Are all functions that have a primitive differentiable?
$begingroup$
Are all functions that have a primitive differentiable?
For some background, I know that not all functions that are integrable are differentiable. For example:
$$
f =
begin{cases}
0 & x neq 0 \
1 & x = 0
end{cases}
$$
is integrable over $mathbb{R}$, and $int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) space forall x in mathbb{R}$ does not exist.
But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.
Thanks !
Edit: here is a very important comment from @HenningMakholm that I thought would be useful for other students encountering this question
Also, (real) functions that have a primitive are not necessarily continuous. For example,
$$
f(x) =
begin{cases}
0 & text{for } x = 0 \
2x sin(frac{1}{x}) - cos(frac{1}{x}) & text{otherwise}
end{cases}
$$
is discontinuous at $x=0$, but nevertheless the derivative of
$$
F(x) =
begin{cases}
0 & text{ for } x = 0 \
x^2 sin(frac{1}{x}) & text{ otherwise}
end{cases}
$$
Functions that have a primitive do have the intermediate-value property but that is weaker than being continuous.
calculus
edited Dec 30 '18 at 16:08
amWhy
192k28225439
asked Dec 29 '18 at 13:41
ninivertninivert
687
$endgroup$
|
show 4 more comments
$begingroup$
Are all functions that have a primitive differentiable?
For some background, I know that not all functions that are integrable are differentiable. For example:
$$
f =
begin{cases}
0 & x neq 0 \
1 & x = 0
end{cases}
$$
is integrable over $mathbb{R}$, and $int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) space forall x in mathbb{R}$ does not exist.
But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.
Thanks !
Edit: here is a very important comment from @HenningMakholm that I thought would be useful for other students encountering this question
Also, (real) functions that have a primitive are not necessarily continuous. For example,
$$
f(x) =
begin{cases}
0 & text{for } x = 0 \
2x sin(frac{1}{x}) - cos(frac{1}{x}) & text{otherwise}
end{cases}
$$
is discontinuous at $x=0$, but nevertheless the derivative of
$$
F(x) =
begin{cases}
0 & text{ for } x = 0 \
x^2 sin(frac{1}{x}) & text{ otherwise}
end{cases}
$$
Functions that have a primitive do have the intermediate-value property but that is weaker than being continuous.
calculus
edited Dec 30 '18 at 16:08
amWhy
192k28225439
asked Dec 29 '18 at 13:41
ninivertninivert
687
$endgroup$
$begingroup$
Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:50
$begingroup$
Take a function that is $C^1$ but not $C^2$.
$endgroup$
– dmtri
Dec 29 '18 at 13:51
$begingroup$
@dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:52
3
$begingroup$
Note that, famously, this is true for functions defined on an open subset of $mathbb C$.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:50
2
$begingroup$
Also, (real) functions that have a primitive are not necessarily continuous. For example, $$ f(x) = begin{cases} 0 & text{for }x=0 \ 2xsin(1/x) - cos(1/x) & text{otherwise} end{cases} $$ is discontinuous at $x=0$, but is nevertheless the derivative of $$ F(x) = begin{cases} 0 & text{for }x=0 \ x^2sin(1/x) & text{otherwise} end{cases} $$ Functions that have a primitive do have the intermetdiate-value property but that is weaker than being continuous.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:55
|
show 4 more comments
$begingroup$
Are all functions that have a primitive differentiable?
For some background, I know that not all functions that are integrable are differentiable. For example:
$$
f =
begin{cases}
0 & x neq 0 \
1 & x = 0
end{cases}
$$
is integrable over $mathbb{R}$, and $int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) space forall x in mathbb{R}$ does not exist.
But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.
Thanks !
Edit: here is a very important comment from @HenningMakholm that I thought would be useful for other students encountering this question
Also, (real) functions that have a primitive are not necessarily continuous. For example,
$$
f(x) =
begin{cases}
0 & text{for } x = 0 \
2x sin(frac{1}{x}) - cos(frac{1}{x}) & text{otherwise}
end{cases}
$$
is discontinuous at $x=0$, but nevertheless the derivative of
$$
F(x) =
begin{cases}
0 & text{ for } x = 0 \
x^2 sin(frac{1}{x}) & text{ otherwise}
end{cases}
$$
Functions that have a primitive do have the intermediate-value property but that is weaker than being continuous.
calculus
edited Dec 30 '18 at 16:08
amWhy
192k28225439
asked Dec 29 '18 at 13:41
ninivertninivert
687
$endgroup$
Are all functions that have a primitive differentiable?
