Does $f'(x)=0$ implies Max/min or point of inflection
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Let $f$ be a differentiable function on $(a,b)$ such that $f'(c)=0$ for some $cin (a,b)$. Also there is no maximum or minimum at $c$ .Does this means $c$ is point of inflection?
Consider $f(x)=x^3$ , this is true. But I want to know if it is true in general. I try to find counter example but can't get any. I hope this is true. If it is please give a hint to begin a proof
Edit : hint this question by maxima / minima, I mean strict maxima or minima
real-analysis calculus derivatives maxima-minima
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|
show 4 more comments
$begingroup$
Let $f$ be a differentiable function on $(a,b)$ such that $f'(c)=0$ for some $cin (a,b)$. Also there is no maximum or minimum at $c$ .Does this means $c$ is point of inflection?
Consider $f(x)=x^3$ , this is true. But I want to know if it is true in general. I try to find counter example but can't get any. I hope this is true. If it is please give a hint to begin a proof
Edit : hint this question by maxima / minima, I mean strict maxima or minima
real-analysis calculus derivatives maxima-minima
$endgroup$
$begingroup$
If it's differential (big if) then $f'(c) =0$ means is an extreme point. Geometrically this means the slope is zero. This is either a local min or max or inflection point. As you know it's neither of two of the options, it must be the third option.
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– fleablood
Dec 21 '18 at 17:15
1
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@fleablood: I don't believe you're correct. A (local) extreme point is by definition a (local) maximum or minimum. So $f'(c)=0$ does NOT imply that $c$ is a local extreme point.
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– Ted Shifrin
Dec 21 '18 at 17:19
1
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I think you need to check your definitions carefully, @fleablood.
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– Ted Shifrin
Dec 21 '18 at 17:23
1
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Actually you're right. I was thinking of the term "critical points".
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– fleablood
Dec 21 '18 at 17:25
1
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@fleablood It is true that local extremum implies critical point, I think you confuse this with the converese which is not true. And inflection point is typically defined as a point where the concavity changes.
$endgroup$
– N. S.
Dec 21 '18 at 17:37
|
show 4 more comments
$begingroup$
Let $f$ be a differentiable function on $(a,b)$ such that $f'(c)=0$ for some $cin (a,b)$. Also there is no maximum or minimum at $c$ .Does this means $c$ is point of inflection?
Consider $f(x)=x^3$ , this is true. But I want to know if it is true in general. I try to find counter example but can't get any. I hope this is true. If it is please give a hint to begin a proof
Edit : hint this question by maxima / minima, I mean strict maxima or minima
real-analysis calculus derivatives maxima-minima
$endgroup$
Let $f$ be a differentiable function on $(a,b)$ such that $f'(c)=0$ for some $cin (a,b)$. Also there is no maximum or minimum at $c$ .Does this means $c$ is point of inflection?
Consider $f(x)=x^3$ , this is true. But I want to know if it is true in general. I try to find counter example but can't get any. I hope this is true. If it is please give a hint to begin a proof
Edit : hint this question by maxima / minima, I mean strict maxima or minima
real-analysis calculus derivatives maxima-minima
real-analysis calculus derivatives maxima-minima
edited Dec 21 '18 at 17:23
Cloud JR
asked Dec 21 '18 at 17:10
Cloud JRCloud JR
930518
930518
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If it's differential (big if) then $f'(c) =0$ means is an extreme point. Geometrically this means the slope is zero. This is either a local min or max or inflection point. As you know it's neither of two of the options, it must be the third option.
$endgroup$
– fleablood
Dec 21 '18 at 17:15
1
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@fleablood: I don't believe you're correct. A (local) extreme point is by definition a (local) maximum or minimum. So $f'(c)=0$ does NOT imply that $c$ is a local extreme point.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:19
1
$begingroup$
I think you need to check your definitions carefully, @fleablood.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:23
1
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Actually you're right. I was thinking of the term "critical points".
$endgroup$
– fleablood
Dec 21 '18 at 17:25
1
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@fleablood It is true that local extremum implies critical point, I think you confuse this with the converese which is not true. And inflection point is typically defined as a point where the concavity changes.
