Can $f'(c)$ exist if $f'(x)$ is not continuous at c
$begingroup$
Can a derivative function have a jump discontinuity at a point and the function still be differentiable at that point? I'm pretty sure it cannot, since it means that the secant slopes do not tend to the same number, but I am not sure how to prove it from the definitions. I have read that the derivative function can have an essential discontinuity at a point and the function still be differentiable at that point. i.e.
$$
f(x) = left{
begin{array}{ll}
x^2sinleft(frac{1}{x}right) & quad x ne 0 \
0 & quad x = 0
end{array}
right.
$$.
Are there any other kinds of discontinuity or are these only the two cases. Many thanks.
real-analysis calculus derivatives continuity
edited Jan 11 at 16:28
José Carlos Santos
156k22126227
asked Jan 11 at 15:33
mattmatt
392213
$endgroup$
add a comment |
$begingroup$
Can a derivative function have a jump discontinuity at a point and the function still be differentiable at that point? I'm pretty sure it cannot, since it means that the secant slopes do not tend to the same number, but I am not sure how to prove it from the definitions. I have read that the derivative function can have an essential discontinuity at a point and the function still be differentiable at that point. i.e.
$$
f(x) = left{
begin{array}{ll}
x^2sinleft(frac{1}{x}right) & quad x ne 0 \
0 & quad x = 0
end{array}
right.
$$.
Are there any other kinds of discontinuity or are these only the two cases. Many thanks.
real-analysis calculus derivatives continuity
edited Jan 11 at 16:28
José Carlos Santos
156k22126227
asked Jan 11 at 15:33
mattmatt
392213
$endgroup$
2
$begingroup$
The derivative need not be continuous, but it must satisfy the IVP.
$endgroup$
– Randall
Jan 11 at 15:34
add a comment |
$begingroup$
Can a derivative function have a jump discontinuity at a point and the function still be differentiable at that point? I'm pretty sure it cannot, since it means that the secant slopes do not tend to the same number, but I am not sure how to prove it from the definitions. I have read that the derivative function can have an essential discontinuity at a point and the function still be differentiable at that point. i.e.
$$
f(x) = left{
begin{array}{ll}
x^2sinleft(frac{1}{x}right) & quad x ne 0 \
0 & quad x = 0
end{array}
right.
$$.
Are there any other kinds of discontinuity or are these only the two cases. Many thanks.
real-analysis calculus derivatives continuity
edited Jan 11 at 16:28
José Carlos Santos
156k22126227
asked Jan 11 at 15:33
mattmatt
392213
$endgroup$
Can a derivative function have a jump discontinuity at a point and the function still be differentiable at that point? I'm pretty sure it cannot, since it means that the secant slopes do not tend to the same number, but I am not sure how to prove it from the definitions. I have read that the derivative function can have an essential discontinuity at a point and the function still be differentiable at that point. i.e.
$$
f(x) = left{
begin{array}{ll}
x^2sinleft(frac{1}{x}right) & quad x ne 0 \
0 & quad x = 0
end{array}
right.
$$.
Are there any other kinds of discontinuity or are these only the two cases. Many thanks.
real-analysis calculus derivatives continuity
real-analysis calculus derivatives continuity
edited Jan 11 at 16:28
José Carlos Santos
156k22126227
asked Jan 11 at 15:33
mattmatt
392213
edited Jan 11 at 16:28
José Carlos Santos
156k22126227
asked Jan 11 at 15:33
mattmatt
392213
edited Jan 11 at 16:28
José Carlos Santos
156k22126227
edited Jan 11 at 16:28
José Carlos Santos
156k22126227
edited Jan 11 at 16:28
José Carlos Santos
156k22126227
156k22126227
asked Jan 11 at 15:33
mattmatt
392213
asked Jan 11 at 15:33
mattmatt
392213
asked Jan 11 at 15:33
mattmatt
392213
392213
2
$begingroup$
The derivative need not be continuous, but it must satisfy the IVP.
$endgroup$
– Randall
Jan 11 at 15:34
add a comment |
2
$begingroup$
The derivative need not be continuous, but it must satisfy the IVP.
$endgroup$
– Randall
Jan 11 at 15:34
2
2
$begingroup$
The derivative need not be continuous, but it must satisfy the IVP.
$endgroup$
– Randall
Jan 11 at 15:34
$begingroup$
The derivative need not be continuous, but it must satisfy the IVP.
$endgroup$
– Randall
Jan 11 at 15:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It follows from Darboux's theorem, that if $f$ is a differentiable function and if $f'$ is discontinuous at $ain D_f$, then the only possible cause of that discontinuity is that the limit $lim_{xto a}f'(x)$ doesn't exist.
answered Jan 11 at 15:37
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
$begingroup$
but can f'(a) still exist then? I know intuitively it can't but I am not too sure how to show it. Can the derivative function look like $$ f'(x) = left{ begin{array}{ll} -1 & quad x < 0 \ 0 & quad x=0\ 1 & quad x > 0 end{array} right. $$
$endgroup$
– matt
Jan 11 at 16:04
$begingroup$
No, by Darboux's theorem.
