Find $delta>0$ where $f(x)<f(X_0)0$ for some $x_0$?
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How to find some $delta>0$ such that $f(x)<f(x_0)<f(y)$ whenever $x_0-delta<x<x_0<y<x_0+delta$ if $f^prime(x_0)>0$ for some point $x_0$ in the interior of the domain of $f$?
I found a similar question here, but I am still confused how to find $delta$.
My attempt:
Proof
Assume not. That is, assume $foralldelta>0, f(x)nless f(x_0)nless f(y)$ whenever $x_0-delta<x<x_0<y<x_0+delta$ if $f^prime(x_0)>0$ for some point $xin$dom($f$). Then $f$ must be nonincreasing. So the $f^prime(x_0)leq 0$. But $f^prime(x_0)>0$, so we have a contradiction....
but this is obviously incomplete and perhaps even wrong.
How do I find a $delta>0$ where all the conditions are satisfied? Also, does this mean $f$ is increasing in the interval $(x_0-delta, x_0+delta)$? I say no because the function can switch directions at $x$ and $y$ so that $x_0-delta>x_0+delta$.
calculus derivatives
asked Nov 18 at 17:25
kaisa
717
add a comment |
up vote
0
down vote
favorite
How to find some $delta>0$ such that $f(x)<f(x_0)<f(y)$ whenever $x_0-delta<x<x_0<y<x_0+delta$ if $f^prime(x_0)>0$ for some point $x_0$ in the interior of the domain of $f$?
I found a similar question here, but I am still confused how to find $delta$.
My attempt:
Proof
Assume not. That is, assume $foralldelta>0, f(x)nless f(x_0)nless f(y)$ whenever $x_0-delta<x<x_0<y<x_0+delta$ if $f^prime(x_0)>0$ for some point $xin$dom($f$). Then $f$ must be nonincreasing. So the $f^prime(x_0)leq 0$. But $f^prime(x_0)>0$, so we have a contradiction....
but this is obviously incomplete and perhaps even wrong.
How do I find a $delta>0$ where all the conditions are satisfied? Also, does this mean $f$ is increasing in the interval $(x_0-delta, x_0+delta)$? I say no because the function can switch directions at $x$ and $y$ so that $x_0-delta>x_0+delta$.
calculus derivatives
asked Nov 18 at 17:25
kaisa
717
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to find some $delta>0$ such that $f(x)<f(x_0)<f(y)$ whenever $x_0-delta<x<x_0<y<x_0+delta$ if $f^prime(x_0)>0$ for some point $x_0$ in the interior of the domain of $f$?
I found a similar question here, but I am still confused how to find $delta$.
My attempt:
Proof
Assume not. That is, assume $foralldelta>0, f(x)nless f(x_0)nless f(y)$ whenever $x_0-delta<x<x_0<y<x_0+delta$ if $f^prime(x_0)>0$ for some point $xin$dom($f$). Then $f$ must be nonincreasing. So the $f^prime(x_0)leq 0$. But $f^prime(x_0)>0$, so we have a contradiction....
but this is obviously incomplete and perhaps even wrong.
How do I find a $delta>0$ where all the conditions are satisfied? Also, does this mean $f$ is increasing in the interval $(x_0-delta, x_0+delta)$? I say no because the function can switch directions at $x$ and $y$ so that $x_0-delta>x_0+delta$.
calculus derivatives
asked Nov 18 at 17:25
kaisa
717
How to find some $delta>0$ such that $f(x)<f(x_0)<f(y)$ whenever $x_0-delta<x<x_0<y<x_0+delta$ if $f^prime(x_0)>0$ for some point $x_0$ in the interior of the domain of $f$?
I found a similar question here, but I am still confused how to find $delta$.
My attempt:
Proof
Assume not. That is, assume $foralldelta>0, f(x)nless f(x_0)nless f(y)$ whenever $x_0-delta<x<x_0<y<x_0+delta$ if $f^prime(x_0)>0$ for some point $xin$dom($f$). Then $f$ must be nonincreasing. So the $f^prime(x_0)leq 0$. But $f^prime(x_0)>0$, so we have a contradiction....
but this is obviously incomplete and perhaps even wrong.
How do I find a $delta>0$ where all the conditions are satisfied? Also, does this mean $f$ is increasing in the interval $(x_0-delta, x_0+delta)$? I say no because the function can switch directions at $x$ and $y$ so that $x_0-delta>x_0+delta$.
calculus derivatives
calculus derivatives
asked Nov 18 at 17:25
kaisa
717
asked Nov 18 at 17:25
kaisa
717
asked Nov 18 at 17:25
kaisa
717
asked Nov 18 at 17:25
kaisa
717
asked Nov 18 at 17:25
kaisa
717
717
add a comment |
add a comment |
1 Answer
1
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0
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By the definitions of derivative and limit, there is a $delta > 0$ such that $|x - x_0|<delta$ implies $frac{f(x) - f(x_0)}{x - x_0} > 0$ since $f'(x_0)>0$. Now treat the cases when $x<x_0$ and $x>x_0$.
answered Nov 18 at 17:32
lzralbu
560412
@Izralbu I still don't quite understand. How will this lead to $f(x)<f(x_0)<f(y)$?