For some background, I know that not all functions that are integrable are differentiable. For example:
$$
f =
begin{cases}
0 & x neq 0 \
1 & x = 0
end{cases}
$$
is integrable over $mathbb{R}$, and $int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) space forall x in mathbb{R}$ does not exist.
But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.
Thanks !
Edit: here is a very important comment from @HenningMakholm that I thought would be useful for other students encountering this question
Also, (real) functions that have a primitive are not necessarily continuous. For example,
$$
f(x) =
begin{cases}
0 & text{for } x = 0 \
2x sin(frac{1}{x}) - cos(frac{1}{x}) & text{otherwise}
end{cases}
$$
is discontinuous at $x=0$, but nevertheless the derivative of
$$
F(x) =
begin{cases}
0 & text{ for } x = 0 \
x^2 sin(frac{1}{x}) & text{ otherwise}
end{cases}
$$
Functions that have a primitive do have the intermediate-value property but that is weaker than being continuous.
calculus
calculus
edited Dec 30 '18 at 16:08
amWhy
192k28225439
asked Dec 29 '18 at 13:41
ninivertninivert
687
edited Dec 30 '18 at 16:08
amWhy
192k28225439
asked Dec 29 '18 at 13:41
ninivertninivert
687
edited Dec 30 '18 at 16:08
amWhy
192k28225439
edited Dec 30 '18 at 16:08
amWhy
192k28225439
edited Dec 30 '18 at 16:08
amWhy
192k28225439
192k28225439
asked Dec 29 '18 at 13:41
ninivertninivert
687
asked Dec 29 '18 at 13:41
ninivertninivert
687
asked Dec 29 '18 at 13:41
ninivertninivert
687
687
$begingroup$
Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:50
$begingroup$
Take a function that is $C^1$ but not $C^2$.
$endgroup$
– dmtri
Dec 29 '18 at 13:51
$begingroup$
@dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:52
3
$begingroup$
Note that, famously, this is true for functions defined on an open subset of $mathbb C$.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:50
2
$begingroup$
Also, (real) functions that have a primitive are not necessarily continuous. For example, $$ f(x) = begin{cases} 0 & text{for }x=0 \ 2xsin(1/x) - cos(1/x) & text{otherwise} end{cases} $$ is discontinuous at $x=0$, but is nevertheless the derivative of $$ F(x) = begin{cases} 0 & text{for }x=0 \ x^2sin(1/x) & text{otherwise} end{cases} $$ Functions that have a primitive do have the intermetdiate-value property but that is weaker than being continuous.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:55
|
show 4 more comments
$begingroup$
Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:50
$begingroup$
Take a function that is $C^1$ but not $C^2$.
$endgroup$
– dmtri
Dec 29 '18 at 13:51
$begingroup$
@dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:52
3
$begingroup$
Note that, famously, this is true for functions defined on an open subset of $mathbb C$.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:50
2
$begingroup$
Also, (real) functions that have a primitive are not necessarily continuous. For example, $$ f(x) = begin{cases} 0 & text{for }x=0 \ 2xsin(1/x) - cos(1/x) & text{otherwise} end{cases} $$ is discontinuous at $x=0$, but is nevertheless the derivative of $$ F(x) = begin{cases} 0 & text{for }x=0 \ x^2sin(1/x) & text{otherwise} end{cases} $$ Functions that have a primitive do have the intermetdiate-value property but that is weaker than being continuous.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:55
$begingroup$
Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:50
$begingroup$
Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:50
$begingroup$
Take a function that is $C^1$ but not $C^2$.
$endgroup$
– dmtri
Dec 29 '18 at 13:51
$begingroup$
Take a function that is $C^1$ but not $C^2$.