$endgroup$
– José Carlos Santos
Jan 11 at 16:08
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
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oldest
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active
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$begingroup$
It follows from Darboux's theorem, that if $f$ is a differentiable function and if $f'$ is discontinuous at $ain D_f$, then the only possible cause of that discontinuity is that the limit $lim_{xto a}f'(x)$ doesn't exist.
answered Jan 11 at 15:37
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
$begingroup$
but can f'(a) still exist then? I know intuitively it can't but I am not too sure how to show it. Can the derivative function look like $$ f'(x) = left{ begin{array}{ll} -1 & quad x < 0 \ 0 & quad x=0\ 1 & quad x > 0 end{array} right. $$
$endgroup$
– matt
Jan 11 at 16:04
$begingroup$
No, by Darboux's theorem.
$endgroup$
– José Carlos Santos
Jan 11 at 16:08
add a comment |
$begingroup$
It follows from Darboux's theorem, that if $f$ is a differentiable function and if $f'$ is discontinuous at $ain D_f$, then the only possible cause of that discontinuity is that the limit $lim_{xto a}f'(x)$ doesn't exist.
answered Jan 11 at 15:37
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
$begingroup$
but can f'(a) still exist then? I know intuitively it can't but I am not too sure how to show it. Can the derivative function look like $$ f'(x) = left{ begin{array}{ll} -1 & quad x < 0 \ 0 & quad x=0\ 1 & quad x > 0 end{array} right. $$
$endgroup$
– matt
Jan 11 at 16:04
$begingroup$
No, by Darboux's theorem.
$endgroup$
– José Carlos Santos
Jan 11 at 16:08
add a comment |
$begingroup$
It follows from Darboux's theorem, that if $f$ is a differentiable function and if $f'$ is discontinuous at $ain D_f$, then the only possible cause of that discontinuity is that the limit $lim_{xto a}f'(x)$ doesn't exist.
answered Jan 11 at 15:37
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
It follows from Darboux's theorem, that if $f$ is a differentiable function and if $f'$ is discontinuous at $ain D_f$, then the only possible cause of that discontinuity is that the limit $lim_{xto a}f'(x)$ doesn't exist.
answered Jan 11 at 15:37
José Carlos SantosJosé Carlos Santos
156k22126227
edited Jan 11 at 16:28
answered Jan 11 at 15:37
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 11 at 15:37
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 11 at 15:37
José Carlos SantosJosé Carlos Santos
156k22126227
156k22126227
$begingroup$
but can f'(a) still exist then? I know intuitively it can't but I am not too sure how to show it. Can the derivative function look like $$ f'(x) = left{ begin{array}{ll} -1 & quad x < 0 \ 0 & quad x=0\ 1 & quad x > 0 end{array} right. $$
$endgroup$
– matt
Jan 11 at 16:04
$begingroup$
No, by Darboux's theorem.
$endgroup$
– José Carlos Santos
Jan 11 at 16:08
add a comment |
$begingroup$
but can f'(a) still exist then? I know intuitively it can't but I am not too sure how to show it. Can the derivative function look like $$ f'(x) = left{ begin{array}{ll} -1 & quad x < 0 \ 0 & quad x=0\ 1 & quad x > 0 end{array} right. $$
$endgroup$
– matt
Jan 11 at 16:04
$begingroup$
No, by Darboux's theorem.
$endgroup$
– José Carlos Santos
Jan 11 at 16:08
$begingroup$
but can f'(a) still exist then? I know intuitively it can't but I am not too sure how to show it. Can the derivative function look like $$ f'(x) = left{ begin{array}{ll} -1 & quad x < 0 \ 0 & quad x=0\ 1 & quad x > 0 end{array} right. $$
$endgroup$
– matt
Jan 11 at 16:04
$begingroup$
but can f'(a) still exist then? I know intuitively it can't but I am not too sure how to show it. Can the derivative function look like $$ f'(x) = left{ begin{array}{ll} -1 & quad x < 0 \ 0 & quad x=0\ 1 & quad x > 0 end{array} right. $$
$endgroup$
– matt
Jan 11 at 16:04
$begingroup$
No, by Darboux's theorem.
$endgroup$
– José Carlos Santos
Jan 11 at 16:08
$begingroup$
No, by Darboux's theorem.
$endgroup$
– José Carlos Santos
Jan 11 at 16:08
add a comment |
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$begingroup$
The derivative need not be continuous, but it must satisfy the IVP.
$endgroup$
– Randall
Jan 11 at 15:34