Mixing
can a quartic graph have 2 turning points?
$begingroup$
Is it possible for a quartic equation when graphed to have 2 turning points and 1 point of inflection. If so what is an example of such equation? I am asking this because i have seen many quartic equations that have 3 or 1 turning point but never seen a quartic graph with 2 turning points.
polynomials quartic-equations
asked May 10 '16 at 8:31
Son JermSon Jerm
64
$endgroup$
|
show 1 more comment
$begingroup$
Is it possible for a quartic equation when graphed to have 2 turning points and 1 point of inflection. If so what is an example of such equation? I am asking this because i have seen many quartic equations that have 3 or 1 turning point but never seen a quartic graph with 2 turning points.
polynomials quartic-equations
asked May 10 '16 at 8:31
Son JermSon Jerm
64
$endgroup$
$begingroup$
Not really sure what you mean there.
$endgroup$
– Son Jerm
May 10 '16 at 8:40
1
$begingroup$
What do you mean with turning points? Local extrema?
$endgroup$
– ccorn
May 10 '16 at 8:41
1
$begingroup$
Yes turning point meaning local extrema.
$endgroup$
– Son Jerm
May 10 '16 at 8:44
$begingroup$
Any polynomial of even degree must have an odd number of local extrema, because local extrema are precisely the points at which the gradient changes sign.
$endgroup$
– almagest
May 10 '16 at 8:46
$begingroup$
So it would not be possible for a quartic equation to have 2 local extrema? as quartic is of even degree, right?
$endgroup$
– Son Jerm
May 10 '16 at 8:49
|
show 1 more comment
$begingroup$
Is it possible for a quartic equation when graphed to have 2 turning points and 1 point of inflection. If so what is an example of such equation? I am asking this because i have seen many quartic equations that have 3 or 1 turning point but never seen a quartic graph with 2 turning points.
polynomials quartic-equations
asked May 10 '16 at 8:31
Son JermSon Jerm
64
$endgroup$
Is it possible for a quartic equation when graphed to have 2 turning points and 1 point of inflection. If so what is an example of such equation? I am asking this because i have seen many quartic equations that have 3 or 1 turning point but never seen a quartic graph with 2 turning points.
polynomials quartic-equations
polynomials quartic-equations
asked May 10 '16 at 8:31
Son JermSon Jerm
64
asked May 10 '16 at 8:31
Son JermSon Jerm
64
asked May 10 '16 at 8:31
Son JermSon Jerm
64
asked May 10 '16 at 8:31
Son JermSon Jerm
64
asked May 10 '16 at 8:31
Son JermSon Jerm
64
64
$begingroup$
Not really sure what you mean there.
$endgroup$
– Son Jerm
May 10 '16 at 8:40
1
$begingroup$
What do you mean with turning points? Local extrema?
$endgroup$
– ccorn
May 10 '16 at 8:41
1
$begingroup$
Yes turning point meaning local extrema.
$endgroup$
– Son Jerm
May 10 '16 at 8:44
$begingroup$
Any polynomial of even degree must have an odd number of local extrema, because local extrema are precisely the points at which the gradient changes sign.
$endgroup$
– almagest
May 10 '16 at 8:46
$begingroup$
So it would not be possible for a quartic equation to have 2 local extrema? as quartic is of even degree, right?
$endgroup$
– Son Jerm
May 10 '16 at 8:49
|
show 1 more comment
$begingroup$
Not really sure what you mean there.
$endgroup$
– Son Jerm
May 10 '16 at 8:40
1
$begingroup$
What do you mean with turning points? Local extrema?
$endgroup$
– ccorn
May 10 '16 at 8:41
1
$begingroup$
Yes turning point meaning local extrema.
$endgroup$
– Son Jerm
May 10 '16 at 8:44
$begingroup$
Any polynomial of even degree must have an odd number of local extrema, because local extrema are precisely the points at which the gradient changes sign.
$endgroup$
– almagest
May 10 '16 at 8:46
$begingroup$
So it would not be possible for a quartic equation to have 2 local extrema? as quartic is of even degree, right?
$endgroup$
– Son Jerm
May 10 '16 at 8:49
$begingroup$
Not really sure what you mean there.
$endgroup$
– Son Jerm
May 10 '16 at 8:40
$begingroup$
Not really sure what you mean there.
$endgroup$
– Son Jerm
May 10 '16 at 8:40
1
1
$begingroup$
What do you mean with turning points? Local extrema?
$endgroup$
– ccorn
May 10 '16 at 8:41
$begingroup$
What do you mean with turning points? Local extrema?
$endgroup$
– ccorn
May 10 '16 at 8:41
1
1
$begingroup$
Yes turning point meaning local extrema.