$endgroup$
– N. S.
Dec 21 '18 at 17:37
|
show 4 more comments
$begingroup$
If it's differential (big if) then $f'(c) =0$ means is an extreme point. Geometrically this means the slope is zero. This is either a local min or max or inflection point. As you know it's neither of two of the options, it must be the third option.
$endgroup$
– fleablood
Dec 21 '18 at 17:15
1
$begingroup$
@fleablood: I don't believe you're correct. A (local) extreme point is by definition a (local) maximum or minimum. So $f'(c)=0$ does NOT imply that $c$ is a local extreme point.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:19
1
$begingroup$
I think you need to check your definitions carefully, @fleablood.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:23
1
$begingroup$
Actually you're right. I was thinking of the term "critical points".
$endgroup$
– fleablood
Dec 21 '18 at 17:25
1
$begingroup$
@fleablood It is true that local extremum implies critical point, I think you confuse this with the converese which is not true. And inflection point is typically defined as a point where the concavity changes.
$endgroup$
– N. S.
Dec 21 '18 at 17:37
$begingroup$
If it's differential (big if) then $f'(c) =0$ means is an extreme point. Geometrically this means the slope is zero. This is either a local min or max or inflection point. As you know it's neither of two of the options, it must be the third option.
$endgroup$
– fleablood
Dec 21 '18 at 17:15
$begingroup$
If it's differential (big if) then $f'(c) =0$ means is an extreme point. Geometrically this means the slope is zero. This is either a local min or max or inflection point. As you know it's neither of two of the options, it must be the third option.
$endgroup$
– fleablood
Dec 21 '18 at 17:15
1
1
$begingroup$
@fleablood: I don't believe you're correct. A (local) extreme point is by definition a (local) maximum or minimum. So $f'(c)=0$ does NOT imply that $c$ is a local extreme point.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:19
$begingroup$
@fleablood: I don't believe you're correct. A (local) extreme point is by definition a (local) maximum or minimum. So $f'(c)=0$ does NOT imply that $c$ is a local extreme point.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:19
1
1
$begingroup$
I think you need to check your definitions carefully, @fleablood.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:23
$begingroup$
I think you need to check your definitions carefully, @fleablood.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:23
1
1
$begingroup$
Actually you're right. I was thinking of the term "critical points".
$endgroup$
– fleablood
Dec 21 '18 at 17:25
$begingroup$
Actually you're right. I was thinking of the term "critical points".
$endgroup$
– fleablood
Dec 21 '18 at 17:25
1
1
$begingroup$
@fleablood It is true that local extremum implies critical point, I think you confuse this with the converese which is not true. And inflection point is typically defined as a point where the concavity changes.
$endgroup$
– N. S.
Dec 21 '18 at 17:37
$begingroup$
@fleablood It is true that local extremum implies critical point, I think you confuse this with the converese which is not true. And inflection point is typically defined as a point where the concavity changes.
$endgroup$
– N. S.
Dec 21 '18 at 17:37
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Consider the function
$$g(x)=xsin(frac{1}{x})$$
with $g(0)=0$.
This function is continuous on $mathbb R$.
Let $F(x)$ be any antiderivative of $g(x)$. Then $F'(0)=0$.
Now, since $g(x)$ is an even function, $F$ is odd, and hence it cannot have a local max/min at $x=0$.
Moreover, $F(x)$ cannot have an inflection point at $x=0$, since this would imply that for some $(0,a)$ the function $g(x)$ would be monotonic.
Added: if $ngeq 2$, $f$ is $n$ times differentiable, $f^{(n)}$ is continuous at $c$ and
$$f'(c)=...=f^{(n-1)}(c)=0 \
f^{(n)}(c) neq 0$$
then
- If $n$ is odd $c$ is an inflection point
- If $n$ is even and $f^{(n)}(c) >0$ then $c$ is a local min.
- If $n$ is even and $f^{(n)}(c) <0$ then $c$ is a local max.
Sketch Proof: Since $f^{(n)}(c) neq 0$ it is either positive or negative.
