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Is there a relationship between the stationary points and the inflection point of a cubic polynomial function?
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Determine the stationary points, A and B, and point of inflection, G, for each of the following cubic polynomials.
(a) $y=x^3 -3x^2 -9x+7$
(b) $y=x^3 -12x^2 +21x-14$
(c) $y=x^3 +9x^2 -12$
Is there any common relationship between A, B and G?
calculus
edited 4 hours ago
Blue
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Determine the stationary points, A and B, and point of inflection, G, for each of the following cubic polynomials.
(a) $y=x^3 -3x^2 -9x+7$
(b) $y=x^3 -12x^2 +21x-14$
(c) $y=x^3 +9x^2 -12$
Is there any common relationship between A, B and G?
calculus
edited 4 hours ago
Blue
45k868142
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
1
To augment the previous comment: Explaining what you know about a problem, and where exactly you got stuck, helps answerers tailor their responses to your skill level, while avoiding wasting time telling you things you already know. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for you.) In this case, do you know what stationary and inflection points are? Can you find them? (If so, tell us what you think they are, even if you're wrong.) Are you only having trouble finding the "common relationship"?
– Blue
4 hours ago
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Determine the stationary points, A and B, and point of inflection, G, for each of the following cubic polynomials.
(a) $y=x^3 -3x^2 -9x+7$
(b) $y=x^3 -12x^2 +21x-14$
(c) $y=x^3 +9x^2 -12$
Is there any common relationship between A, B and G?
calculus
edited 4 hours ago
Blue
45k868142
Determine the stationary points, A and B, and point of inflection, G, for each of the following cubic polynomials.
(a) $y=x^3 -3x^2 -9x+7$
(b) $y=x^3 -12x^2 +21x-14$
(c) $y=x^3 +9x^2 -12$
Is there any common relationship between A, B and G?
calculus
calculus
edited 4 hours ago
Blue
45k868142
edited 4 hours ago
Blue
45k868142
edited 4 hours ago
Blue
45k868142
edited 4 hours ago
Blue
45k868142
edited 4 hours ago
Blue
45k868142
45k868142
asked 4 hours ago
user603345
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
1
To augment the previous comment: Explaining what you know about a problem, and where exactly you got stuck, helps answerers tailor their responses to your skill level, while avoiding wasting time telling you things you already know. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for you.) In this case, do you know what stationary and inflection points are? Can you find them? (If so, tell us what you think they are, even if you're wrong.) Are you only having trouble finding the "common relationship"?
– Blue
4 hours ago
add a comment |Â
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
1
To augment the previous comment: Explaining what you know about a problem, and where exactly you got stuck, helps answerers tailor their responses to your skill level, while avoiding wasting time telling you things you already know. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for you.) In this case, do you know what stationary and inflection points are? Can you find them? (If so, tell us what you think they are, even if you're wrong.) Are you only having trouble finding the "common relationship"?
– Blue
4 hours ago
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
1
1
To augment the previous comment: Explaining what you know about a problem, and where exactly you got stuck, helps answerers tailor their responses to your skill level, while avoiding wasting time telling you things you already know. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for you.) In this case, do you know what stationary and inflection points are? Can you find them? (If so, tell us what you think they are, even if you're wrong.) Are you only having trouble finding the "common relationship"?
– Blue
4 hours ago
To augment the previous comment: Explaining what you know about a problem, and where exactly you got stuck, helps answerers tailor their responses to your skill level, while avoiding wasting time telling you things you already know. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for you.) In this case, do you know what stationary and inflection points are? Can you find them? (If so, tell us what you think they are, even if you're wrong.) Are you only having trouble finding the "common relationship"?
– Blue
4 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
The first derivative of a cubic polynomial is a quadratic polynomial $q$, and as such is even with respect to its stationary point $xi$. It follows that the real zeros of $q$, if there are any, are at equal distance on both sides of $xi$. This implies $G=A+Bover2$.
answered 3 hours ago
Christian Blatter
167k7111316
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up vote
2
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For any cubic, by translating the independent variable, you can make the quadratic term disappear and obtain a form
$$x^3+px+q.$$
You can also translate vertically and reduce to
$$x^3+px.$$
Then by rescaling the variable and the function,
$$frac(sx)^3+psxs^3=x^3+frac ps^2x=x^3pm ,x$$ by setting $s^2=|p|$. And if $p=0$, we simply have $x^3$.
These transformation are affine and preserve ratios, and in particular the relative positions of the extrema and the inflection.
Without further computation, as the function is odd the single inflection must lie midway of the two extrema.
