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looking at the piecewise definition why isn't the absolute value of x differentiable at 0?
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I understand how simple it is to see that the absolute value function is not differentiable at zero however by definition we have that the absolute value of $x$ is equal to $x$ when $x$ is greater than or equal to $0$, but that why is the derivative at zero not the derivative of $x$, $1$?
calculus derivatives absolute-value
edited Jul 15 at 21:27
José Carlos Santos
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asked Jul 15 at 21:22
ojd
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add a comment |Â
up vote
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I understand how simple it is to see that the absolute value function is not differentiable at zero however by definition we have that the absolute value of $x$ is equal to $x$ when $x$ is greater than or equal to $0$, but that why is the derivative at zero not the derivative of $x$, $1$?
calculus derivatives absolute-value
edited Jul 15 at 21:27
José Carlos Santos
114k1698177
asked Jul 15 at 21:22
ojd
424
Can you find the left-hand derivative at zero? Is it equal to $1$?
– user539887
Jul 15 at 21:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I understand how simple it is to see that the absolute value function is not differentiable at zero however by definition we have that the absolute value of $x$ is equal to $x$ when $x$ is greater than or equal to $0$, but that why is the derivative at zero not the derivative of $x$, $1$?
calculus derivatives absolute-value
edited Jul 15 at 21:27
José Carlos Santos
114k1698177
asked Jul 15 at 21:22
ojd
424
I understand how simple it is to see that the absolute value function is not differentiable at zero however by definition we have that the absolute value of $x$ is equal to $x$ when $x$ is greater than or equal to $0$, but that why is the derivative at zero not the derivative of $x$, $1$?
calculus derivatives absolute-value
edited Jul 15 at 21:27
José Carlos Santos
114k1698177
asked Jul 15 at 21:22
ojd
424
edited Jul 15 at 21:27
José Carlos Santos
114k1698177
edited Jul 15 at 21:27
José Carlos Santos
114k1698177
edited Jul 15 at 21:27
José Carlos Santos
114k1698177
114k1698177
asked Jul 15 at 21:22
ojd
424
asked Jul 15 at 21:22
ojd
424
asked Jul 15 at 21:22
ojd
424
424
Can you find the left-hand derivative at zero? Is it equal to $1$?
– user539887
Jul 15 at 21:24
add a comment |Â
Can you find the left-hand derivative at zero? Is it equal to $1$?
– user539887
Jul 15 at 21:24
Can you find the left-hand derivative at zero? Is it equal to $1$?
– user539887
Jul 15 at 21:24
Can you find the left-hand derivative at zero? Is it equal to $1$?
– user539887
Jul 15 at 21:24
add a comment |Â
2 Answers
2
active
oldest
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up vote
0
down vote
accepted
Because from the fact that $xgeqslant0implies|x|=x$, the only thing that you can deduce about the differentiablity of the aboslute function at $0$ is that the right derivative at $0$ is $1$. And $xleqslant0implies|x|=-x$, from which you can deduce that the left derivative at $0$ is $-1$. Since the left and the right derivatives at $0$ are different, the absolute value function is not differentiable at $0$.
answered Jul 15 at 21:26
José Carlos Santos
114k1698177
why is it that we can only make deductions about the left and right derivatives? if the case includes zero then surely you could write it as an interval, say ( ∞ , 0 ] and this would include the derivative at 0.
– ojd
Jul 15 at 21:54
Because the assertion $xgeqslant0implies|x|=x$ only contains information about what happens on $[0,+infty)$. SInce $0$ is the extreme left of this interval, the onlt information we can extract is therefore about the right derivative at $0$. That's all.
– José Carlos Santos
Jul 15 at 21:57
add a comment |Â
up vote
0
down vote
You could just as easily have said:
however by definition we have that the absolute value of $x$ is equal to $-x$ when $x$ is less than or equal to 0, but then why is the derivative at zero not the derivative of $-x$, $-1$?
In more detail: the derivative is (by definition) a limit of a difference quotient, i.e. with $f(x)=|x|$ we have $f'(0)$ defined by
$$f'(0) =lim_hto 0frach$$
I think many of us, especially when we are new to Calculus, tend to have an unconscious "right-handed bias": when we think about an expression like $f(a+h)$ we naturally think of it as calculating the value of $f(x)$ at a point slightly to the right of $x=a$. But the limit is a two-sided operation, and there's no reason to think of $h$ as a positive number. In this case, it's true that as you approach $h=0$ from the right, the difference quotient is $1$. However, as you approach $h=0$ from the left, the difference quotient is $-1$. Because these values don't agree, the limit at $h=0$ does not exist.
answered Jul 15 at 21:39
mweiss
17.1k23268
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Because from the fact that $xgeqslant0implies|x|=x$, the only thing that you can deduce about the differentiablity of the aboslute function at $0$ is that the right derivative at $0$ is $1$. And $xleqslant0implies|x|=-x$, from which you can deduce that the left derivative at $0$ is $-1$. Since the left and the right derivatives at $0$ are different, the absolute value function is not differentiable at $0$.
answered Jul 15 at 21:26
José Carlos Santos
114k1698177
why is it that we can only make deductions about the left and right derivatives? if the case includes zero then surely you could write it as an interval, say ( ∞ , 0 ] and this would include the derivative at 0.