If $f^{(n)}(c)<0$ then replace $f$ by $-f$. Note that in the case $n$ even thiwill change local max to min.
By continuity, there exists some $a>0$ such that $f^{(n)}>0$ on $(c-a, c+a)$.
Now you do an inductive argument. $f^{(n)}>0$ on $(c-a, c+a)$ means $f^{(n-1)}$ is strictly increasing on $(c-a, c+a)$ and thus, since $f^{(n-1)}(c)=0$ you get
$$f^{n-1}(x) <0 forall x in (c-a,c) \
f^{n-1}(x) >0 forall x in (c,c-a) (*)\$$
This gives that $f^{n-2}$ is decreasing on (c-a,c)$ and increasing on $(c,c+a)$. Therefore
$$f^{n-2}(x) >0 forall x in (c-a,c)
f^{n-2}(x) >0 forall x in (c,c-a) (**)$$
Then, same argument shows that
$$f^{n-3}(x) <0 forall x in (c-a,c) \
f^{n-3}(x) >0 forall x in (c,c-a) (*)\$$
and so on, with (*) and (**)alternating. Based of $n$ being odd or even, either $f'$ or $f''$ will satisfy $(*)$ [this is why we need $n geq 2$].
Now, if $f'$ satisfies $(*)$ then it is easy to see that $c$ is a local max.
If $f''$ satisfies $(*)$ theb it is easy to see that $c$ is an inflection point.
QED
P.P.S. The key for the proof is the existence and continuity of the first derivative which doesn't vanish at $c$. The above counterexample fails this :)
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Thanks it helps a lot
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– Cloud JR
Dec 21 '18 at 17:38
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@CloudJR Check my P.S. too, the claim is actually true for "nice" functions.
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– N. S.
Dec 21 '18 at 17:41
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@CloudJR If you need the details I can sketch the idea of proof for analytic functions..."Analyticity" can actually be replaced by $f$ is $n$ times differentiable, $f^{(n)}$ is continuous and $$f'(c)=....f^{(n-1)}(c)=0 \f^{(n)}(c) neq 0$$
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– N. S.
Dec 21 '18 at 17:46
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Where can I find more about this....like proof and more ..any book?
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– Cloud JR
Dec 21 '18 at 17:46
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@CloudJR see the new edit.
$endgroup$
– N. S.
Dec 21 '18 at 18:06
|
show 4 more comments
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The simplest counter example is the constant function
$$
fequiv 0
$$
on any interval $(a,b)$. Clearly $f'(c)=0$ for any $cin(a,b)$ but it's not a point of inflection.
Edit: I interpreted the words max/min to mean strict minima/maxima, if the question refers to nonstrict ones then this is not a counter example.
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But every point is a local min/max.
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– N. S.
Dec 21 '18 at 17:17
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But every point is maximum or minimum
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– Cloud JR
Dec 21 '18 at 17:17
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Yeah, I just realized that I probably misinterpreted the question. Sorry for that.
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– BigbearZzz
Dec 21 '18 at 17:18
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Consider the function
$$g(x)=xsin(frac{1}{x})$$
with $g(0)=0$.
This function is continuous on $mathbb R$.
Let $F(x)$ be any antiderivative of $g(x)$. Then $F'(0)=0$.
Now, since $g(x)$ is an even function, $F$ is odd, and hence it cannot have a local max/min at $x=0$.
Moreover, $F(x)$ cannot have an inflection point at $x=0$, since this would imply that for some $(0,a)$ the function $g(x)$ would be monotonic.
Added: if $ngeq 2$, $f$ is $n$ times differentiable, $f^{(n)}$ is continuous at $c$ and
$$f'(c)=...=f^{(n-1)}(c)=0 \
f^{(n)}(c) neq 0$$
then
- If $n$ is odd $c$ is an inflection point
- If $n$ is even and $f^{(n)}(c) >0$ then $c$ is a local min.
- If $n$ is even and $f^{(n)}(c) <0$ then $c$ is a local max.
Sketch Proof: Since $f^{(n)}(c) neq 0$ it is either positive or negative.