$endgroup$
– Son Jerm
May 10 '16 at 8:44
$begingroup$
Yes turning point meaning local extrema.
$endgroup$
– Son Jerm
May 10 '16 at 8:44
$begingroup$
Any polynomial of even degree must have an odd number of local extrema, because local extrema are precisely the points at which the gradient changes sign.
$endgroup$
– almagest
May 10 '16 at 8:46
$begingroup$
Any polynomial of even degree must have an odd number of local extrema, because local extrema are precisely the points at which the gradient changes sign.
$endgroup$
– almagest
May 10 '16 at 8:46
$begingroup$
So it would not be possible for a quartic equation to have 2 local extrema? as quartic is of even degree, right?
$endgroup$
– Son Jerm
May 10 '16 at 8:49
$begingroup$
So it would not be possible for a quartic equation to have 2 local extrema? as quartic is of even degree, right?
$endgroup$
– Son Jerm
May 10 '16 at 8:49
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
You can tell the behavior of a polynomial $p(x)$ at $pminfty$ by looking at the term of highest degree, because for large $|x|$ the polynomial
$$p(x)=c_nx^n+c_{n-1}x^{n-1}+cdots +c_0$$
is very close, in a relative sense, to $c_nx^n$.
This means that if $n$ is even we must have that the sign as $xtoinfty$ is the same as $xto -infty$ because this is true of $c_nx^n$. This can only happen if there are an odd number of turning points. Similarly, for odd $n$ the sign as $xtoinfty$ is the opposite of the sign as $xto -infty$, so there must be an even number of turning points.
answered May 10 '16 at 9:03
Matt SamuelMatt Samuel
38.6k63768
$endgroup$
$begingroup$
I have a question here. For a quartic function, setting the first order derivative equal to zero should give us the local extrema. Since the first derivative is a cubic function, which can have three real roots, shouldn't the number of turning points for quartic be 1 or 2 or 3? Similarly, the maximum number of turning points in a cubic function should be 2 (coming from solving the quadratic).
$endgroup$
– PGupta
Aug 5 '18 at 14:51
$begingroup$
Is it because the solution to the cubic will give potential extrema (including inflection points)--so even if the cubic has two roots, one point will be a turning point and another will be the inflection point?
$endgroup$
– PGupta
Aug 5 '18 at 14:55
$begingroup$
@pgupta A cubic can't have exactly two real roots. Complex roots come in conjugate pairs.
$endgroup$
– Matt Samuel
Aug 5 '18 at 14:57
$begingroup$
If the cubic is $(x-3)(x+1)^2$ doesn't it have two real roots, 3 and -1, where -1 has a multiplicity of 2?
$endgroup$
– PGupta
Aug 6 '18 at 18:40
$begingroup$
@Pgupta Yes, you're right. Not sure what I was thinking.
$endgroup$
– Matt Samuel
Aug 7 '18 at 21:56
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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$begingroup$
You can tell the behavior of a polynomial $p(x)$ at $pminfty$ by looking at the term of highest degree, because for large $|x|$ the polynomial
$$p(x)=c_nx^n+c_{n-1}x^{n-1}+cdots +c_0$$
is very close, in a relative sense, to $c_nx^n$.
This means that if $n$ is even we must have that the sign as $xtoinfty$ is the same as $xto -infty$ because this is true of $c_nx^n$. This can only happen if there are an odd number of turning points. Similarly, for odd $n$ the sign as $xtoinfty$ is the opposite of the sign as $xto -infty$, so there must be an even number of turning points.
answered May 10 '16 at 9:03
Matt SamuelMatt Samuel
38.6k63768
$endgroup$
$begingroup$
I have a question here. For a quartic function, setting the first order derivative equal to zero should give us the local extrema. Since the first derivative is a cubic function, which can have three real roots, shouldn't the number of turning points for quartic be 1 or 2 or 3? Similarly, the maximum number of turning points in a cubic function should be 2 (coming from solving the quadratic).
$endgroup$
– PGupta
Aug 5 '18 at 14:51
$begingroup$
Is it because the solution to the cubic will give potential extrema (including inflection points)--so even if the cubic has two roots, one point will be a turning point and another will be the inflection point?
$endgroup$
– PGupta
Aug 5 '18 at 14:55
$begingroup$
@pgupta A cubic can't have exactly two real roots. Complex roots come in conjugate pairs.
$endgroup$
– Matt Samuel
Aug 5 '18 at 14:57
$begingroup$
If the cubic is $(x-3)(x+1)^2$ doesn't it have two real roots, 3 and -1, where -1 has a multiplicity of 2?
$endgroup$
– PGupta
Aug 6 '18 at 18:40
$begingroup$
@Pgupta Yes, you're right. Not sure what I was thinking.