– ojd
Jul 15 at 21:54
Because the assertion $xgeqslant0implies|x|=x$ only contains information about what happens on $[0,+infty)$. SInce $0$ is the extreme left of this interval, the onlt information we can extract is therefore about the right derivative at $0$. That's all.
– José Carlos Santos
Jul 15 at 21:57
add a comment |Â
up vote
0
down vote
accepted
Because from the fact that $xgeqslant0implies|x|=x$, the only thing that you can deduce about the differentiablity of the aboslute function at $0$ is that the right derivative at $0$ is $1$. And $xleqslant0implies|x|=-x$, from which you can deduce that the left derivative at $0$ is $-1$. Since the left and the right derivatives at $0$ are different, the absolute value function is not differentiable at $0$.
answered Jul 15 at 21:26
José Carlos Santos
114k1698177
why is it that we can only make deductions about the left and right derivatives? if the case includes zero then surely you could write it as an interval, say ( ∞ , 0 ] and this would include the derivative at 0.
– ojd
Jul 15 at 21:54
Because the assertion $xgeqslant0implies|x|=x$ only contains information about what happens on $[0,+infty)$. SInce $0$ is the extreme left of this interval, the onlt information we can extract is therefore about the right derivative at $0$. That's all.
– José Carlos Santos
Jul 15 at 21:57
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Because from the fact that $xgeqslant0implies|x|=x$, the only thing that you can deduce about the differentiablity of the aboslute function at $0$ is that the right derivative at $0$ is $1$. And $xleqslant0implies|x|=-x$, from which you can deduce that the left derivative at $0$ is $-1$. Since the left and the right derivatives at $0$ are different, the absolute value function is not differentiable at $0$.
answered Jul 15 at 21:26
José Carlos Santos
114k1698177
Because from the fact that $xgeqslant0implies|x|=x$, the only thing that you can deduce about the differentiablity of the aboslute function at $0$ is that the right derivative at $0$ is $1$. And $xleqslant0implies|x|=-x$, from which you can deduce that the left derivative at $0$ is $-1$. Since the left and the right derivatives at $0$ are different, the absolute value function is not differentiable at $0$.
answered Jul 15 at 21:26
José Carlos Santos
114k1698177
answered Jul 15 at 21:26
José Carlos Santos
114k1698177
answered Jul 15 at 21:26
José Carlos Santos
114k1698177
answered Jul 15 at 21:26
José Carlos Santos
114k1698177
114k1698177
why is it that we can only make deductions about the left and right derivatives? if the case includes zero then surely you could write it as an interval, say ( ∞ , 0 ] and this would include the derivative at 0.
– ojd
Jul 15 at 21:54
Because the assertion $xgeqslant0implies|x|=x$ only contains information about what happens on $[0,+infty)$. SInce $0$ is the extreme left of this interval, the onlt information we can extract is therefore about the right derivative at $0$. That's all.
– José Carlos Santos
Jul 15 at 21:57
add a comment |Â
why is it that we can only make deductions about the left and right derivatives? if the case includes zero then surely you could write it as an interval, say ( ∞ , 0 ] and this would include the derivative at 0.
– ojd
Jul 15 at 21:54
Because the assertion $xgeqslant0implies|x|=x$ only contains information about what happens on $[0,+infty)$. SInce $0$ is the extreme left of this interval, the onlt information we can extract is therefore about the right derivative at $0$. That's all.
– José Carlos Santos
Jul 15 at 21:57
why is it that we can only make deductions about the left and right derivatives? if the case includes zero then surely you could write it as an interval, say ( ∞ , 0 ] and this would include the derivative at 0.
– ojd
Jul 15 at 21:54
why is it that we can only make deductions about the left and right derivatives? if the case includes zero then surely you could write it as an interval, say ( ∞ , 0 ] and this would include the derivative at 0.
– ojd
Jul 15 at 21:54
Because the assertion $xgeqslant0implies|x|=x$ only contains information about what happens on $[0,+infty)$. SInce $0$ is the extreme left of this interval, the onlt information we can extract is therefore about the right derivative at $0$. That's all.