$endgroup$
– dmtri
Dec 29 '18 at 13:51
$begingroup$
@dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:52
$begingroup$
@dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:52
3
3
$begingroup$
Note that, famously, this is true for functions defined on an open subset of $mathbb C$.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:50
$begingroup$
Note that, famously, this is true for functions defined on an open subset of $mathbb C$.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:50
2
2
$begingroup$
Also, (real) functions that have a primitive are not necessarily continuous. For example, $$ f(x) = begin{cases} 0 & text{for }x=0 \ 2xsin(1/x) - cos(1/x) & text{otherwise} end{cases} $$ is discontinuous at $x=0$, but is nevertheless the derivative of $$ F(x) = begin{cases} 0 & text{for }x=0 \ x^2sin(1/x) & text{otherwise} end{cases} $$ Functions that have a primitive do have the intermetdiate-value property but that is weaker than being continuous.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:55
$begingroup$
Also, (real) functions that have a primitive are not necessarily continuous. For example, $$ f(x) = begin{cases} 0 & text{for }x=0 \ 2xsin(1/x) - cos(1/x) & text{otherwise} end{cases} $$ is discontinuous at $x=0$, but is nevertheless the derivative of $$ F(x) = begin{cases} 0 & text{for }x=0 \ x^2sin(1/x) & text{otherwise} end{cases} $$ Functions that have a primitive do have the intermetdiate-value property but that is weaker than being continuous.
$endgroup$
– Henning Makholm
Dec 29 '18 at 15:55
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$
answered Dec 29 '18 at 13:51
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
$endgroup$
$begingroup$
Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:55
2
$begingroup$
@AndrésE.Caicedo I edited to be more clarifying
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:02
add a comment |
$begingroup$
Even more simply: $f(x) := |x|$.
answered Dec 29 '18 at 14:55
The_SympathizerThe_Sympathizer
7,4852245
$endgroup$
3
$begingroup$
Of course, how did I not think about that ? I already accepted the @Martin's answer, but thanks for the example ! For further reference, I'll just state here that $int |x| dx = frac{1}{2}x|x|$
$endgroup$
– ninivert
Dec 29 '18 at 15:26
add a comment |
$begingroup$
Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.
$endgroup$
$begingroup$
@Martin has justified my statement.
$endgroup$
– user630002
Dec 29 '18 at 13:54
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$
answered Dec 29 '18 at 13:51
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
$endgroup$
$begingroup$
Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:55
2
$begingroup$
@AndrésE.Caicedo I edited to be more clarifying
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:02
add a comment |
$begingroup$
A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$
answered Dec 29 '18 at 13:51
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
$endgroup$
$begingroup$
Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:55
2
$begingroup$
@AndrésE.Caicedo I edited to be more clarifying
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:02
add a comment |
$begingroup$
A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$
answered Dec 29 '18 at 13:51
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
$endgroup$
A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$
answered Dec 29 '18 at 13:51
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
edited Dec 29 '18 at 14:01
answered Dec 29 '18 at 13:51
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
answered Dec 29 '18 at 13:51
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
answered Dec 29 '18 at 13:51
Martín Vacas VignoloMartín Vacas Vignolo
3,810623
3,810623
$begingroup$
Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:55
2
$begingroup$
@AndrésE.Caicedo I edited to be more clarifying
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:02
add a comment |
$begingroup$
Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:55
2
$begingroup$
@AndrésE.Caicedo I edited to be more clarifying
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:02
$begingroup$
Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:55
$begingroup$
Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
$endgroup$
– Andrés E. Caicedo
Dec 29 '18 at 13:55
2
2
$begingroup$
@AndrésE.Caicedo I edited to be more clarifying
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:02
$begingroup$
@AndrésE.Caicedo I edited to be more clarifying
$endgroup$
– Martín Vacas Vignolo
Dec 29 '18 at 14:02
add a comment |
$begingroup$
Even more simply: $f(x) := |x|$.
answered Dec 29 '18 at 14:55
The_SympathizerThe_Sympathizer
7,4852245
$endgroup$
3
$begingroup$
Of course, how did I not think about that ? I already accepted the @Martin's answer, but thanks for the example ! For further reference, I'll just state here that $int |x| dx = frac{1}{2}x|x|$
$endgroup$
– ninivert
Dec 29 '18 at 15:26
add a comment |
$begingroup$
Even more simply: $f(x) := |x|$.