We can in fact say that, as regards the remarkable points, there are essentially three cubics: $x^3,x^3-x,x^3+x$.
(If the cubic coefficient was negative, the graph must be flipped and one may distinguish six cases, $pm x^3,pm x^3pm x$.)
answered 2 hours ago
Yves Daoust
117k667213
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up vote
1
down vote
You are in fact asking if there is a relation between the real roots and the extremum of a quadratic function. By symmetry of the parabola, yes, $G$ is the mid-point of $AB$.
answered 2 hours ago
Yves Daoust
117k667213
add a comment |Â
up vote
0
down vote
Hint: $fracmathrm dymathrm dx = 0$ to find the $x$-values of the stationary points, then sub these back into the original equation to find the y-values. As for the points of inflection, these occur when $fracmathrm d^2ymathrm dx^2 = 0$, similarly find the x-values and sub these back into the original equation for the y-values of the points of inflection.
Eg for a)
$$fracmathrm dymathrm dx = 3x^2 - 6x - 9 = 0$$
$$x^2-2x-3=0$$
$$(x-3)(x+1)=0 implies x=-1, x=3$$
$$y(-1) = 12$$ $$y(3) = -20$$
$$fracmathrm d^2ymathrm dx^2 = 6x-6 = 0 implies x = 1$$
$$y(1) = -4$$
So the two stationary points are $(-1,12)$ and $(3,-20)$ and the point of inflection is $(1,-4)$
answered 4 hours ago
Patrick Jankowski
21010
1
I am afraid that this does not answer the question.
– Yves Daoust
2 hours ago
Ah indeed haha, I didn't notice he asked about the relationship between the different points. Thanks for pointing it out
– Patrick Jankowski
43 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The first derivative of a cubic polynomial is a quadratic polynomial $q$, and as such is even with respect to its stationary point $xi$. It follows that the real zeros of $q$, if there are any, are at equal distance on both sides of $xi$. This implies $G=A+Bover2$.
answered 3 hours ago
Christian Blatter
167k7111316
add a comment |Â
up vote
2
down vote
The first derivative of a cubic polynomial is a quadratic polynomial $q$, and as such is even with respect to its stationary point $xi$. It follows that the real zeros of $q$, if there are any, are at equal distance on both sides of $xi$. This implies $G=A+Bover2$.
answered 3 hours ago
Christian Blatter
167k7111316
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The first derivative of a cubic polynomial is a quadratic polynomial $q$, and as such is even with respect to its stationary point $xi$. It follows that the real zeros of $q$, if there are any, are at equal distance on both sides of $xi$. This implies $G=A+Bover2$.
answered 3 hours ago
Christian Blatter
167k7111316
The first derivative of a cubic polynomial is a quadratic polynomial $q$, and as such is even with respect to its stationary point $xi$. It follows that the real zeros of $q$, if there are any, are at equal distance on both sides of $xi$. This implies $G=A+Bover2$.
answered 3 hours ago
Christian Blatter
167k7111316
answered 3 hours ago
Christian Blatter
167k7111316
answered 3 hours ago
Christian Blatter
167k7111316
answered 3 hours ago
Christian Blatter
167k7111316
167k7111316
add a comment |Â
add a comment |Â
up vote
2
down vote
For any cubic, by translating the independent variable, you can make the quadratic term disappear and obtain a form
$$x^3+px+q.$$
You can also translate vertically and reduce to
$$x^3+px.$$
Then by rescaling the variable and the function,
$$frac(sx)^3+psxs^3=x^3+frac ps^2x=x^3pm ,x$$ by setting $s^2=|p|$. And if $p=0$, we simply have $x^3$.
These transformation are affine and preserve ratios, and in particular the relative positions of the extrema and the inflection.
Without further computation, as the function is odd the single inflection must lie midway of the two extrema.
We can in fact say that, as regards the remarkable points, there are essentially three cubics: $x^3,x^3-x,x^3+x$.
(If the cubic coefficient was negative, the graph must be flipped and one may distinguish six cases, $pm x^3,pm x^3pm x$.)
answered 2 hours ago
Yves Daoust
117k667213
add a comment |Â
up vote
2
down vote
For any cubic, by translating the independent variable, you can make the quadratic term disappear and obtain a form
$$x^3+px+q.$$
You can also translate vertically and reduce to
$$x^3+px.$$
Then by rescaling the variable and the function,
$$frac(sx)^3+psxs^3=x^3+frac ps^2x=x^3pm ,x$$ by setting $s^2=|p|$. And if $p=0$, we simply have $x^3$.