If $f^{(n)}(c)<0$ then replace $f$ by $-f$. Note that in the case $n$ even thiwill change local max to min.
By continuity, there exists some $a>0$ such that $f^{(n)}>0$ on $(c-a, c+a)$.
Now you do an inductive argument. $f^{(n)}>0$ on $(c-a, c+a)$ means $f^{(n-1)}$ is strictly increasing on $(c-a, c+a)$ and thus, since $f^{(n-1)}(c)=0$ you get
$$f^{n-1}(x) <0 forall x in (c-a,c) \
f^{n-1}(x) >0 forall x in (c,c-a) (*)\$$
This gives that $f^{n-2}$ is decreasing on (c-a,c)$ and increasing on $(c,c+a)$. Therefore
$$f^{n-2}(x) >0 forall x in (c-a,c)
f^{n-2}(x) >0 forall x in (c,c-a) (**)$$
Then, same argument shows that
$$f^{n-3}(x) <0 forall x in (c-a,c) \
f^{n-3}(x) >0 forall x in (c,c-a) (*)\$$
and so on, with (*) and (**)alternating. Based of $n$ being odd or even, either $f'$ or $f''$ will satisfy $(*)$ [this is why we need $n geq 2$].
Now, if $f'$ satisfies $(*)$ then it is easy to see that $c$ is a local max.
If $f''$ satisfies $(*)$ theb it is easy to see that $c$ is an inflection point.
QED
P.P.S. The key for the proof is the existence and continuity of the first derivative which doesn't vanish at $c$. The above counterexample fails this :)
$endgroup$
$begingroup$
Thanks it helps a lot
$endgroup$
– Cloud JR
Dec 21 '18 at 17:38
$begingroup$
@CloudJR Check my P.S. too, the claim is actually true for "nice" functions.
$endgroup$
– N. S.
Dec 21 '18 at 17:41
$begingroup$
@CloudJR If you need the details I can sketch the idea of proof for analytic functions..."Analyticity" can actually be replaced by $f$ is $n$ times differentiable, $f^{(n)}$ is continuous and $$f'(c)=....f^{(n-1)}(c)=0 \f^{(n)}(c) neq 0$$
$endgroup$
– N. S.
Dec 21 '18 at 17:46
$begingroup$
Where can I find more about this....like proof and more ..any book?
$endgroup$
– Cloud JR
Dec 21 '18 at 17:46
$begingroup$
@CloudJR see the new edit.
$endgroup$
– N. S.
Dec 21 '18 at 18:06
|
show 4 more comments
$begingroup$
Consider the function
$$g(x)=xsin(frac{1}{x})$$
with $g(0)=0$.
This function is continuous on $mathbb R$.
Let $F(x)$ be any antiderivative of $g(x)$. Then $F'(0)=0$.
Now, since $g(x)$ is an even function, $F$ is odd, and hence it cannot have a local max/min at $x=0$.
Moreover, $F(x)$ cannot have an inflection point at $x=0$, since this would imply that for some $(0,a)$ the function $g(x)$ would be monotonic.
Added: if $ngeq 2$, $f$ is $n$ times differentiable, $f^{(n)}$ is continuous at $c$ and
$$f'(c)=...=f^{(n-1)}(c)=0 \
f^{(n)}(c) neq 0$$
then
- If $n$ is odd $c$ is an inflection point
- If $n$ is even and $f^{(n)}(c) >0$ then $c$ is a local min.
- If $n$ is even and $f^{(n)}(c) <0$ then $c$ is a local max.
Sketch Proof: Since $f^{(n)}(c) neq 0$ it is either positive or negative.
If $f^{(n)}(c)<0$ then replace $f$ by $-f$. Note that in the case $n$ even thiwill change local max to min.
By continuity, there exists some $a>0$ such that $f^{(n)}>0$ on $(c-a, c+a)$.
