Mixing
No. of critical points
$begingroup$
Domain: $mathopen]1,4mathclose[$, $f(x)= 3x^2 - 6x$
How many critical points exist?
(Zero, 1, 3, 4)
By diff.($x$):
$f'(x)= 6x-6$
Then $x=1$ is local point of minimum, but according to the interval should I consider it or not as a critical point?
calculus derivatives
edited Jan 21 at 23:01
egreg
183k1486205
asked Jan 21 at 22:31
AmgeeedAmgeeed
11
$endgroup$
add a comment |
$begingroup$
Domain: $mathopen]1,4mathclose[$, $f(x)= 3x^2 - 6x$
How many critical points exist?
(Zero, 1, 3, 4)
By diff.($x$):
$f'(x)= 6x-6$
Then $x=1$ is local point of minimum, but according to the interval should I consider it or not as a critical point?
calculus derivatives
edited Jan 21 at 23:01
egreg
183k1486205
asked Jan 21 at 22:31
AmgeeedAmgeeed
11
$endgroup$
$begingroup$
Welcome to StackExchange, Amgeeed! I'm afraid I can't make sense of your post, please view math.meta.stackexchange.com/questions/5020/… for a quick guide to MathJax
$endgroup$
– bounceback
Jan 21 at 22:41
2
$begingroup$
Accordingly, it seems the OP is using the French-style of interval notation, meaning that $]1,4[$ is equivalent to writing the interval $(1,4) = {xinmathbb R ;;|;; 1 < x < 4}$. In particular, @Amgeeed, since the function is not defined at $x=1$, that critical point must be excluded, showing you that there are zero critical points.
$endgroup$
– Decaf-Math
Jan 21 at 22:48
add a comment |
$begingroup$
Domain: $mathopen]1,4mathclose[$, $f(x)= 3x^2 - 6x$
How many critical points exist?
(Zero, 1, 3, 4)
By diff.($x$):
$f'(x)= 6x-6$
Then $x=1$ is local point of minimum, but according to the interval should I consider it or not as a critical point?
calculus derivatives
edited Jan 21 at 23:01
egreg
183k1486205
asked Jan 21 at 22:31
AmgeeedAmgeeed
11
$endgroup$
Domain: $mathopen]1,4mathclose[$, $f(x)= 3x^2 - 6x$
How many critical points exist?
(Zero, 1, 3, 4)
By diff.($x$):
$f'(x)= 6x-6$
Then $x=1$ is local point of minimum, but according to the interval should I consider it or not as a critical point?
calculus derivatives
calculus derivatives
edited Jan 21 at 23:01
egreg
183k1486205
asked Jan 21 at 22:31
AmgeeedAmgeeed
11
edited Jan 21 at 23:01
egreg
183k1486205
asked Jan 21 at 22:31
AmgeeedAmgeeed
11
edited Jan 21 at 23:01
egreg
183k1486205
edited Jan 21 at 23:01
egreg
183k1486205
edited Jan 21 at 23:01
egreg
183k1486205
183k1486205
asked Jan 21 at 22:31
AmgeeedAmgeeed
11
asked Jan 21 at 22:31
AmgeeedAmgeeed
11
asked Jan 21 at 22:31
AmgeeedAmgeeed
11
11
$begingroup$
Welcome to StackExchange, Amgeeed! I'm afraid I can't make sense of your post, please view math.meta.stackexchange.com/questions/5020/… for a quick guide to MathJax
$endgroup$
– bounceback
Jan 21 at 22:41
2
$begingroup$
Accordingly, it seems the OP is using the French-style of interval notation, meaning that $]1,4[$ is equivalent to writing the interval $(1,4) = {xinmathbb R ;;|;; 1 < x < 4}$. In particular, @Amgeeed, since the function is not defined at $x=1$, that critical point must be excluded, showing you that there are zero critical points.
$endgroup$
– Decaf-Math
Jan 21 at 22:48
add a comment |
$begingroup$
Welcome to StackExchange, Amgeeed! I'm afraid I can't make sense of your post, please view math.meta.stackexchange.com/questions/5020/… for a quick guide to MathJax
$endgroup$
– bounceback
Jan 21 at 22:41
2
$begingroup$
Accordingly, it seems the OP is using the French-style of interval notation, meaning that $]1,4[$ is equivalent to writing the interval $(1,4) = {xinmathbb R ;;|;; 1 < x < 4}$. In particular, @Amgeeed, since the function is not defined at $x=1$, that critical point must be excluded, showing you that there are zero critical points.
