santos
113k1697176
answered Jul 24 at 16:48
José Carlos Santos
113k1697176
answered Jul 24 at 16:48
José Carlos Santos
113k1697176
answered Jul 24 at 16:48
José Carlos Santos
113k1697176
113k1697176
add a comment |Â
add a comment |Â
this can be easily proven by using the definition of limit and derivative at $x_0$
– Vasya
Jul 24 at 16:41