Now you do an inductive argument. $f^{(n)}>0$ on $(c-a, c+a)$ means $f^{(n-1)}$ is strictly increasing on $(c-a, c+a)$ and thus, since $f^{(n-1)}(c)=0$ you get
$$f^{n-1}(x) <0 forall x in (c-a,c) \
f^{n-1}(x) >0 forall x in (c,c-a) (*)\$$
This gives that $f^{n-2}$ is decreasing on (c-a,c)$ and increasing on $(c,c+a)$. Therefore
$$f^{n-2}(x) >0 forall x in (c-a,c)
f^{n-2}(x) >0 forall x in (c,c-a) (**)$$
Then, same argument shows that
$$f^{n-3}(x) <0 forall x in (c-a,c) \
f^{n-3}(x) >0 forall x in (c,c-a) (*)\$$
and so on, with (*) and (**)alternating. Based of $n$ being odd or even, either $f'$ or $f''$ will satisfy $(*)$ [this is why we need $n geq 2$].
Now, if $f'$ satisfies $(*)$ then it is easy to see that $c$ is a local max.
If $f''$ satisfies $(*)$ theb it is easy to see that $c$ is an inflection point.
QED
P.P.S. The key for the proof is the existence and continuity of the first derivative which doesn't vanish at $c$. The above counterexample fails this :)
$endgroup$
$begingroup$
Thanks it helps a lot
$endgroup$
– Cloud JR
Dec 21 '18 at 17:38
$begingroup$
@CloudJR Check my P.S. too, the claim is actually true for "nice" functions.
$endgroup$
– N. S.
Dec 21 '18 at 17:41
$begingroup$
@CloudJR If you need the details I can sketch the idea of proof for analytic functions..."Analyticity" can actually be replaced by $f$ is $n$ times differentiable, $f^{(n)}$ is continuous and $$f'(c)=....f^{(n-1)}(c)=0 \f^{(n)}(c) neq 0$$
$endgroup$
– N. S.
Dec 21 '18 at 17:46
$begingroup$
Where can I find more about this....like proof and more ..any book?
$endgroup$
– Cloud JR
Dec 21 '18 at 17:46
$begingroup$
@CloudJR see the new edit.
$endgroup$
– N. S.
Dec 21 '18 at 18:06
|
show 4 more comments
$begingroup$
Consider the function
$$g(x)=xsin(frac{1}{x})$$
with $g(0)=0$.
This function is continuous on $mathbb R$.
Let $F(x)$ be any antiderivative of $g(x)$. Then $F'(0)=0$.
Now, since $g(x)$ is an even function, $F$ is odd, and hence it cannot have a local max/min at $x=0$.
Moreover, $F(x)$ cannot have an inflection point at $x=0$, since this would imply that for some $(0,a)$ the function $g(x)$ would be monotonic.
Added: if $ngeq 2$, $f$ is $n$ times differentiable, $f^{(n)}$ is continuous at $c$ and
$$f'(c)=...=f^{(n-1)}(c)=0 \
f^{(n)}(c) neq 0$$
then
- If $n$ is odd $c$ is an inflection point
- If $n$ is even and $f^{(n)}(c) >0$ then $c$ is a local min.
- If $n$ is even and $f^{(n)}(c) <0$ then $c$ is a local max.
Sketch Proof: Since $f^{(n)}(c) neq 0$ it is either positive or negative.
If $f^{(n)}(c)<0$ then replace $f$ by $-f$. Note that in the case $n$ even thiwill change local max to min.
By continuity, there exists some $a>0$ such that $f^{(n)}>0$ on $(c-a, c+a)$.
Now you do an inductive argument. $f^{(n)}>0$ on $(c-a, c+a)$ means $f^{(n-1)}$ is strictly increasing on $(c-a, c+a)$ and thus, since $f^{(n-1)}(c)=0$ you get
$$f^{n-1}(x) <0 forall x in (c-a,c) \
f^{n-1}(x) >0 forall x in (c,c-a) (*)\$$
This gives that $f^{n-2}$ is decreasing on (c-a,c)$ and increasing on $(c,c+a)$. Therefore
$$f^{n-2}(x) >0 forall x in (c-a,c)
f^{n-2}(x) >0 forall x in (c,c-a) (**)$$
Then, same argument shows that
$$f^{n-3}(x) <0 forall x in (c-a,c) \
f^{n-3}(x) >0 forall x in (c,c-a) (*)\$$
and so on, with (*) and (**)alternating. Based of $n$ being odd or even, either $f'$ or $f''$ will satisfy $(*)$ [this is why we need $n geq 2$].
