Zeros of a continuously differentiable function
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Let $f: Drightarrow mathbb{R}$ be a continuously differentiable function defined on a domain $D$. Is it true that if $x$ is a non critical point of $f$, then there is a neighborhood of $x$ which contains no accumulation point of the set of zeros of $f$?
Thanks in advance!
analysis
asked Dec 5 at 13:20
Jiu
446110
add a comment |
up vote
0
down vote
favorite
Let $f: Drightarrow mathbb{R}$ be a continuously differentiable function defined on a domain $D$. Is it true that if $x$ is a non critical point of $f$, then there is a neighborhood of $x$ which contains no accumulation point of the set of zeros of $f$?
Thanks in advance!
analysis
asked Dec 5 at 13:20
Jiu
446110
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f: Drightarrow mathbb{R}$ be a continuously differentiable function defined on a domain $D$. Is it true that if $x$ is a non critical point of $f$, then there is a neighborhood of $x$ which contains no accumulation point of the set of zeros of $f$?
Thanks in advance!
analysis
asked Dec 5 at 13:20
Jiu
446110
Let $f: Drightarrow mathbb{R}$ be a continuously differentiable function defined on a domain $D$. Is it true that if $x$ is a non critical point of $f$, then there is a neighborhood of $x$ which contains no accumulation point of the set of zeros of $f$?
Thanks in advance!
analysis
analysis
asked Dec 5 at 13:20
Jiu
446110
asked Dec 5 at 13:20
Jiu
446110
asked Dec 5 at 13:20
Jiu
446110
asked Dec 5 at 13:20
Jiu
446110
asked Dec 5 at 13:20
Jiu
446110
446110
add a comment |
add a comment |
1 Answer
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If $f(x)ne0$, then there is a neighborhood of $x_0$ with no zeros of $f$. Without loss of generality, asume $fcolon(-a,a)toBbb R$, $f(0)=0$ and $f'(0)ne0$. Then
$$
f(x)=f'(0),x+h(x)quadtext{with}quad lim_{xto0}frac{h(x)}{x}=0.
$$
There exists $delta>0$ such that
$$
|x|<deltaimpliesBigl|frac{h(x)}{x}Bigr|lefrac{|f'(0)|}{2}.
$$
Then, if $0<|x|<delta$,
$$
|f(x)|ge|f'(0)|,|x|-|h(x)|ge|f'(0)|,|x|-frac{|f'(0)|}{2},|x|gefrac{|f'(0)|}{2},|x|>0.
$$
answered Dec 5 at 15:19
Julián Aguirre
67.3k24094
Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$?
– Jiu
Dec 5 at 15:22
No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$.
– Julián Aguirre
Dec 5 at 15:26
you’re right. Thanks!
– Jiu
Dec 5 at 15:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $f(x)ne0$, then there is a neighborhood of $x_0$ with no zeros of $f$. Without loss of generality, asume $fcolon(-a,a)toBbb R$, $f(0)=0$ and $f'(0)ne0$. Then
$$
f(x)=f'(0),x+h(x)quadtext{with}quad lim_{xto0}frac{h(x)}{x}=0.
$$
There exists $delta>0$ such that
$$
|x|<deltaimpliesBigl|frac{h(x)}{x}Bigr|lefrac{|f'(0)|}{2}.
$$
Then, if $0<|x|<delta$,
$$
|f(x)|ge|f'(0)|,|x|-|h(x)|ge|f'(0)|,|x|-frac{|f'(0)|}{2},|x|gefrac{|f'(0)|}{2},|x|>0.
$$
answered Dec 5 at 15:19
Julián Aguirre
67.3k24094
Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$?
– Jiu
Dec 5 at 15:22
No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$.
– Julián Aguirre
Dec 5 at 15:26
you’re right. Thanks!