These transformation are affine and preserve ratios, and in particular the relative positions of the extrema and the inflection.
Without further computation, as the function is odd the single inflection must lie midway of the two extrema.
We can in fact say that, as regards the remarkable points, there are essentially three cubics: $x^3,x^3-x,x^3+x$.
(If the cubic coefficient was negative, the graph must be flipped and one may distinguish six cases, $pm x^3,pm x^3pm x$.)
answered 2 hours ago
Yves Daoust
117k667213
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For any cubic, by translating the independent variable, you can make the quadratic term disappear and obtain a form
$$x^3+px+q.$$
You can also translate vertically and reduce to
$$x^3+px.$$
Then by rescaling the variable and the function,
$$frac(sx)^3+psxs^3=x^3+frac ps^2x=x^3pm ,x$$ by setting $s^2=|p|$. And if $p=0$, we simply have $x^3$.
These transformation are affine and preserve ratios, and in particular the relative positions of the extrema and the inflection.
Without further computation, as the function is odd the single inflection must lie midway of the two extrema.
We can in fact say that, as regards the remarkable points, there are essentially three cubics: $x^3,x^3-x,x^3+x$.
(If the cubic coefficient was negative, the graph must be flipped and one may distinguish six cases, $pm x^3,pm x^3pm x$.)
answered 2 hours ago
Yves Daoust
117k667213
For any cubic, by translating the independent variable, you can make the quadratic term disappear and obtain a form
$$x^3+px+q.$$
You can also translate vertically and reduce to
$$x^3+px.$$
Then by rescaling the variable and the function,
$$frac(sx)^3+psxs^3=x^3+frac ps^2x=x^3pm ,x$$ by setting $s^2=|p|$. And if $p=0$, we simply have $x^3$.
These transformation are affine and preserve ratios, and in particular the relative positions of the extrema and the inflection.
Without further computation, as the function is odd the single inflection must lie midway of the two extrema.
We can in fact say that, as regards the remarkable points, there are essentially three cubics: $x^3,x^3-x,x^3+x$.
(If the cubic coefficient was negative, the graph must be flipped and one may distinguish six cases, $pm x^3,pm x^3pm x$.)
answered 2 hours ago
Yves Daoust
117k667213
edited 2 hours ago
answered 2 hours ago
Yves Daoust
117k667213
answered 2 hours ago
Yves Daoust
117k667213
answered 2 hours ago
Yves Daoust
117k667213
117k667213
add a comment |Â
add a comment |Â
up vote
1
down vote
You are in fact asking if there is a relation between the real roots and the extremum of a quadratic function. By symmetry of the parabola, yes, $G$ is the mid-point of $AB$.
answered 2 hours ago
Yves Daoust
117k667213
add a comment |Â
up vote
1
down vote
You are in fact asking if there is a relation between the real roots and the extremum of a quadratic function. By symmetry of the parabola, yes, $G$ is the mid-point of $AB$.
answered 2 hours ago
Yves Daoust
117k667213
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are in fact asking if there is a relation between the real roots and the extremum of a quadratic function. By symmetry of the parabola, yes, $G$ is the mid-point of $AB$.
answered 2 hours ago
Yves Daoust
117k667213
You are in fact asking if there is a relation between the real roots and the extremum of a quadratic function. By symmetry of the parabola, yes, $G$ is the mid-point of $AB$.
answered 2 hours ago
Yves Daoust
117k667213
answered 2 hours ago
Yves Daoust
117k667213
answered 2 hours ago
Yves Daoust
117k667213
answered 2 hours ago
Yves Daoust
117k667213
117k667213
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint: $fracmathrm dymathrm dx = 0$ to find the $x$-values of the stationary points, then sub these back into the original equation to find the y-values. As for the points of inflection, these occur when $fracmathrm d^2ymathrm dx^2 = 0$, similarly find the x-values and sub these back into the original equation for the y-values of the points of inflection.
Eg for a)
$$fracmathrm dymathrm dx = 3x^2 - 6x - 9 = 0$$
$$x^2-2x-3=0$$
$$(x-3)(x+1)=0 implies x=-1, x=3$$
$$y(-1) = 12$$ $$y(3) = -20$$
$$fracmathrm d^2ymathrm dx^2 = 6x-6 = 0 implies x = 1$$
$$y(1) = -4$$
So the two stationary points are $(-1,12)$ and $(3,-20)$ and the point of inflection is $(1,-4)$
answered 4 hours ago
Patrick Jankowski
21010
1
I am afraid that this does not answer the question.
