Differentiability Vs. Continuity
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Let $$f(x)=frac{x^2+x-6}{x^2+5x+6}$$ Is $f$ continuous at the point $x=-3$? Is $f$ differentiable at point $x=-3$? Is it really possible that a function is differentiable on a point (not in domain) but not continuous on that point?
calculus
edited Jan 18 at 8:40
idriskameni
749321
asked Jan 18 at 8:34
perfectnumberperfectnumber
1
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add a comment |
$begingroup$
Let $$f(x)=frac{x^2+x-6}{x^2+5x+6}$$ Is $f$ continuous at the point $x=-3$? Is $f$ differentiable at point $x=-3$? Is it really possible that a function is differentiable on a point (not in domain) but not continuous on that point?
calculus
edited Jan 18 at 8:40
idriskameni
749321
asked Jan 18 at 8:34
perfectnumberperfectnumber
1
$endgroup$
$begingroup$
Differentiability implies continuity. The contrapositive is also true: discontinuity implies non-differentiability. What can you say about $f(-3)$? Does it exist?
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– Shubham Johri
Jan 18 at 8:36
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$f(x)=frac {x-2} {x+2}$. If you define the value of $f$ at $-3$ to be $5$ then it is differentiable (and hence continuous). As it is, it is neither continuous nor differentiable
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– Kavi Rama Murthy
Jan 18 at 8:36
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The left and right hand limits are both 5 @shubham johri
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– perfectnumber
Jan 18 at 8:40
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Im confuse about continuity because -3 is not part of its domain. So its means that f is continuous everywhere except -3 and -2.
$endgroup$
– perfectnumber
Jan 18 at 8:43
add a comment |
$begingroup$
Let $$f(x)=frac{x^2+x-6}{x^2+5x+6}$$ Is $f$ continuous at the point $x=-3$? Is $f$ differentiable at point $x=-3$? Is it really possible that a function is differentiable on a point (not in domain) but not continuous on that point?
calculus
edited Jan 18 at 8:40
idriskameni
749321
asked Jan 18 at 8:34
perfectnumberperfectnumber
1
$endgroup$
Let $$f(x)=frac{x^2+x-6}{x^2+5x+6}$$ Is $f$ continuous at the point $x=-3$? Is $f$ differentiable at point $x=-3$? Is it really possible that a function is differentiable on a point (not in domain) but not continuous on that point?
calculus
calculus
edited Jan 18 at 8:40
idriskameni
749321
asked Jan 18 at 8:34
perfectnumberperfectnumber
1
edited Jan 18 at 8:40
idriskameni
749321
asked Jan 18 at 8:34
perfectnumberperfectnumber
1
edited Jan 18 at 8:40
idriskameni
749321
edited Jan 18 at 8:40
idriskameni
749321
edited Jan 18 at 8:40
idriskameni
749321
749321
asked Jan 18 at 8:34
perfectnumberperfectnumber
1
asked Jan 18 at 8:34
perfectnumberperfectnumber
1
asked Jan 18 at 8:34
perfectnumberperfectnumber
1
1
$begingroup$
Differentiability implies continuity. The contrapositive is also true: discontinuity implies non-differentiability. What can you say about $f(-3)$? Does it exist?
$endgroup$
– Shubham Johri
Jan 18 at 8:36
$begingroup$
$f(x)=frac {x-2} {x+2}$. If you define the value of $f$ at $-3$ to be $5$ then it is differentiable (and hence continuous). As it is, it is neither continuous nor differentiable
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:36
$begingroup$
The left and right hand limits are both 5 @shubham johri
$endgroup$
– perfectnumber
Jan 18 at 8:40
$begingroup$
Im confuse about continuity because -3 is not part of its domain. So its means that f is continuous everywhere except -3 and -2.
$endgroup$
– perfectnumber
Jan 18 at 8:43
add a comment |
$begingroup$
Differentiability implies continuity. The contrapositive is also true: discontinuity implies non-differentiability. What can you say about $f(-3)$? Does it exist?
$endgroup$
– Shubham Johri
Jan 18 at 8:36
$begingroup$
$f(x)=frac {x-2} {x+2}$. If you define the value of $f$ at $-3$ to be $5$ then it is differentiable (and hence continuous). As it is, it is neither continuous nor differentiable
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:36
$begingroup$
The left and right hand limits are both 5 @shubham johri
$endgroup$
– perfectnumber
Jan 18 at 8:40
$begingroup$
Im confuse about continuity because -3 is not part of its domain. So its means that f is continuous everywhere except -3 and -2.
$endgroup$
– perfectnumber
Jan 18 at 8:43
$begingroup$
Differentiability implies continuity. The contrapositive is also true: discontinuity implies non-differentiability. What can you say about $f(-3)$? Does it exist?