$endgroup$
– Decaf-Math
Jan 21 at 22:48
$begingroup$
Welcome to StackExchange, Amgeeed! I'm afraid I can't make sense of your post, please view math.meta.stackexchange.com/questions/5020/… for a quick guide to MathJax
$endgroup$
– bounceback
Jan 21 at 22:41
$begingroup$
Welcome to StackExchange, Amgeeed! I'm afraid I can't make sense of your post, please view math.meta.stackexchange.com/questions/5020/… for a quick guide to MathJax
$endgroup$
– bounceback
Jan 21 at 22:41
2
2
$begingroup$
Accordingly, it seems the OP is using the French-style of interval notation, meaning that $]1,4[$ is equivalent to writing the interval $(1,4) = {xinmathbb R ;;|;; 1 < x < 4}$. In particular, @Amgeeed, since the function is not defined at $x=1$, that critical point must be excluded, showing you that there are zero critical points.
$endgroup$
– Decaf-Math
Jan 21 at 22:48
$begingroup$
Accordingly, it seems the OP is using the French-style of interval notation, meaning that $]1,4[$ is equivalent to writing the interval $(1,4) = {xinmathbb R ;;|;; 1 < x < 4}$. In particular, @Amgeeed, since the function is not defined at $x=1$, that critical point must be excluded, showing you that there are zero critical points.
$endgroup$
– Decaf-Math
Jan 21 at 22:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A function maps a domain (a set) to another set. A function is not only defined by how it maps points but also by WHICH points (the domain) are mapped and indicating what the domain is an essential aspect of defining a function.
That means the function $f: ]1,4[ to mathbb R$ via $f(x) = 3x^2 - 6x$ is a DIFFERENT function than $overline{f}:mathbb Rto mathbb R$ via $overline{f}(x) = 3x^2 - 6x$.
So $x = 1$ can not be a critical point of $f$ because $x = 1$ is not a point in $f$'s domain at all.
answered Jan 21 at 23:09
fleabloodfleablood
72k22687
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
A function maps a domain (a set) to another set. A function is not only defined by how it maps points but also by WHICH points (the domain) are mapped and indicating what the domain is an essential aspect of defining a function.
That means the function $f: ]1,4[ to mathbb R$ via $f(x) = 3x^2 - 6x$ is a DIFFERENT function than $overline{f}:mathbb Rto mathbb R$ via $overline{f}(x) = 3x^2 - 6x$.
So $x = 1$ can not be a critical point of $f$ because $x = 1$ is not a point in $f$'s domain at all.
answered Jan 21 at 23:09
fleabloodfleablood
72k22687
$endgroup$
add a comment |
$begingroup$
A function maps a domain (a set) to another set. A function is not only defined by how it maps points but also by WHICH points (the domain) are mapped and indicating what the domain is an essential aspect of defining a function.
That means the function $f: ]1,4[ to mathbb R$ via $f(x) = 3x^2 - 6x$ is a DIFFERENT function than $overline{f}:mathbb Rto mathbb R$ via $overline{f}(x) = 3x^2 - 6x$.
So $x = 1$ can not be a critical point of $f$ because $x = 1$ is not a point in $f$'s domain at all.
answered Jan 21 at 23:09
fleabloodfleablood
72k22687
$endgroup$
add a comment |
$begingroup$
A function maps a domain (a set) to another set. A function is not only defined by how it maps points but also by WHICH points (the domain) are mapped and indicating what the domain is an essential aspect of defining a function.
That means the function $f: ]1,4[ to mathbb R$ via $f(x) = 3x^2 - 6x$ is a DIFFERENT function than $overline{f}:mathbb Rto mathbb R$ via $overline{f}(x) = 3x^2 - 6x$.
So $x = 1$ can not be a critical point of $f$ because $x = 1$ is not a point in $f$'s domain at all.
answered Jan 21 at 23:09
fleabloodfleablood
72k22687
$endgroup$
A function maps a domain (a set) to another set. A function is not only defined by how it maps points but also by WHICH points (the domain) are mapped and indicating what the domain is an essential aspect of defining a function.
That means the function $f: ]1,4[ to mathbb R$ via $f(x) = 3x^2 - 6x$ is a DIFFERENT function than $overline{f}:mathbb Rto mathbb R$ via $overline{f}(x) = 3x^2 - 6x$.
So $x = 1$ can not be a critical point of $f$ because $x = 1$ is not a point in $f$'s domain at all.
answered Jan 21 at 23:09
fleabloodfleablood
72k22687
answered Jan 21 at 23:09
fleabloodfleablood
72k22687
answered Jan 21 at 23:09
fleabloodfleablood
72k22687
answered Jan 21 at 23:09
fleabloodfleablood
72k22687
72k22687
add a comment |
add a comment |
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$begingroup$
Welcome to StackExchange, Amgeeed! I'm afraid I can't make sense of your post, please view math.meta.stackexchange.com/questions/5020/… for a quick guide to MathJax
$endgroup$
– bounceback
Jan 21 at 22:41
2
$begingroup$
Accordingly, it seems the OP is using the French-style of interval notation, meaning that $]1,4[$ is equivalent to writing the interval $(1,4) = {xinmathbb R ;;|;; 1 < x < 4}$. In particular, @Amgeeed, since the function is not defined at $x=1$, that critical point must be excluded, showing you that there are zero critical points.
$endgroup$
– Decaf-Math
Jan 21 at 22:48