$endgroup$
– Matt Samuel
Aug 7 '18 at 21:56
|
show 1 more comment
$begingroup$
You can tell the behavior of a polynomial $p(x)$ at $pminfty$ by looking at the term of highest degree, because for large $|x|$ the polynomial
$$p(x)=c_nx^n+c_{n-1}x^{n-1}+cdots +c_0$$
is very close, in a relative sense, to $c_nx^n$.
This means that if $n$ is even we must have that the sign as $xtoinfty$ is the same as $xto -infty$ because this is true of $c_nx^n$. This can only happen if there are an odd number of turning points. Similarly, for odd $n$ the sign as $xtoinfty$ is the opposite of the sign as $xto -infty$, so there must be an even number of turning points.
answered May 10 '16 at 9:03
Matt SamuelMatt Samuel
38.6k63768
$endgroup$
$begingroup$
I have a question here. For a quartic function, setting the first order derivative equal to zero should give us the local extrema. Since the first derivative is a cubic function, which can have three real roots, shouldn't the number of turning points for quartic be 1 or 2 or 3? Similarly, the maximum number of turning points in a cubic function should be 2 (coming from solving the quadratic).
$endgroup$
– PGupta
Aug 5 '18 at 14:51
$begingroup$
Is it because the solution to the cubic will give potential extrema (including inflection points)--so even if the cubic has two roots, one point will be a turning point and another will be the inflection point?
$endgroup$
– PGupta
Aug 5 '18 at 14:55
$begingroup$
@pgupta A cubic can't have exactly two real roots. Complex roots come in conjugate pairs.
$endgroup$
– Matt Samuel
Aug 5 '18 at 14:57
$begingroup$
If the cubic is $(x-3)(x+1)^2$ doesn't it have two real roots, 3 and -1, where -1 has a multiplicity of 2?
$endgroup$
– PGupta
Aug 6 '18 at 18:40
$begingroup$
@Pgupta Yes, you're right. Not sure what I was thinking.
$endgroup$
– Matt Samuel
Aug 7 '18 at 21:56
|
show 1 more comment
$begingroup$
You can tell the behavior of a polynomial $p(x)$ at $pminfty$ by looking at the term of highest degree, because for large $|x|$ the polynomial
$$p(x)=c_nx^n+c_{n-1}x^{n-1}+cdots +c_0$$
is very close, in a relative sense, to $c_nx^n$.
This means that if $n$ is even we must have that the sign as $xtoinfty$ is the same as $xto -infty$ because this is true of $c_nx^n$. This can only happen if there are an odd number of turning points. Similarly, for odd $n$ the sign as $xtoinfty$ is the opposite of the sign as $xto -infty$, so there must be an even number of turning points.
answered May 10 '16 at 9:03
Matt SamuelMatt Samuel
38.6k63768
$endgroup$
You can tell the behavior of a polynomial $p(x)$ at $pminfty$ by looking at the term of highest degree, because for large $|x|$ the polynomial
$$p(x)=c_nx^n+c_{n-1}x^{n-1}+cdots +c_0$$
is very close, in a relative sense, to $c_nx^n$.
This means that if $n$ is even we must have that the sign as $xtoinfty$ is the same as $xto -infty$ because this is true of $c_nx^n$. This can only happen if there are an odd number of turning points. Similarly, for odd $n$ the sign as $xtoinfty$ is the opposite of the sign as $xto -infty$, so there must be an even number of turning points.
answered May 10 '16 at 9:03
Matt SamuelMatt Samuel
38.6k63768
answered May 10 '16 at 9:03
Matt SamuelMatt Samuel
38.6k63768
answered May 10 '16 at 9:03
Matt SamuelMatt Samuel
38.6k63768
answered May 10 '16 at 9:03
Matt SamuelMatt Samuel
38.6k63768
38.6k63768
$begingroup$
I have a question here. For a quartic function, setting the first order derivative equal to zero should give us the local extrema. Since the first derivative is a cubic function, which can have three real roots, shouldn't the number of turning points for quartic be 1 or 2 or 3? Similarly, the maximum number of turning points in a cubic function should be 2 (coming from solving the quadratic).
$endgroup$
– PGupta
Aug 5 '18 at 14:51
$begingroup$
Is it because the solution to the cubic will give potential extrema (including inflection points)--so even if the cubic has two roots, one point will be a turning point and another will be the inflection point?
$endgroup$
– PGupta
Aug 5 '18 at 14:55
$begingroup$
@pgupta A cubic can't have exactly two real roots. Complex roots come in conjugate pairs.
$endgroup$
– Matt Samuel
Aug 5 '18 at 14:57
$begingroup$
If the cubic is $(x-3)(x+1)^2$ doesn't it have two real roots, 3 and -1, where -1 has a multiplicity of 2?