$endgroup$
– Shubham Johri
Jan 18 at 8:36
$begingroup$
Differentiability implies continuity. The contrapositive is also true: discontinuity implies non-differentiability. What can you say about $f(-3)$? Does it exist?
$endgroup$
– Shubham Johri
Jan 18 at 8:36
$begingroup$
$f(x)=frac {x-2} {x+2}$. If you define the value of $f$ at $-3$ to be $5$ then it is differentiable (and hence continuous). As it is, it is neither continuous nor differentiable
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:36
$begingroup$
$f(x)=frac {x-2} {x+2}$. If you define the value of $f$ at $-3$ to be $5$ then it is differentiable (and hence continuous). As it is, it is neither continuous nor differentiable
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:36
$begingroup$
The left and right hand limits are both 5 @shubham johri
$endgroup$
– perfectnumber
Jan 18 at 8:40
$begingroup$
The left and right hand limits are both 5 @shubham johri
$endgroup$
– perfectnumber
Jan 18 at 8:40
$begingroup$
Im confuse about continuity because -3 is not part of its domain. So its means that f is continuous everywhere except -3 and -2.
$endgroup$
– perfectnumber
Jan 18 at 8:43
$begingroup$
Im confuse about continuity because -3 is not part of its domain. So its means that f is continuous everywhere except -3 and -2.
$endgroup$
– perfectnumber
Jan 18 at 8:43
add a comment |
1 Answer
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A function cannot be continuous and it can't be differentiable at a point where it isn't defined. Your function $f$ is only defined on $Bbb Rsetminus {-3, -2}$, and thus asking whether it's continuous at $x = -3$ is meaningless. Note that the function isn't discontinuous either. Asking whether it's discontinuous at $x = -3$ is equally meaningless.
That being said, for all other values of $x$, we have that $f(x)$ is equal to $frac{x-2}{x+2}$, and that function is defined and continuous at $x = -3$. Thus we way that $f$ has a removable singularity at $x = -3$. If we remove this singularity, then yes, the new function is both continuous and differentiable at $x = -3$.
answered Jan 18 at 8:41
ArthurArthur
123k7122211
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I am still confused especially with the removable singularity. :|
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– perfectnumber
Jan 18 at 8:56
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@perfectnumber You can remove the singularity, but that would result in a new, different function. The original function isn't defined at $x = -3$, and thus it can't be continuous there. It doesn't matter what kind of "undefinability" it has. Removable or not, the function simply isn't defined, and that's the end of it.
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– Arthur
Jan 18 at 9:01
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In other words, f is not continuous at x=-3 hence not differentiable?
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– perfectnumber
Jan 18 at 9:04
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@perfectnumber Exactly. I think i would personally use "and" instead of "hence" in this specific case. As I see it, the function not being defined stops both properties, rather than it stopping continuity, which in turn stops differentiability. But that's a matter of philosophy and esthetics, and what you said is entirely correct.
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– Arthur
Jan 18 at 9:08
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One clarification @Arthur, it doesn't mean that the derivative of f is defined at x=-3, f is also differentiable at x=-3?
$endgroup$
– perfectnumber
Jan 18 at 9:17
|
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$begingroup$
A function cannot be continuous and it can't be differentiable at a point where it isn't defined. Your function $f$ is only defined on $Bbb Rsetminus {-3, -2}$, and thus asking whether it's continuous at $x = -3$ is meaningless. Note that the function isn't discontinuous either. Asking whether it's discontinuous at $x = -3$ is equally meaningless.
That being said, for all other values of $x$, we have that $f(x)$ is equal to $frac{x-2}{x+2}$, and that function is defined and continuous at $x = -3$. Thus we way that $f$ has a removable singularity at $x = -3$. If we remove this singularity, then yes, the new function is both continuous and differentiable at $x = -3$.
answered Jan 18 at 8:41
ArthurArthur
123k7122211
$endgroup$
$begingroup$
I am still confused especially with the removable singularity. :|
$endgroup$
– perfectnumber
Jan 18 at 8:56
$begingroup$
@perfectnumber You can remove the singularity, but that would result in a new, different function. The original function isn't defined at $x = -3$, and thus it can't be continuous there. It doesn't matter what kind of "undefinability" it has. Removable or not, the function simply isn't defined, and that's the end of it.
$endgroup$
– Arthur
Jan 18 at 9:01
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In other words, f is not continuous at x=-3 hence not differentiable?
$endgroup$
– perfectnumber
Jan 18 at 9:04
$begingroup$
@perfectnumber Exactly. I think i would personally use "and" instead of "hence" in this specific case. As I see it, the function not being defined stops both properties, rather than it stopping continuity, which in turn stops differentiability. But that's a matter of philosophy and esthetics, and what you said is entirely correct.