answered Dec 29 '18 at 14:55
The_SympathizerThe_Sympathizer
7,4852245
$endgroup$
3
$begingroup$
Of course, how did I not think about that ? I already accepted the @Martin's answer, but thanks for the example ! For further reference, I'll just state here that $int |x| dx = frac{1}{2}x|x|$
$endgroup$
– ninivert
Dec 29 '18 at 15:26
add a comment |
$begingroup$
Even more simply: $f(x) := |x|$.
answered Dec 29 '18 at 14:55
The_SympathizerThe_Sympathizer
7,4852245
$endgroup$
Even more simply: $f(x) := |x|$.
answered Dec 29 '18 at 14:55
The_SympathizerThe_Sympathizer
7,4852245
answered Dec 29 '18 at 14:55
The_SympathizerThe_Sympathizer
7,4852245
answered Dec 29 '18 at 14:55
The_SympathizerThe_Sympathizer
7,4852245
answered Dec 29 '18 at 14:55
The_SympathizerThe_Sympathizer
7,4852245
7,4852245
3
$begingroup$
Of course, how did I not think about that ? I already accepted the @Martin's answer, but thanks for the example ! For further reference, I'll just state here that $int |x| dx = frac{1}{2}x|x|$
$endgroup$
– ninivert
Dec 29 '18 at 15:26
add a comment |
3
$begingroup$
Of course, how did I not think about that ? I already accepted the @Martin's answer, but thanks for the example ! For further reference, I'll just state here that $int |x| dx = frac{1}{2}x|x|$
$endgroup$
– ninivert
Dec 29 '18 at 15:26
3
3
$begingroup$
Of course, how did I not think about that ? I already accepted the @Martin's answer, but thanks for the example ! For further reference, I'll just state here that $int |x| dx = frac{1}{2}x|x|$
$endgroup$
– ninivert
Dec 29 '18 at 15:26
$begingroup$
Of course, how did I not think about that ? I already accepted the @Martin's answer, but thanks for the example ! For further reference, I'll just state here that $int |x| dx = frac{1}{2}x|x|$
$endgroup$
– ninivert
Dec 29 '18 at 15:26
add a comment |
$begingroup$
Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.
$endgroup$
$begingroup$
@Martin has justified my statement.
$endgroup$
– user630002
Dec 29 '18 at 13:54
add a comment |
$begingroup$
Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.
$endgroup$
$begingroup$
@Martin has justified my statement.
$endgroup$
– user630002
Dec 29 '18 at 13:54
add a comment |
$begingroup$
Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.
$endgroup$
Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.
answered Dec 29 '18 at 13:53
user630002
$begingroup$
@Martin has justified my statement.
$endgroup$
– user630002
Dec 29 '18 at 13:54
add a comment |
$begingroup$
@Martin has justified my statement.
$endgroup$
– user630002
Dec 29 '18 at 13:54
$begingroup$
@Martin has justified my statement.
$endgroup$
– user630002
Dec 29 '18 at 13:54
$begingroup$
@Martin has justified my statement.
$endgroup$
– user630002
Dec 29 '18 at 13:54
add a comment |
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Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
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– Andrés E. Caicedo
Dec 29 '18 at 13:50
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Take a function that is $C^1$ but not $C^2$.
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– dmtri
Dec 29 '18 at 13:51
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@dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
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– Andrés E. Caicedo
Dec 29 '18 at 13:52
3
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Note that, famously, this is true for functions defined on an open subset of $mathbb C$.
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– Henning Makholm
Dec 29 '18 at 15:50
2
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Also, (real) functions that have a primitive are not necessarily continuous. For example, $$ f(x) = begin{cases} 0 & text{for }x=0 \ 2xsin(1/x) - cos(1/x) & text{otherwise} end{cases} $$ is discontinuous at $x=0$, but is nevertheless the derivative of $$ F(x) = begin{cases} 0 & text{for }x=0 \ x^2sin(1/x) & text{otherwise} end{cases} $$ Functions that have a primitive do have the intermetdiate-value property but that is weaker than being continuous.
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– Henning Makholm
Dec 29 '18 at 15:55