$endgroup$
– PGupta
Aug 6 '18 at 18:40
$begingroup$
@Pgupta Yes, you're right. Not sure what I was thinking.
$endgroup$
– Matt Samuel
Aug 7 '18 at 21:56
|
show 1 more comment
$begingroup$
I have a question here. For a quartic function, setting the first order derivative equal to zero should give us the local extrema. Since the first derivative is a cubic function, which can have three real roots, shouldn't the number of turning points for quartic be 1 or 2 or 3? Similarly, the maximum number of turning points in a cubic function should be 2 (coming from solving the quadratic).
$endgroup$
– PGupta
Aug 5 '18 at 14:51
$begingroup$
Is it because the solution to the cubic will give potential extrema (including inflection points)--so even if the cubic has two roots, one point will be a turning point and another will be the inflection point?
$endgroup$
– PGupta
Aug 5 '18 at 14:55
$begingroup$
@pgupta A cubic can't have exactly two real roots. Complex roots come in conjugate pairs.
$endgroup$
– Matt Samuel
Aug 5 '18 at 14:57
$begingroup$
If the cubic is $(x-3)(x+1)^2$ doesn't it have two real roots, 3 and -1, where -1 has a multiplicity of 2?
$endgroup$
– PGupta
Aug 6 '18 at 18:40
$begingroup$
@Pgupta Yes, you're right. Not sure what I was thinking.
$endgroup$
– Matt Samuel
Aug 7 '18 at 21:56
$begingroup$
I have a question here. For a quartic function, setting the first order derivative equal to zero should give us the local extrema. Since the first derivative is a cubic function, which can have three real roots, shouldn't the number of turning points for quartic be 1 or 2 or 3? Similarly, the maximum number of turning points in a cubic function should be 2 (coming from solving the quadratic).
$endgroup$
– PGupta
Aug 5 '18 at 14:51
$begingroup$
I have a question here. For a quartic function, setting the first order derivative equal to zero should give us the local extrema. Since the first derivative is a cubic function, which can have three real roots, shouldn't the number of turning points for quartic be 1 or 2 or 3? Similarly, the maximum number of turning points in a cubic function should be 2 (coming from solving the quadratic).
$endgroup$
– PGupta
Aug 5 '18 at 14:51
$begingroup$
Is it because the solution to the cubic will give potential extrema (including inflection points)--so even if the cubic has two roots, one point will be a turning point and another will be the inflection point?
$endgroup$
– PGupta
Aug 5 '18 at 14:55
$begingroup$
Is it because the solution to the cubic will give potential extrema (including inflection points)--so even if the cubic has two roots, one point will be a turning point and another will be the inflection point?
$endgroup$
– PGupta
Aug 5 '18 at 14:55
$begingroup$
@pgupta A cubic can't have exactly two real roots. Complex roots come in conjugate pairs.
$endgroup$
– Matt Samuel
Aug 5 '18 at 14:57
$begingroup$
@pgupta A cubic can't have exactly two real roots. Complex roots come in conjugate pairs.
$endgroup$
– Matt Samuel
Aug 5 '18 at 14:57
$begingroup$
If the cubic is $(x-3)(x+1)^2$ doesn't it have two real roots, 3 and -1, where -1 has a multiplicity of 2?
$endgroup$
– PGupta
Aug 6 '18 at 18:40
$begingroup$
If the cubic is $(x-3)(x+1)^2$ doesn't it have two real roots, 3 and -1, where -1 has a multiplicity of 2?
$endgroup$
– PGupta
Aug 6 '18 at 18:40
$begingroup$
@Pgupta Yes, you're right. Not sure what I was thinking.
$endgroup$
– Matt Samuel
Aug 7 '18 at 21:56
$begingroup$
@Pgupta Yes, you're right. Not sure what I was thinking.
$endgroup$
– Matt Samuel
Aug 7 '18 at 21:56
|
show 1 more comment
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$begingroup$
Not really sure what you mean there.
$endgroup$
– Son Jerm
May 10 '16 at 8:40
1
$begingroup$
What do you mean with turning points? Local extrema?
$endgroup$
– ccorn
May 10 '16 at 8:41
1
$begingroup$
Yes turning point meaning local extrema.
$endgroup$
– Son Jerm
May 10 '16 at 8:44
$begingroup$
Any polynomial of even degree must have an odd number of local extrema, because local extrema are precisely the points at which the gradient changes sign.
$endgroup$
– almagest
May 10 '16 at 8:46
$begingroup$
So it would not be possible for a quartic equation to have 2 local extrema? as quartic is of even degree, right?
$endgroup$
– Son Jerm
May 10 '16 at 8:49