$endgroup$
– Arthur
Jan 18 at 9:08
$begingroup$
One clarification @Arthur, it doesn't mean that the derivative of f is defined at x=-3, f is also differentiable at x=-3?
$endgroup$
– perfectnumber
Jan 18 at 9:17
|
show 1 more comment
$begingroup$
A function cannot be continuous and it can't be differentiable at a point where it isn't defined. Your function $f$ is only defined on $Bbb Rsetminus {-3, -2}$, and thus asking whether it's continuous at $x = -3$ is meaningless. Note that the function isn't discontinuous either. Asking whether it's discontinuous at $x = -3$ is equally meaningless.
That being said, for all other values of $x$, we have that $f(x)$ is equal to $frac{x-2}{x+2}$, and that function is defined and continuous at $x = -3$. Thus we way that $f$ has a removable singularity at $x = -3$. If we remove this singularity, then yes, the new function is both continuous and differentiable at $x = -3$.
answered Jan 18 at 8:41
ArthurArthur
123k7122211
$endgroup$
$begingroup$
I am still confused especially with the removable singularity. :|
$endgroup$
– perfectnumber
Jan 18 at 8:56
$begingroup$
@perfectnumber You can remove the singularity, but that would result in a new, different function. The original function isn't defined at $x = -3$, and thus it can't be continuous there. It doesn't matter what kind of "undefinability" it has. Removable or not, the function simply isn't defined, and that's the end of it.
$endgroup$
– Arthur
Jan 18 at 9:01
$begingroup$
In other words, f is not continuous at x=-3 hence not differentiable?
$endgroup$
– perfectnumber
Jan 18 at 9:04
$begingroup$
@perfectnumber Exactly. I think i would personally use "and" instead of "hence" in this specific case. As I see it, the function not being defined stops both properties, rather than it stopping continuity, which in turn stops differentiability. But that's a matter of philosophy and esthetics, and what you said is entirely correct.
$endgroup$
– Arthur
Jan 18 at 9:08
$begingroup$
One clarification @Arthur, it doesn't mean that the derivative of f is defined at x=-3, f is also differentiable at x=-3?
$endgroup$
– perfectnumber
Jan 18 at 9:17
|
show 1 more comment
$begingroup$
A function cannot be continuous and it can't be differentiable at a point where it isn't defined. Your function $f$ is only defined on $Bbb Rsetminus {-3, -2}$, and thus asking whether it's continuous at $x = -3$ is meaningless. Note that the function isn't discontinuous either. Asking whether it's discontinuous at $x = -3$ is equally meaningless.
That being said, for all other values of $x$, we have that $f(x)$ is equal to $frac{x-2}{x+2}$, and that function is defined and continuous at $x = -3$. Thus we way that $f$ has a removable singularity at $x = -3$. If we remove this singularity, then yes, the new function is both continuous and differentiable at $x = -3$.
answered Jan 18 at 8:41
ArthurArthur
123k7122211
$endgroup$
A function cannot be continuous and it can't be differentiable at a point where it isn't defined. Your function $f$ is only defined on $Bbb Rsetminus {-3, -2}$, and thus asking whether it's continuous at $x = -3$ is meaningless. Note that the function isn't discontinuous either. Asking whether it's discontinuous at $x = -3$ is equally meaningless.
That being said, for all other values of $x$, we have that $f(x)$ is equal to $frac{x-2}{x+2}$, and that function is defined and continuous at $x = -3$. Thus we way that $f$ has a removable singularity at $x = -3$. If we remove this singularity, then yes, the new function is both continuous and differentiable at $x = -3$.
answered Jan 18 at 8:41
ArthurArthur
123k7122211
answered Jan 18 at 8:41
ArthurArthur
123k7122211
answered Jan 18 at 8:41
ArthurArthur
123k7122211
answered Jan 18 at 8:41
ArthurArthur
123k7122211
123k7122211
$begingroup$
I am still confused especially with the removable singularity. :|
$endgroup$
– perfectnumber
Jan 18 at 8:56
$begingroup$
@perfectnumber You can remove the singularity, but that would result in a new, different function. The original function isn't defined at $x = -3$, and thus it can't be continuous there. It doesn't matter what kind of "undefinability" it has. Removable or not, the function simply isn't defined, and that's the end of it.
$endgroup$
– Arthur
Jan 18 at 9:01
$begingroup$
In other words, f is not continuous at x=-3 hence not differentiable?
$endgroup$
– perfectnumber
Jan 18 at 9:04
$begingroup$
@perfectnumber Exactly. I think i would personally use "and" instead of "hence" in this specific case. As I see it, the function not being defined stops both properties, rather than it stopping continuity, which in turn stops differentiability. But that's a matter of philosophy and esthetics, and what you said is entirely correct.