– José Carlos Santos
Jul 15 at 21:57
Because the assertion $xgeqslant0implies|x|=x$ only contains information about what happens on $[0,+infty)$. SInce $0$ is the extreme left of this interval, the onlt information we can extract is therefore about the right derivative at $0$. That's all.
– José Carlos Santos
Jul 15 at 21:57
add a comment |Â
up vote
0
down vote
You could just as easily have said:
however by definition we have that the absolute value of $x$ is equal to $-x$ when $x$ is less than or equal to 0, but then why is the derivative at zero not the derivative of $-x$, $-1$?
In more detail: the derivative is (by definition) a limit of a difference quotient, i.e. with $f(x)=|x|$ we have $f'(0)$ defined by
$$f'(0) =lim_hto 0frach$$
I think many of us, especially when we are new to Calculus, tend to have an unconscious "right-handed bias": when we think about an expression like $f(a+h)$ we naturally think of it as calculating the value of $f(x)$ at a point slightly to the right of $x=a$. But the limit is a two-sided operation, and there's no reason to think of $h$ as a positive number. In this case, it's true that as you approach $h=0$ from the right, the difference quotient is $1$. However, as you approach $h=0$ from the left, the difference quotient is $-1$. Because these values don't agree, the limit at $h=0$ does not exist.
answered Jul 15 at 21:39
mweiss
17.1k23268
add a comment |Â
up vote
0
down vote
You could just as easily have said:
however by definition we have that the absolute value of $x$ is equal to $-x$ when $x$ is less than or equal to 0, but then why is the derivative at zero not the derivative of $-x$, $-1$?
In more detail: the derivative is (by definition) a limit of a difference quotient, i.e. with $f(x)=|x|$ we have $f'(0)$ defined by
$$f'(0) =lim_hto 0frach$$
I think many of us, especially when we are new to Calculus, tend to have an unconscious "right-handed bias": when we think about an expression like $f(a+h)$ we naturally think of it as calculating the value of $f(x)$ at a point slightly to the right of $x=a$. But the limit is a two-sided operation, and there's no reason to think of $h$ as a positive number. In this case, it's true that as you approach $h=0$ from the right, the difference quotient is $1$. However, as you approach $h=0$ from the left, the difference quotient is $-1$. Because these values don't agree, the limit at $h=0$ does not exist.
answered Jul 15 at 21:39
mweiss
17.1k23268
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You could just as easily have said:
however by definition we have that the absolute value of $x$ is equal to $-x$ when $x$ is less than or equal to 0, but then why is the derivative at zero not the derivative of $-x$, $-1$?
In more detail: the derivative is (by definition) a limit of a difference quotient, i.e. with $f(x)=|x|$ we have $f'(0)$ defined by
$$f'(0) =lim_hto 0frach$$
I think many of us, especially when we are new to Calculus, tend to have an unconscious "right-handed bias": when we think about an expression like $f(a+h)$ we naturally think of it as calculating the value of $f(x)$ at a point slightly to the right of $x=a$. But the limit is a two-sided operation, and there's no reason to think of $h$ as a positive number. In this case, it's true that as you approach $h=0$ from the right, the difference quotient is $1$. However, as you approach $h=0$ from the left, the difference quotient is $-1$. Because these values don't agree, the limit at $h=0$ does not exist.
answered Jul 15 at 21:39
mweiss
17.1k23268
You could just as easily have said:
however by definition we have that the absolute value of $x$ is equal to $-x$ when $x$ is less than or equal to 0, but then why is the derivative at zero not the derivative of $-x$, $-1$?
In more detail: the derivative is (by definition) a limit of a difference quotient, i.e. with $f(x)=|x|$ we have $f'(0)$ defined by
$$f'(0) =lim_hto 0frach$$
I think many of us, especially when we are new to Calculus, tend to have an unconscious "right-handed bias": when we think about an expression like $f(a+h)$ we naturally think of it as calculating the value of $f(x)$ at a point slightly to the right of $x=a$. But the limit is a two-sided operation, and there's no reason to think of $h$ as a positive number. In this case, it's true that as you approach $h=0$ from the right, the difference quotient is $1$. However, as you approach $h=0$ from the left, the difference quotient is $-1$. Because these values don't agree, the limit at $h=0$ does not exist.
answered Jul 15 at 21:39
mweiss
17.1k23268
answered Jul 15 at 21:39
mweiss
17.1k23268
answered Jul 15 at 21:39
mweiss
17.1k23268
answered Jul 15 at 21:39
mweiss
17.1k23268
17.1k23268
add a comment |Â
add a comment |Â
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Can you find the left-hand derivative at zero? Is it equal to $1$?
– user539887
Jul 15 at 21:24