– Jiu
Dec 5 at 15:28
add a comment |
up vote
1
down vote
accepted
If $f(x)ne0$, then there is a neighborhood of $x_0$ with no zeros of $f$. Without loss of generality, asume $fcolon(-a,a)toBbb R$, $f(0)=0$ and $f'(0)ne0$. Then
$$
f(x)=f'(0),x+h(x)quadtext{with}quad lim_{xto0}frac{h(x)}{x}=0.
$$
There exists $delta>0$ such that
$$
|x|<deltaimpliesBigl|frac{h(x)}{x}Bigr|lefrac{|f'(0)|}{2}.
$$
Then, if $0<|x|<delta$,
$$
|f(x)|ge|f'(0)|,|x|-|h(x)|ge|f'(0)|,|x|-frac{|f'(0)|}{2},|x|gefrac{|f'(0)|}{2},|x|>0.
$$
answered Dec 5 at 15:19
Julián Aguirre
67.3k24094
Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$?
– Jiu
Dec 5 at 15:22
No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$.
– Julián Aguirre
Dec 5 at 15:26
you’re right. Thanks!
– Jiu
Dec 5 at 15:28
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $f(x)ne0$, then there is a neighborhood of $x_0$ with no zeros of $f$. Without loss of generality, asume $fcolon(-a,a)toBbb R$, $f(0)=0$ and $f'(0)ne0$. Then
$$
f(x)=f'(0),x+h(x)quadtext{with}quad lim_{xto0}frac{h(x)}{x}=0.
$$
There exists $delta>0$ such that
$$
|x|<deltaimpliesBigl|frac{h(x)}{x}Bigr|lefrac{|f'(0)|}{2}.
$$
Then, if $0<|x|<delta$,
$$
|f(x)|ge|f'(0)|,|x|-|h(x)|ge|f'(0)|,|x|-frac{|f'(0)|}{2},|x|gefrac{|f'(0)|}{2},|x|>0.
$$
answered Dec 5 at 15:19
Julián Aguirre
67.3k24094
If $f(x)ne0$, then there is a neighborhood of $x_0$ with no zeros of $f$. Without loss of generality, asume $fcolon(-a,a)toBbb R$, $f(0)=0$ and $f'(0)ne0$. Then
$$
f(x)=f'(0),x+h(x)quadtext{with}quad lim_{xto0}frac{h(x)}{x}=0.
$$
There exists $delta>0$ such that
$$
|x|<deltaimpliesBigl|frac{h(x)}{x}Bigr|lefrac{|f'(0)|}{2}.
$$
Then, if $0<|x|<delta$,
$$
|f(x)|ge|f'(0)|,|x|-|h(x)|ge|f'(0)|,|x|-frac{|f'(0)|}{2},|x|gefrac{|f'(0)|}{2},|x|>0.
$$
answered Dec 5 at 15:19
Julián Aguirre
67.3k24094
answered Dec 5 at 15:19
Julián Aguirre
67.3k24094
answered Dec 5 at 15:19
Julián Aguirre
67.3k24094
answered Dec 5 at 15:19
Julián Aguirre
67.3k24094
67.3k24094
Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$?
– Jiu
Dec 5 at 15:22
No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$.
– Julián Aguirre
Dec 5 at 15:26
you’re right. Thanks!
– Jiu
Dec 5 at 15:28
add a comment |
Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$?
– Jiu
Dec 5 at 15:22
No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$.
– Julián Aguirre
Dec 5 at 15:26
you’re right. Thanks!
– Jiu
Dec 5 at 15:28
Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$?
– Jiu
Dec 5 at 15:22
Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$?
– Jiu
Dec 5 at 15:22
No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$.
– Julián Aguirre
Dec 5 at 15:26
No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$.
– Julián Aguirre
Dec 5 at 15:26
you’re right. Thanks!
– Jiu
Dec 5 at 15:28
you’re right. Thanks!
– Jiu
Dec 5 at 15:28
add a comment |
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