Proportionality without linear restriction?Binary relation, reflexive, symmetric and transitiveWhat's $fracdydx$ of function $y=frac 12sin x$?Does type-theory have a concept of “relation”?Proving a change of variables is a variable over a ringThere exists a notation to “equations” where the unknown to solve is a kind of relation?does a univariate polynomial have a unique factorization?A property holds generically for polynomials - meaning?Can you turn a well-founded relation into a well-quasi-ordering?Laplace Transforms: How Do I Explain/Phrase What is Wrong with This Textbook Explanation?Can this notation be used to describe relations that are not functions?
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Proportionality without linear restriction?
Binary relation, reflexive, symmetric and transitiveWhat's $fracdydx$ of function $y=frac 12sin x$?Does type-theory have a concept of “relation”?Proving a change of variables is a variable over a ringThere exists a notation to “equations” where the unknown to solve is a kind of relation?does a univariate polynomial have a unique factorization?A property holds generically for polynomials - meaning?Can you turn a well-founded relation into a well-quasi-ordering?Laplace Transforms: How Do I Explain/Phrase What is Wrong with This Textbook Explanation?Can this notation be used to describe relations that are not functions?
$begingroup$
Is there a word to express the property $fracdydx>0$ of a relation?
I've heard the word "proportional" used to express this colloquially, although it is incorrect when $y=kx$ does not hold and it's another yet unknown polynomial over some restricted domain.
functions derivatives polynomials relations
asked 2 days ago
Jorge BarriosJorge Barrios
1054
$endgroup$
add a comment |
$begingroup$
Is there a word to express the property $fracdydx>0$ of a relation?
I've heard the word "proportional" used to express this colloquially, although it is incorrect when $y=kx$ does not hold and it's another yet unknown polynomial over some restricted domain.
functions derivatives polynomials relations
asked 2 days ago
Jorge BarriosJorge Barrios
1054
$endgroup$
add a comment |
$begingroup$
Is there a word to express the property $fracdydx>0$ of a relation?
I've heard the word "proportional" used to express this colloquially, although it is incorrect when $y=kx$ does not hold and it's another yet unknown polynomial over some restricted domain.
functions derivatives polynomials relations
asked 2 days ago
Jorge BarriosJorge Barrios
1054
$endgroup$
Is there a word to express the property $fracdydx>0$ of a relation?
I've heard the word "proportional" used to express this colloquially, although it is incorrect when $y=kx$ does not hold and it's another yet unknown polynomial over some restricted domain.
functions derivatives polynomials relations
functions derivatives polynomials relations
asked 2 days ago
Jorge BarriosJorge Barrios
1054
asked 2 days ago
Jorge BarriosJorge Barrios
1054
asked 2 days ago
Jorge BarriosJorge Barrios
1054
asked 2 days ago
Jorge BarriosJorge Barrios
1054
asked 2 days ago
Jorge BarriosJorge Barrios
1054
1054
add a comment |
add a comment |
1 Answer
1
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$begingroup$
A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.
answered 2 days ago
TravisTravis
63k767150
$endgroup$
add a comment |
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1 Answer
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$begingroup$
A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.
answered 2 days ago
TravisTravis
63k767150
$endgroup$
add a comment |
$begingroup$
A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.
answered 2 days ago
TravisTravis
63k767150
$endgroup$
add a comment |
$begingroup$
A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.
answered 2 days ago
TravisTravis
63k767150
$endgroup$
A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.
answered 2 days ago
TravisTravis
63k767150
answered 2 days ago
TravisTravis
63k767150
answered 2 days ago
TravisTravis
63k767150
answered 2 days ago
TravisTravis
63k767150
63k767150
add a comment |
add a comment |
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