Monotonically increasing functions
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In Baby Rudin, Theorem 5.11 says, Suppose $f$ is differentiable in $(a,b)$. If $f'(x) geq 0$ for all $x in (a,b)$, then $f$ is monotonically increasing, but this is an if and only if, right?
If we analyze the behavior of a monotonically increasing differentiable function, then we realize that $frac{f(y) - f(x)}{y-x}$ is always nonnegative. So, if $f'(x)$ exists, then $f'(x) geq 0$, right?
real-analysis analysis derivatives partial-derivative
asked Dec 2 '18 at 15:35
PeterPeter
864
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add a comment |
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In Baby Rudin, Theorem 5.11 says, Suppose $f$ is differentiable in $(a,b)$. If $f'(x) geq 0$ for all $x in (a,b)$, then $f$ is monotonically increasing, but this is an if and only if, right?
If we analyze the behavior of a monotonically increasing differentiable function, then we realize that $frac{f(y) - f(x)}{y-x}$ is always nonnegative. So, if $f'(x)$ exists, then $f'(x) geq 0$, right?
real-analysis analysis derivatives partial-derivative
asked Dec 2 '18 at 15:35
PeterPeter
864
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1
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Yes, you are right.
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– José Carlos Santos
Dec 2 '18 at 15:37
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thanks @JoséCarlosSantos!
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– Peter
Dec 2 '18 at 15:40
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In definitions only the “if” part is necessary most of the time.
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– molarmass
Dec 2 '18 at 17:07
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yes @molarmass, but this is not a definition.
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– Peter
Dec 2 '18 at 17:10
add a comment |
$begingroup$
In Baby Rudin, Theorem 5.11 says, Suppose $f$ is differentiable in $(a,b)$. If $f'(x) geq 0$ for all $x in (a,b)$, then $f$ is monotonically increasing, but this is an if and only if, right?
If we analyze the behavior of a monotonically increasing differentiable function, then we realize that $frac{f(y) - f(x)}{y-x}$ is always nonnegative. So, if $f'(x)$ exists, then $f'(x) geq 0$, right?
real-analysis analysis derivatives partial-derivative
asked Dec 2 '18 at 15:35
PeterPeter
864
$endgroup$
In Baby Rudin, Theorem 5.11 says, Suppose $f$ is differentiable in $(a,b)$. If $f'(x) geq 0$ for all $x in (a,b)$, then $f$ is monotonically increasing, but this is an if and only if, right?
If we analyze the behavior of a monotonically increasing differentiable function, then we realize that $frac{f(y) - f(x)}{y-x}$ is always nonnegative. So, if $f'(x)$ exists, then $f'(x) geq 0$, right?
real-analysis analysis derivatives partial-derivative
real-analysis analysis derivatives partial-derivative
asked Dec 2 '18 at 15:35
PeterPeter
864
asked Dec 2 '18 at 15:35
PeterPeter
864
asked Dec 2 '18 at 15:35
PeterPeter
864
asked Dec 2 '18 at 15:35
PeterPeter
864
asked Dec 2 '18 at 15:35
PeterPeter
864
864
1
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 15:37
$begingroup$
thanks @JoséCarlosSantos!
$endgroup$
– Peter
Dec 2 '18 at 15:40
$begingroup$
In definitions only the “if” part is necessary most of the time.
$endgroup$
– molarmass
Dec 2 '18 at 17:07
$begingroup$
yes @molarmass, but this is not a definition.
$endgroup$
– Peter
Dec 2 '18 at 17:10
add a comment |
1
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 15:37
$begingroup$
thanks @JoséCarlosSantos!
$endgroup$
– Peter
Dec 2 '18 at 15:40
$begingroup$
In definitions only the “if” part is necessary most of the time.
$endgroup$
– molarmass
Dec 2 '18 at 17:07
$begingroup$
yes @molarmass, but this is not a definition.
$endgroup$
– Peter
Dec 2 '18 at 17:10
1
1
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 15:37
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 15:37
$begingroup$
thanks @JoséCarlosSantos!
$endgroup$
– Peter
Dec 2 '18 at 15:40
$begingroup$
thanks @JoséCarlosSantos!
$endgroup$
– Peter
Dec 2 '18 at 15:40
$begingroup$
In definitions only the “if” part is necessary most of the time.
$endgroup$
– molarmass
Dec 2 '18 at 17:07
$begingroup$
In definitions only the “if” part is necessary most of the time.
$endgroup$
– molarmass
Dec 2 '18 at 17:07
$begingroup$
yes @molarmass, but this is not a definition.
$endgroup$
– Peter
Dec 2 '18 at 17:10
$begingroup$
yes @molarmass, but this is not a definition.
$endgroup$
– Peter
Dec 2 '18 at 17:10
add a comment |
1 Answer
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Yes dude you are right. Here’s an extended problem: what’s the behavior of $f’(x)$ for strictly monotonically increasing function?
answered Dec 2 '18 at 15:43
DiaryofNewtonDiaryofNewton
355
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add a comment |
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Yes dude you are right. Here’s an extended problem: what’s the behavior of $f’(x)$ for strictly monotonically increasing function?
answered Dec 2 '18 at 15:43
DiaryofNewtonDiaryofNewton
355
$endgroup$
add a comment |
$begingroup$
Yes dude you are right. Here’s an extended problem: what’s the behavior of $f’(x)$ for strictly monotonically increasing function?
answered Dec 2 '18 at 15:43
DiaryofNewtonDiaryofNewton
355
$endgroup$
add a comment |
$begingroup$
Yes dude you are right. Here’s an extended problem: what’s the behavior of $f’(x)$ for strictly monotonically increasing function?
answered Dec 2 '18 at 15:43
DiaryofNewtonDiaryofNewton
355
$endgroup$
Yes dude you are right. Here’s an extended problem: what’s the behavior of $f’(x)$ for strictly monotonically increasing function?
answered Dec 2 '18 at 15:43
DiaryofNewtonDiaryofNewton
355
answered Dec 2 '18 at 15:43
DiaryofNewtonDiaryofNewton
355
answered Dec 2 '18 at 15:43
DiaryofNewtonDiaryofNewton
355
answered Dec 2 '18 at 15:43
DiaryofNewtonDiaryofNewton
355
355
add a comment |
add a comment |
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1
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Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 15:37
$begingroup$
thanks @JoséCarlosSantos!
$endgroup$
– Peter
Dec 2 '18 at 15:40
$begingroup$
In definitions only the “if” part is necessary most of the time.
$endgroup$
– molarmass
Dec 2 '18 at 17:07
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yes @molarmass, but this is not a definition.
$endgroup$
– Peter
Dec 2 '18 at 17:10