$endgroup$
– Arthur
Jan 18 at 9:08
$begingroup$
One clarification @Arthur, it doesn't mean that the derivative of f is defined at x=-3, f is also differentiable at x=-3?
$endgroup$
– perfectnumber
Jan 18 at 9:17
|
show 1 more comment
$begingroup$
I am still confused especially with the removable singularity. :|
$endgroup$
– perfectnumber
Jan 18 at 8:56
$begingroup$
@perfectnumber You can remove the singularity, but that would result in a new, different function. The original function isn't defined at $x = -3$, and thus it can't be continuous there. It doesn't matter what kind of "undefinability" it has. Removable or not, the function simply isn't defined, and that's the end of it.
$endgroup$
– Arthur
Jan 18 at 9:01
$begingroup$
In other words, f is not continuous at x=-3 hence not differentiable?
$endgroup$
– perfectnumber
Jan 18 at 9:04
$begingroup$
@perfectnumber Exactly. I think i would personally use "and" instead of "hence" in this specific case. As I see it, the function not being defined stops both properties, rather than it stopping continuity, which in turn stops differentiability. But that's a matter of philosophy and esthetics, and what you said is entirely correct.
$endgroup$
– Arthur
Jan 18 at 9:08
$begingroup$
One clarification @Arthur, it doesn't mean that the derivative of f is defined at x=-3, f is also differentiable at x=-3?
$endgroup$
– perfectnumber
Jan 18 at 9:17
$begingroup$
I am still confused especially with the removable singularity. :|
$endgroup$
– perfectnumber
Jan 18 at 8:56
$begingroup$
I am still confused especially with the removable singularity. :|
$endgroup$
– perfectnumber
Jan 18 at 8:56
$begingroup$
@perfectnumber You can remove the singularity, but that would result in a new, different function. The original function isn't defined at $x = -3$, and thus it can't be continuous there. It doesn't matter what kind of "undefinability" it has. Removable or not, the function simply isn't defined, and that's the end of it.
$endgroup$
– Arthur
Jan 18 at 9:01
$begingroup$
@perfectnumber You can remove the singularity, but that would result in a new, different function. The original function isn't defined at $x = -3$, and thus it can't be continuous there. It doesn't matter what kind of "undefinability" it has. Removable or not, the function simply isn't defined, and that's the end of it.
$endgroup$
– Arthur
Jan 18 at 9:01
$begingroup$
In other words, f is not continuous at x=-3 hence not differentiable?
$endgroup$
– perfectnumber
Jan 18 at 9:04
$begingroup$
In other words, f is not continuous at x=-3 hence not differentiable?
$endgroup$
– perfectnumber
Jan 18 at 9:04
$begingroup$
@perfectnumber Exactly. I think i would personally use "and" instead of "hence" in this specific case. As I see it, the function not being defined stops both properties, rather than it stopping continuity, which in turn stops differentiability. But that's a matter of philosophy and esthetics, and what you said is entirely correct.
$endgroup$
– Arthur
Jan 18 at 9:08
$begingroup$
@perfectnumber Exactly. I think i would personally use "and" instead of "hence" in this specific case. As I see it, the function not being defined stops both properties, rather than it stopping continuity, which in turn stops differentiability. But that's a matter of philosophy and esthetics, and what you said is entirely correct.
$endgroup$
– Arthur
Jan 18 at 9:08
$begingroup$
One clarification @Arthur, it doesn't mean that the derivative of f is defined at x=-3, f is also differentiable at x=-3?
$endgroup$
– perfectnumber
Jan 18 at 9:17
$begingroup$
One clarification @Arthur, it doesn't mean that the derivative of f is defined at x=-3, f is also differentiable at x=-3?
$endgroup$
– perfectnumber
Jan 18 at 9:17
|
show 1 more comment
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$begingroup$
Differentiability implies continuity. The contrapositive is also true: discontinuity implies non-differentiability. What can you say about $f(-3)$? Does it exist?
$endgroup$
– Shubham Johri
Jan 18 at 8:36
$begingroup$
$f(x)=frac {x-2} {x+2}$. If you define the value of $f$ at $-3$ to be $5$ then it is differentiable (and hence continuous). As it is, it is neither continuous nor differentiable
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:36
$begingroup$
The left and right hand limits are both 5 @shubham johri
$endgroup$
– perfectnumber
Jan 18 at 8:40
$begingroup$
Im confuse about continuity because -3 is not part of its domain. So its means that f is continuous everywhere except -3 and -2.
$endgroup$
– perfectnumber
Jan 18 at 8:43