Now, if $f'$ satisfies $(*)$ then it is easy to see that $c$ is a local max.
If $f''$ satisfies $(*)$ theb it is easy to see that $c$ is an inflection point.
QED
P.P.S. The key for the proof is the existence and continuity of the first derivative which doesn't vanish at $c$. The above counterexample fails this :)
$endgroup$
Consider the function
$$g(x)=xsin(frac{1}{x})$$
with $g(0)=0$.
This function is continuous on $mathbb R$.
Let $F(x)$ be any antiderivative of $g(x)$. Then $F'(0)=0$.
Now, since $g(x)$ is an even function, $F$ is odd, and hence it cannot have a local max/min at $x=0$.
Moreover, $F(x)$ cannot have an inflection point at $x=0$, since this would imply that for some $(0,a)$ the function $g(x)$ would be monotonic.
Added: if $ngeq 2$, $f$ is $n$ times differentiable, $f^{(n)}$ is continuous at $c$ and
$$f'(c)=...=f^{(n-1)}(c)=0 \
f^{(n)}(c) neq 0$$
then
- If $n$ is odd $c$ is an inflection point
- If $n$ is even and $f^{(n)}(c) >0$ then $c$ is a local min.
- If $n$ is even and $f^{(n)}(c) <0$ then $c$ is a local max.
Sketch Proof: Since $f^{(n)}(c) neq 0$ it is either positive or negative.
If $f^{(n)}(c)<0$ then replace $f$ by $-f$. Note that in the case $n$ even thiwill change local max to min.
By continuity, there exists some $a>0$ such that $f^{(n)}>0$ on $(c-a, c+a)$.
Now you do an inductive argument. $f^{(n)}>0$ on $(c-a, c+a)$ means $f^{(n-1)}$ is strictly increasing on $(c-a, c+a)$ and thus, since $f^{(n-1)}(c)=0$ you get
$$f^{n-1}(x) <0 forall x in (c-a,c) \
f^{n-1}(x) >0 forall x in (c,c-a) (*)\$$
This gives that $f^{n-2}$ is decreasing on (c-a,c)$ and increasing on $(c,c+a)$. Therefore
$$f^{n-2}(x) >0 forall x in (c-a,c)
f^{n-2}(x) >0 forall x in (c,c-a) (**)$$
Then, same argument shows that
$$f^{n-3}(x) <0 forall x in (c-a,c) \
f^{n-3}(x) >0 forall x in (c,c-a) (*)\$$
and so on, with (*) and (**)alternating. Based of $n$ being odd or even, either $f'$ or $f''$ will satisfy $(*)$ [this is why we need $n geq 2$].
Now, if $f'$ satisfies $(*)$ then it is easy to see that $c$ is a local max.
If $f''$ satisfies $(*)$ theb it is easy to see that $c$ is an inflection point.
QED
P.P.S. The key for the proof is the existence and continuity of the first derivative which doesn't vanish at $c$. The above counterexample fails this :)
edited Dec 21 '18 at 18:08
answered Dec 21 '18 at 17:16
N. S.N. S.
105k7115210
105k7115210
$begingroup$
Thanks it helps a lot
$endgroup$
– Cloud JR
Dec 21 '18 at 17:38
$begingroup$
@CloudJR Check my P.S. too, the claim is actually true for "nice" functions.
$endgroup$
– N. S.
Dec 21 '18 at 17:41
$begingroup$
@CloudJR If you need the details I can sketch the idea of proof for analytic functions..."Analyticity" can actually be replaced by $f$ is $n$ times differentiable, $f^{(n)}$ is continuous and $$f'(c)=....f^{(n-1)}(c)=0 \f^{(n)}(c) neq 0$$
$endgroup$
– N. S.
Dec 21 '18 at 17:46
$begingroup$
Where can I find more about this....like proof and more ..any book?
