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Proving the non-differentiability of $|x| + x$. [duplicate]
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Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable
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For the function $f(x) = |x| + x$, how can one prove that it is not differentiable at the point $x = 0$? Specifically, I wish to use the Definition of the Derivative
$$lim_Delta x to 0fracf(x + Delta x) - f(x)Delta x$$
to attain two non-equal limiting values by taking $Delta x$ to be first positive, and then negative.
It was my original thought to rewrite $f(x)$ as
$$ f(x) =
begincases
2x&textif, xgeq 0\
0&textif, x < 0
endcases
$$
and then take the derivative for each separate case and show that they aren't equal, but I wish to use the above definition with both a positive and a negative $Delta x$, and so the cases method would fall out of line with that.
Thank you.
limits derivatives absolute-value
asked Jul 29 at 18:26
Jamie Corkhill
1158
marked as duplicate by amWhy, John Ma, Lord Shark the Unknown, user 108128, Rhys Steele Jul 30 at 12:27
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.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable
4 answers
For the function $f(x) = |x| + x$, how can one prove that it is not differentiable at the point $x = 0$? Specifically, I wish to use the Definition of the Derivative
$$lim_Delta x to 0fracf(x + Delta x) - f(x)Delta x$$
to attain two non-equal limiting values by taking $Delta x$ to be first positive, and then negative.
It was my original thought to rewrite $f(x)$ as
$$ f(x) =
begincases
2x&textif, xgeq 0\
0&textif, x < 0
endcases
$$
and then take the derivative for each separate case and show that they aren't equal, but I wish to use the above definition with both a positive and a negative $Delta x$, and so the cases method would fall out of line with that.
Thank you.
limits derivatives absolute-value
asked Jul 29 at 18:26
Jamie Corkhill
1158
marked as duplicate by amWhy, John Ma, Lord Shark the Unknown, user 108128, Rhys Steele Jul 30 at 12:27
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.
1
It's not differentiable at $0$ because it's a sum of differentiable ($x$) and not differentiable ($|x|$) function
– Rumpelstiltskin
Jul 29 at 18:41
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable
4 answers
For the function $f(x) = |x| + x$, how can one prove that it is not differentiable at the point $x = 0$? Specifically, I wish to use the Definition of the Derivative
$$lim_Delta x to 0fracf(x + Delta x) - f(x)Delta x$$
to attain two non-equal limiting values by taking $Delta x$ to be first positive, and then negative.
It was my original thought to rewrite $f(x)$ as
$$ f(x) =
begincases
2x&textif, xgeq 0\
0&textif, x < 0
endcases
$$
and then take the derivative for each separate case and show that they aren't equal, but I wish to use the above definition with both a positive and a negative $Delta x$, and so the cases method would fall out of line with that.
Thank you.
limits derivatives absolute-value
asked Jul 29 at 18:26
Jamie Corkhill
1158
This question already has an answer here:
Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable
4 answers
For the function $f(x) = |x| + x$, how can one prove that it is not differentiable at the point $x = 0$? Specifically, I wish to use the Definition of the Derivative
$$lim_Delta x to 0fracf(x + Delta x) - f(x)Delta x$$
to attain two non-equal limiting values by taking $Delta x$ to be first positive, and then negative.
It was my original thought to rewrite $f(x)$ as
$$ f(x) =
begincases
2x&textif, xgeq 0\
0&textif, x < 0
endcases
$$
and then take the derivative for each separate case and show that they aren't equal, but I wish to use the above definition with both a positive and a negative $Delta x$, and so the cases method would fall out of line with that.
Thank you.
This question already has an answer here:
Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable
4 answers
limits derivatives absolute-value
asked Jul 29 at 18:26
Jamie Corkhill
1158
asked Jul 29 at 18:26
Jamie Corkhill
1158
asked Jul 29 at 18:26
Jamie Corkhill
1158
asked Jul 29 at 18:26
Jamie Corkhill
1158
1158
marked as duplicate by amWhy, John Ma, Lord Shark the Unknown, user 108128, Rhys Steele Jul 30 at 12:27
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.
marked as duplicate by amWhy, John Ma, Lord Shark the Unknown, user 108128, Rhys Steele Jul 30 at 12:27
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.