– kaisa
Nov 18 at 21:03
Note that $x - x_0$ and $f(x) - f(x_0)$ must be both positive or both negative since their quotient is positive. If $x < x_0$ then $f(x) < f(x_0)$ else if $y > x_0$ then $f(y) > f(x_0)$.
– lzralbu
Nov 19 at 15:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
By the definitions of derivative and limit, there is a $delta > 0$ such that $|x - x_0|<delta$ implies $frac{f(x) - f(x_0)}{x - x_0} > 0$ since $f'(x_0)>0$. Now treat the cases when $x<x_0$ and $x>x_0$.
answered Nov 18 at 17:32
lzralbu
560412
@Izralbu I still don't quite understand. How will this lead to $f(x)<f(x_0)<f(y)$?
– kaisa
Nov 18 at 21:03
Note that $x - x_0$ and $f(x) - f(x_0)$ must be both positive or both negative since their quotient is positive. If $x < x_0$ then $f(x) < f(x_0)$ else if $y > x_0$ then $f(y) > f(x_0)$.
– lzralbu
Nov 19 at 15:36
add a comment |
up vote
0
down vote
By the definitions of derivative and limit, there is a $delta > 0$ such that $|x - x_0|<delta$ implies $frac{f(x) - f(x_0)}{x - x_0} > 0$ since $f'(x_0)>0$. Now treat the cases when $x<x_0$ and $x>x_0$.
answered Nov 18 at 17:32
lzralbu
560412
@Izralbu I still don't quite understand. How will this lead to $f(x)<f(x_0)<f(y)$?
– kaisa
Nov 18 at 21:03
Note that $x - x_0$ and $f(x) - f(x_0)$ must be both positive or both negative since their quotient is positive. If $x < x_0$ then $f(x) < f(x_0)$ else if $y > x_0$ then $f(y) > f(x_0)$.
– lzralbu
Nov 19 at 15:36
add a comment |
up vote
0
down vote
up vote
0
down vote
By the definitions of derivative and limit, there is a $delta > 0$ such that $|x - x_0|<delta$ implies $frac{f(x) - f(x_0)}{x - x_0} > 0$ since $f'(x_0)>0$. Now treat the cases when $x<x_0$ and $x>x_0$.
answered Nov 18 at 17:32
lzralbu
560412
By the definitions of derivative and limit, there is a $delta > 0$ such that $|x - x_0|<delta$ implies $frac{f(x) - f(x_0)}{x - x_0} > 0$ since $f'(x_0)>0$. Now treat the cases when $x<x_0$ and $x>x_0$.
answered Nov 18 at 17:32
lzralbu
560412
answered Nov 18 at 17:32
lzralbu
560412
answered Nov 18 at 17:32
lzralbu
560412
answered Nov 18 at 17:32
lzralbu
560412
560412
@Izralbu I still don't quite understand. How will this lead to $f(x)<f(x_0)<f(y)$?
– kaisa
Nov 18 at 21:03
Note that $x - x_0$ and $f(x) - f(x_0)$ must be both positive or both negative since their quotient is positive. If $x < x_0$ then $f(x) < f(x_0)$ else if $y > x_0$ then $f(y) > f(x_0)$.
– lzralbu
Nov 19 at 15:36
add a comment |
@Izralbu I still don't quite understand. How will this lead to $f(x)<f(x_0)<f(y)$?
– kaisa
Nov 18 at 21:03
Note that $x - x_0$ and $f(x) - f(x_0)$ must be both positive or both negative since their quotient is positive. If $x < x_0$ then $f(x) < f(x_0)$ else if $y > x_0$ then $f(y) > f(x_0)$.
– lzralbu
Nov 19 at 15:36
@Izralbu I still don't quite understand. How will this lead to $f(x)<f(x_0)<f(y)$?
– kaisa
Nov 18 at 21:03
@Izralbu I still don't quite understand. How will this lead to $f(x)<f(x_0)<f(y)$?
– kaisa
Nov 18 at 21:03
Note that $x - x_0$ and $f(x) - f(x_0)$ must be both positive or both negative since their quotient is positive. If $x < x_0$ then $f(x) < f(x_0)$ else if $y > x_0$ then $f(y) > f(x_0)$.
– lzralbu
Nov 19 at 15:36
Note that $x - x_0$ and $f(x) - f(x_0)$ must be both positive or both negative since their quotient is positive. If $x < x_0$ then $f(x) < f(x_0)$ else if $y > x_0$ then $f(y) > f(x_0)$.
– lzralbu
Nov 19 at 15:36
add a comment |
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