$endgroup$
– Cloud JR
Dec 21 '18 at 17:46
$begingroup$
@CloudJR see the new edit.
$endgroup$
– N. S.
Dec 21 '18 at 18:06
|
show 4 more comments
$begingroup$
Thanks it helps a lot
$endgroup$
– Cloud JR
Dec 21 '18 at 17:38
$begingroup$
@CloudJR Check my P.S. too, the claim is actually true for "nice" functions.
$endgroup$
– N. S.
Dec 21 '18 at 17:41
$begingroup$
@CloudJR If you need the details I can sketch the idea of proof for analytic functions..."Analyticity" can actually be replaced by $f$ is $n$ times differentiable, $f^{(n)}$ is continuous and $$f'(c)=....f^{(n-1)}(c)=0 \f^{(n)}(c) neq 0$$
$endgroup$
– N. S.
Dec 21 '18 at 17:46
$begingroup$
Where can I find more about this....like proof and more ..any book?
$endgroup$
– Cloud JR
Dec 21 '18 at 17:46
$begingroup$
@CloudJR see the new edit.
$endgroup$
– N. S.
Dec 21 '18 at 18:06
$begingroup$
Thanks it helps a lot
$endgroup$
– Cloud JR
Dec 21 '18 at 17:38
$begingroup$
Thanks it helps a lot
$endgroup$
– Cloud JR
Dec 21 '18 at 17:38
$begingroup$
@CloudJR Check my P.S. too, the claim is actually true for "nice" functions.
$endgroup$
– N. S.
Dec 21 '18 at 17:41
$begingroup$
@CloudJR Check my P.S. too, the claim is actually true for "nice" functions.
$endgroup$
– N. S.
Dec 21 '18 at 17:41
$begingroup$
@CloudJR If you need the details I can sketch the idea of proof for analytic functions..."Analyticity" can actually be replaced by $f$ is $n$ times differentiable, $f^{(n)}$ is continuous and $$f'(c)=....f^{(n-1)}(c)=0 \f^{(n)}(c) neq 0$$
$endgroup$
– N. S.
Dec 21 '18 at 17:46
$begingroup$
@CloudJR If you need the details I can sketch the idea of proof for analytic functions..."Analyticity" can actually be replaced by $f$ is $n$ times differentiable, $f^{(n)}$ is continuous and $$f'(c)=....f^{(n-1)}(c)=0 \f^{(n)}(c) neq 0$$
$endgroup$
– N. S.
Dec 21 '18 at 17:46
$begingroup$
Where can I find more about this....like proof and more ..any book?
$endgroup$
– Cloud JR
Dec 21 '18 at 17:46
$begingroup$
Where can I find more about this....like proof and more ..any book?
$endgroup$
– Cloud JR
Dec 21 '18 at 17:46
$begingroup$
@CloudJR see the new edit.
$endgroup$
– N. S.
Dec 21 '18 at 18:06
$begingroup$
@CloudJR see the new edit.
$endgroup$
– N. S.
Dec 21 '18 at 18:06
|
show 4 more comments
$begingroup$
The simplest counter example is the constant function
$$
fequiv 0
$$
on any interval $(a,b)$. Clearly $f'(c)=0$ for any $cin(a,b)$ but it's not a point of inflection.
Edit: I interpreted the words max/min to mean strict minima/maxima, if the question refers to nonstrict ones then this is not a counter example.
$endgroup$
$begingroup$
But every point is a local min/max.
$endgroup$
– N. S.
Dec 21 '18 at 17:17
$begingroup$
But every point is maximum or minimum
$endgroup$
– Cloud JR
Dec 21 '18 at 17:17
$begingroup$
Yeah, I just realized that I probably misinterpreted the question. Sorry for that.
$endgroup$
– BigbearZzz
Dec 21 '18 at 17:18
add a comment |
$begingroup$
The simplest counter example is the constant function
$$
fequiv 0
$$
on any interval $(a,b)$. Clearly $f'(c)=0$ for any $cin(a,b)$ but it's not a point of inflection.