– Yves Daoust
2 hours ago
Ah indeed haha, I didn't notice he asked about the relationship between the different points. Thanks for pointing it out
– Patrick Jankowski
43 mins ago
add a comment |Â
up vote
0
down vote
Hint: $fracmathrm dymathrm dx = 0$ to find the $x$-values of the stationary points, then sub these back into the original equation to find the y-values. As for the points of inflection, these occur when $fracmathrm d^2ymathrm dx^2 = 0$, similarly find the x-values and sub these back into the original equation for the y-values of the points of inflection.
Eg for a)
$$fracmathrm dymathrm dx = 3x^2 - 6x - 9 = 0$$
$$x^2-2x-3=0$$
$$(x-3)(x+1)=0 implies x=-1, x=3$$
$$y(-1) = 12$$ $$y(3) = -20$$
$$fracmathrm d^2ymathrm dx^2 = 6x-6 = 0 implies x = 1$$
$$y(1) = -4$$
So the two stationary points are $(-1,12)$ and $(3,-20)$ and the point of inflection is $(1,-4)$
answered 4 hours ago
Patrick Jankowski
21010
1
I am afraid that this does not answer the question.
– Yves Daoust
2 hours ago
Ah indeed haha, I didn't notice he asked about the relationship between the different points. Thanks for pointing it out
– Patrick Jankowski
43 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: $fracmathrm dymathrm dx = 0$ to find the $x$-values of the stationary points, then sub these back into the original equation to find the y-values. As for the points of inflection, these occur when $fracmathrm d^2ymathrm dx^2 = 0$, similarly find the x-values and sub these back into the original equation for the y-values of the points of inflection.
Eg for a)
$$fracmathrm dymathrm dx = 3x^2 - 6x - 9 = 0$$
$$x^2-2x-3=0$$
$$(x-3)(x+1)=0 implies x=-1, x=3$$
$$y(-1) = 12$$ $$y(3) = -20$$
$$fracmathrm d^2ymathrm dx^2 = 6x-6 = 0 implies x = 1$$
$$y(1) = -4$$
So the two stationary points are $(-1,12)$ and $(3,-20)$ and the point of inflection is $(1,-4)$
answered 4 hours ago
Patrick Jankowski
21010
Hint: $fracmathrm dymathrm dx = 0$ to find the $x$-values of the stationary points, then sub these back into the original equation to find the y-values. As for the points of inflection, these occur when $fracmathrm d^2ymathrm dx^2 = 0$, similarly find the x-values and sub these back into the original equation for the y-values of the points of inflection.
Eg for a)
$$fracmathrm dymathrm dx = 3x^2 - 6x - 9 = 0$$
$$x^2-2x-3=0$$
$$(x-3)(x+1)=0 implies x=-1, x=3$$
$$y(-1) = 12$$ $$y(3) = -20$$
$$fracmathrm d^2ymathrm dx^2 = 6x-6 = 0 implies x = 1$$
$$y(1) = -4$$
So the two stationary points are $(-1,12)$ and $(3,-20)$ and the point of inflection is $(1,-4)$
answered 4 hours ago
Patrick Jankowski
21010
answered 4 hours ago
Patrick Jankowski
21010
answered 4 hours ago
Patrick Jankowski
21010
answered 4 hours ago
Patrick Jankowski
21010
21010
1
I am afraid that this does not answer the question.
– Yves Daoust
2 hours ago
Ah indeed haha, I didn't notice he asked about the relationship between the different points. Thanks for pointing it out
– Patrick Jankowski
43 mins ago
add a comment |Â
1
I am afraid that this does not answer the question.
– Yves Daoust
2 hours ago
Ah indeed haha, I didn't notice he asked about the relationship between the different points. Thanks for pointing it out
– Patrick Jankowski
43 mins ago
1
1
I am afraid that this does not answer the question.
– Yves Daoust
2 hours ago
I am afraid that this does not answer the question.
– Yves Daoust
2 hours ago
Ah indeed haha, I didn't notice he asked about the relationship between the different points. Thanks for pointing it out
– Patrick Jankowski
43 mins ago
Ah indeed haha, I didn't notice he asked about the relationship between the different points. Thanks for pointing it out
– Patrick Jankowski
43 mins ago
add a comment |Â
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
1
To augment the previous comment: Explaining what you know about a problem, and where exactly you got stuck, helps answerers tailor their responses to your skill level, while avoiding wasting time telling you things you already know. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for you.) In this case, do you know what stationary and inflection points are? Can you find them? (If so, tell us what you think they are, even if you're wrong.) Are you only having trouble finding the "common relationship"?
– Blue
4 hours ago