if $f'(c)<0$ then there exists a nbd $I$ of $c$ such that $f'(x)<0$ for all $xin I$
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I have been trying to prove the statement:
Let $Dsubset mathbb{R}$ and $fcolon Dtomathbb{R}$.If $f'(c)<0$ then there exists a nbd $I$ of $c$ such that $f'(x)<0$
for all $xin I$.
Here's my attempt:
Let $S:={|x-c| : f'(x)ge 0 text{for some } xin D}$. Clearly, $S$ is bounded below by $0$. Now if $S$ is empty, then for all $x in D$, we have $f'(x) <0$ and we are done. Now, suppose that $S$ is not empty. Then let $delta := inf S$. Now, we have that for all $|x-c| < delta$, $|f'(x)|<0$.
Is this proof okay? Or did I go wrong? Alternative proofs?
real-analysis proof-verification alternative-proof
add a comment |
up vote
-1
down vote
favorite
I have been trying to prove the statement:
Let $Dsubset mathbb{R}$ and $fcolon Dtomathbb{R}$.If $f'(c)<0$ then there exists a nbd $I$ of $c$ such that $f'(x)<0$
for all $xin I$.
Here's my attempt:
Let $S:={|x-c| : f'(x)ge 0 text{for some } xin D}$. Clearly, $S$ is bounded below by $0$. Now if $S$ is empty, then for all $x in D$, we have $f'(x) <0$ and we are done. Now, suppose that $S$ is not empty. Then let $delta := inf S$. Now, we have that for all $|x-c| < delta$, $|f'(x)|<0$.
Is this proof okay? Or did I go wrong? Alternative proofs?
real-analysis proof-verification alternative-proof
You need to specify what $D$ is, and any assumptions on $f$
– user25959
Nov 24 at 16:30
What if $delta = 0$?
– xbh
Nov 24 at 16:32
@xbh it seems i did not consider that. The answer below disprove my statement.
– Ashish K
Nov 24 at 16:49
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have been trying to prove the statement:
Let $Dsubset mathbb{R}$ and $fcolon Dtomathbb{R}$.If $f'(c)<0$ then there exists a nbd $I$ of $c$ such that $f'(x)<0$
for all $xin I$.
Here's my attempt:
Let $S:={|x-c| : f'(x)ge 0 text{for some } xin D}$. Clearly, $S$ is bounded below by $0$. Now if $S$ is empty, then for all $x in D$, we have $f'(x) <0$ and we are done. Now, suppose that $S$ is not empty. Then let $delta := inf S$. Now, we have that for all $|x-c| < delta$, $|f'(x)|<0$.
Is this proof okay? Or did I go wrong? Alternative proofs?
real-analysis proof-verification alternative-proof
I have been trying to prove the statement:
Let $Dsubset mathbb{R}$ and $fcolon Dtomathbb{R}$.If $f'(c)<0$ then there exists a nbd $I$ of $c$ such that $f'(x)<0$
for all $xin I$.
Here's my attempt:
Let $S:={|x-c| : f'(x)ge 0 text{for some } xin D}$. Clearly, $S$ is bounded below by $0$. Now if $S$ is empty, then for all $x in D$, we have $f'(x) <0$ and we are done. Now, suppose that $S$ is not empty. Then let $delta := inf S$. Now, we have that for all $|x-c| < delta$, $|f'(x)|<0$.
Is this proof okay? Or did I go wrong? Alternative proofs?
real-analysis proof-verification alternative-proof
real-analysis proof-verification alternative-proof
edited Nov 24 at 16:57
asked Nov 24 at 16:22
Ashish K
776513
776513
You need to specify what $D$ is, and any assumptions on $f$
– user25959
Nov 24 at 16:30
What if $delta = 0$?
– xbh
Nov 24 at 16:32
@xbh it seems i did not consider that. The answer below disprove my statement.
– Ashish K
Nov 24 at 16:49
add a comment |
You need to specify what $D$ is, and any assumptions on $f$
– user25959
Nov 24 at 16:30
What if $delta = 0$?
– xbh
Nov 24 at 16:32
@xbh it seems i did not consider that. The answer below disprove my statement.
– Ashish K
Nov 24 at 16:49
You need to specify what $D$ is, and any assumptions on $f$
– user25959
Nov 24 at 16:30
You need to specify what $D$ is, and any assumptions on $f$
– user25959
Nov 24 at 16:30
What if $delta = 0$?
– xbh
Nov 24 at 16:32
What if $delta = 0$?
– xbh
Nov 24 at 16:32
@xbh it seems i did not consider that. The answer below disprove my statement.
– Ashish K
Nov 24 at 16:49
@xbh it seems i did not consider that. The answer below disprove my statement.
– Ashish K
Nov 24 at 16:49
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
It's not true. Take
$$
f(x)=cases{x^2sin(1/x)+x/2& if $xneq0$\0& otherwise}
$$
Then $f'(0)=0.5$, but for there are $x$ arbitrarily close to $0$ such that $f'(x)=-0.5$.
Hey, can you expand it a bit more? My statement seeks $f'(c)<0$ but for $f'(0)=0.5>0$?
– Ashish K
Nov 24 at 16:53
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
2
down vote
accepted
It's not true. Take
$$
f(x)=cases{x^2sin(1/x)+x/2& if $xneq0$\0& otherwise}
$$
Then $f'(0)=0.5$, but for there are $x$ arbitrarily close to $0$ such that $f'(x)=-0.5$.
Hey, can you expand it a bit more? My statement seeks $f'(c)<0$ but for $f'(0)=0.5>0$?
– Ashish K
Nov 24 at 16:53
add a comment |
up vote
2
down vote
accepted
It's not true. Take
$$
f(x)=cases{x^2sin(1/x)+x/2& if $xneq0$\0& otherwise}
$$
Then $f'(0)=0.5$, but for there are $x$ arbitrarily close to $0$ such that $f'(x)=-0.5$.
Hey, can you expand it a bit more? My statement seeks $f'(c)<0$ but for $f'(0)=0.5>0$?
– Ashish K
Nov 24 at 16:53
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It's not true. Take
$$
f(x)=cases{x^2sin(1/x)+x/2& if $xneq0$\0& otherwise}
$$
Then $f'(0)=0.5$, but for there are $x$ arbitrarily close to $0$ such that $f'(x)=-0.5$.
It's not true. Take
$$
f(x)=cases{x^2sin(1/x)+x/2& if $xneq0$\0& otherwise}
$$
Then $f'(0)=0.5$, but for there are $x$ arbitrarily close to $0$ such that $f'(x)=-0.5$.
answered Nov 24 at 16:31
Arthur
110k7105186
110k7105186
Hey, can you expand it a bit more? My statement seeks $f'(c)<0$ but for $f'(0)=0.5>0$?
– Ashish K
Nov 24 at 16:53
add a comment |
Hey, can you expand it a bit more? My statement seeks $f'(c)<0$ but for $f'(0)=0.5>0$?
– Ashish K
Nov 24 at 16:53
Hey, can you expand it a bit more? My statement seeks $f'(c)<0$ but for $f'(0)=0.5>0$?
– Ashish K
Nov 24 at 16:53
Hey, can you expand it a bit more? My statement seeks $f'(c)<0$ but for $f'(0)=0.5>0$?
– Ashish K
Nov 24 at 16:53
add a comment |
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You need to specify what $D$ is, and any assumptions on $f$
– user25959
Nov 24 at 16:30
What if $delta = 0$?
– xbh
Nov 24 at 16:32
@xbh it seems i did not consider that. The answer below disprove my statement.
– Ashish K
Nov 24 at 16:49