Edit: I interpreted the words max/min to mean strict minima/maxima, if the question refers to nonstrict ones then this is not a counter example.
$endgroup$
$begingroup$
But every point is a local min/max.
$endgroup$
– N. S.
Dec 21 '18 at 17:17
$begingroup$
But every point is maximum or minimum
$endgroup$
– Cloud JR
Dec 21 '18 at 17:17
$begingroup$
Yeah, I just realized that I probably misinterpreted the question. Sorry for that.
$endgroup$
– BigbearZzz
Dec 21 '18 at 17:18
add a comment |
$begingroup$
The simplest counter example is the constant function
$$
fequiv 0
$$
on any interval $(a,b)$. Clearly $f'(c)=0$ for any $cin(a,b)$ but it's not a point of inflection.
Edit: I interpreted the words max/min to mean strict minima/maxima, if the question refers to nonstrict ones then this is not a counter example.
$endgroup$
The simplest counter example is the constant function
$$
fequiv 0
$$
on any interval $(a,b)$. Clearly $f'(c)=0$ for any $cin(a,b)$ but it's not a point of inflection.
Edit: I interpreted the words max/min to mean strict minima/maxima, if the question refers to nonstrict ones then this is not a counter example.
edited Dec 21 '18 at 17:18
answered Dec 21 '18 at 17:15
BigbearZzzBigbearZzz
9,07421653
9,07421653
$begingroup$
But every point is a local min/max.
$endgroup$
– N. S.
Dec 21 '18 at 17:17
$begingroup$
But every point is maximum or minimum
$endgroup$
– Cloud JR
Dec 21 '18 at 17:17
$begingroup$
Yeah, I just realized that I probably misinterpreted the question. Sorry for that.
$endgroup$
– BigbearZzz
Dec 21 '18 at 17:18
add a comment |
$begingroup$
But every point is a local min/max.
$endgroup$
– N. S.
Dec 21 '18 at 17:17
$begingroup$
But every point is maximum or minimum
$endgroup$
– Cloud JR
Dec 21 '18 at 17:17
$begingroup$
Yeah, I just realized that I probably misinterpreted the question. Sorry for that.
$endgroup$
– BigbearZzz
Dec 21 '18 at 17:18
$begingroup$
But every point is a local min/max.
$endgroup$
– N. S.
Dec 21 '18 at 17:17
$begingroup$
But every point is a local min/max.
$endgroup$
– N. S.
Dec 21 '18 at 17:17
$begingroup$
But every point is maximum or minimum
$endgroup$
– Cloud JR
Dec 21 '18 at 17:17
$begingroup$
But every point is maximum or minimum
$endgroup$
– Cloud JR
Dec 21 '18 at 17:17
$begingroup$
Yeah, I just realized that I probably misinterpreted the question. Sorry for that.
$endgroup$
– BigbearZzz
Dec 21 '18 at 17:18
$begingroup$
Yeah, I just realized that I probably misinterpreted the question. Sorry for that.
$endgroup$
– BigbearZzz
Dec 21 '18 at 17:18
add a comment |
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$begingroup$
If it's differential (big if) then $f'(c) =0$ means is an extreme point. Geometrically this means the slope is zero. This is either a local min or max or inflection point. As you know it's neither of two of the options, it must be the third option.
$endgroup$
– fleablood
Dec 21 '18 at 17:15
1
$begingroup$
@fleablood: I don't believe you're correct. A (local) extreme point is by definition a (local) maximum or minimum. So $f'(c)=0$ does NOT imply that $c$ is a local extreme point.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:19
1
$begingroup$
I think you need to check your definitions carefully, @fleablood.
$endgroup$
– Ted Shifrin
Dec 21 '18 at 17:23
1
$begingroup$
Actually you're right. I was thinking of the term "critical points".
$endgroup$
– fleablood
Dec 21 '18 at 17:25
1
$begingroup$
@fleablood It is true that local extremum implies critical point, I think you confuse this with the converese which is not true. And inflection point is typically defined as a point where the concavity changes.
$endgroup$
– N. S.
Dec 21 '18 at 17:37