1
It's not differentiable at $0$ because it's a sum of differentiable ($x$) and not differentiable ($|x|$) function
– Rumpelstiltskin
Jul 29 at 18:41
add a comment |Â
1
It's not differentiable at $0$ because it's a sum of differentiable ($x$) and not differentiable ($|x|$) function
– Rumpelstiltskin
Jul 29 at 18:41
1
1
It's not differentiable at $0$ because it's a sum of differentiable ($x$) and not differentiable ($|x|$) function
– Rumpelstiltskin
Jul 29 at 18:41
It's not differentiable at $0$ because it's a sum of differentiable ($x$) and not differentiable ($|x|$) function
– Rumpelstiltskin
Jul 29 at 18:41
add a comment |Â
2 Answers
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accepted
If $Delta x>0$, then $$fracf(Delta x)-f(0)Delta x=2,$$whereas if $Delta x<0$, then $$fracf(Delta x)-f(0)Delta x=0.$$Therefore$$lim_Delta xto0^+fracf(Delta x)-f(0)Delta x=2neq0=lim_Delta xto0^-fracf(Delta x)-f(0)Delta x.$$
answered Jul 29 at 18:29
José Carlos Santos
112k1696173
How do you which case to use for $Delta x$ greater than or less than 0? In the definition, $2x$ is used if $x geq 0$, were $x = x + Delta x$. So, knowing not the value of $x$ and only that $Delta x$ is near 0, how do we know if $x + Delta x$ is positive or negative to decide which case to use?
– Jamie Corkhill
Jul 29 at 18:55
2
@JamieCorkhill My mistake. I wrote $x$, but I had $0$ in my mind. I suppose that it is correct now.
– José Carlos Santos
Jul 29 at 18:58
add a comment |Â
up vote
0
down vote
hint: show $f'(0^+) neq f'(0^-)$
answered Jul 29 at 18:29
DeepSea
68.8k54284
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
If $Delta x>0$, then $$fracf(Delta x)-f(0)Delta x=2,$$whereas if $Delta x<0$, then $$fracf(Delta x)-f(0)Delta x=0.$$Therefore$$lim_Delta xto0^+fracf(Delta x)-f(0)Delta x=2neq0=lim_Delta xto0^-fracf(Delta x)-f(0)Delta x.$$
answered Jul 29 at 18:29
José Carlos Santos
112k1696173
How do you which case to use for $Delta x$ greater than or less than 0? In the definition, $2x$ is used if $x geq 0$, were $x = x + Delta x$. So, knowing not the value of $x$ and only that $Delta x$ is near 0, how do we know if $x + Delta x$ is positive or negative to decide which case to use?
– Jamie Corkhill
Jul 29 at 18:55
2
@JamieCorkhill My mistake. I wrote $x$, but I had $0$ in my mind. I suppose that it is correct now.
– José Carlos Santos
Jul 29 at 18:58
add a comment |Â
up vote
5
down vote
accepted
If $Delta x>0$, then $$fracf(Delta x)-f(0)Delta x=2,$$whereas if $Delta x<0$, then $$fracf(Delta x)-f(0)Delta x=0.$$Therefore$$lim_Delta xto0^+fracf(Delta x)-f(0)Delta x=2neq0=lim_Delta xto0^-fracf(Delta x)-f(0)Delta x.$$
answered Jul 29 at 18:29
José Carlos Santos
112k1696173
How do you which case to use for $Delta x$ greater than or less than 0? In the definition, $2x$ is used if $x geq 0$, were $x = x + Delta x$. So, knowing not the value of $x$ and only that $Delta x$ is near 0, how do we know if $x + Delta x$ is positive or negative to decide which case to use?
– Jamie Corkhill
Jul 29 at 18:55
2
@JamieCorkhill My mistake. I wrote $x$, but I had $0$ in my mind. I suppose that it is correct now.
– José Carlos Santos
Jul 29 at 18:58
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
If $Delta x>0$, then $$fracf(Delta x)-f(0)Delta x=2,$$whereas if $Delta x<0$, then $$fracf(Delta x)-f(0)Delta x=0.$$Therefore$$lim_Delta xto0^+fracf(Delta x)-f(0)Delta x=2neq0=lim_Delta xto0^-fracf(Delta x)-f(0)Delta x.$$
answered Jul 29 at 18:29
José Carlos Santos
112k1696173
If $Delta x>0$, then $$fracf(Delta x)-f(0)Delta x=2,$$whereas if $Delta x<0$, then $$fracf(Delta x)-f(0)Delta x=0.$$Therefore$$lim_Delta xto0^+fracf(Delta x)-f(0)Delta x=2neq0=lim_Delta xto0^-fracf(Delta x)-f(0)Delta x.$$
answered Jul 29 at 18:29
José Carlos Santos
112k1696173
edited Jul 29 at 18:57
answered Jul 29 at 18:29
José Carlos Santos
112k1696173
answered Jul 29 at 18:29
José Carlos Santos
112k1696173
answered Jul 29 at 18:29
José Carlos Santos
112k1696173
112k1696173
How do you which case to use for $Delta x$ greater than or less than 0? In the definition, $2x$ is used if $x geq 0$, were $x = x + Delta x$. So, knowing not the value of $x$ and only that $Delta x$ is near 0, how do we know if $x + Delta x$ is positive or negative to decide which case to use?
– Jamie Corkhill
Jul 29 at 18:55
2
@JamieCorkhill My mistake. I wrote $x$, but I had $0$ in my mind. I suppose that it is correct now.
– José Carlos Santos
Jul 29 at 18:58
add a comment |Â
How do you which case to use for $Delta x$ greater than or less than 0? In the definition, $2x$ is used if $x geq 0$, were $x = x + Delta x$. So, knowing not the value of $x$ and only that $Delta x$ is near 0, how do we know if $x + Delta x$ is positive or negative to decide which case to use?
– Jamie Corkhill
Jul 29 at 18:55
2
@JamieCorkhill My mistake. I wrote $x$, but I had $0$ in my mind. I suppose that it is correct now.
– José Carlos Santos
Jul 29 at 18:58
How do you which case to use for $Delta x$ greater than or less than 0? In the definition, $2x$ is used if $x geq 0$, were $x = x + Delta x$. So, knowing not the value of $x$ and only that $Delta x$ is near 0, how do we know if $x + Delta x$ is positive or negative to decide which case to use?
– Jamie Corkhill
Jul 29 at 18:55
How do you which case to use for $Delta x$ greater than or less than 0? In the definition, $2x$ is used if $x geq 0$, were $x = x + Delta x$. So, knowing not the value of $x$ and only that $Delta x$ is near 0, how do we know if $x + Delta x$ is positive or negative to decide which case to use?
– Jamie Corkhill
Jul 29 at 18:55
2
2
@JamieCorkhill My mistake. I wrote $x$, but I had $0$ in my mind. I suppose that it is correct now.
– José Carlos Santos
Jul 29 at 18:58
@JamieCorkhill My mistake. I wrote $x$, but I had $0$ in my mind. I suppose that it is correct now.
– José Carlos Santos
Jul 29 at 18:58
add a comment |Â
up vote
0
down vote
hint: show $f'(0^+) neq f'(0^-)$
answered Jul 29 at 18:29
DeepSea
68.8k54284
add a comment |Â
up vote
0
down vote
hint: show $f'(0^+) neq f'(0^-)$
answered Jul 29 at 18:29
DeepSea
68.8k54284
add a comment |Â
up vote
0
down vote
up vote
0
down vote
hint: show $f'(0^+) neq f'(0^-)$
answered Jul 29 at 18:29
DeepSea
68.8k54284
hint: show $f'(0^+) neq f'(0^-)$
answered Jul 29 at 18:29
DeepSea
68.8k54284
answered Jul 29 at 18:29
DeepSea
68.8k54284
answered Jul 29 at 18:29
DeepSea
68.8k54284
answered Jul 29 at 18:29
DeepSea
68.8k54284
68.8k54284
add a comment |Â
add a comment |Â
1
It's not differentiable at $0$ because it's a sum of differentiable ($x$) and not differentiable ($|x|$) function
– Rumpelstiltskin
Jul 29 at 18:41