Calculus demonstration. Inflex points.
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Demonstrate that all cubic function with three different real zeros, has an inflex point whose coordinate x is the average of the three zeros.
I've been meddling with this problem for a while, but I have not been able to even get close. I don't understand it, but I centainly understand the concepts that are mentioned.
calculus proof-verification
asked Jan 20 at 0:24
ScoofjeerScoofjeer
1335
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add a comment |
$begingroup$
Demonstrate that all cubic function with three different real zeros, has an inflex point whose coordinate x is the average of the three zeros.
I've been meddling with this problem for a while, but I have not been able to even get close. I don't understand it, but I centainly understand the concepts that are mentioned.
calculus proof-verification
asked Jan 20 at 0:24
ScoofjeerScoofjeer
1335
$endgroup$
add a comment |
$begingroup$
Demonstrate that all cubic function with three different real zeros, has an inflex point whose coordinate x is the average of the three zeros.
I've been meddling with this problem for a while, but I have not been able to even get close. I don't understand it, but I centainly understand the concepts that are mentioned.
calculus proof-verification
asked Jan 20 at 0:24
ScoofjeerScoofjeer
1335
$endgroup$
Demonstrate that all cubic function with three different real zeros, has an inflex point whose coordinate x is the average of the three zeros.
I've been meddling with this problem for a while, but I have not been able to even get close. I don't understand it, but I centainly understand the concepts that are mentioned.
calculus proof-verification
calculus proof-verification
asked Jan 20 at 0:24
ScoofjeerScoofjeer
1335
asked Jan 20 at 0:24
ScoofjeerScoofjeer
1335
asked Jan 20 at 0:24
ScoofjeerScoofjeer
1335
asked Jan 20 at 0:24
ScoofjeerScoofjeer
1335
asked Jan 20 at 0:24
ScoofjeerScoofjeer
1335
1335
add a comment |
add a comment |
2 Answers
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Let's call the three different real roots of our cubic $a,b,c$. We can write our cubic as $(x-a)(x-b)(x-c)=(x^2-(a+b)x+ab)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$
Now the derivative of this cubic is $3x^2-2(a+b+c)x+ab+ac+bc$ and the second derivative is $6x-2(a+b+c)$. We have that in a point of inflection the second derivative is $=0$ thus this point has $x$ coordinate equal to $frac{a+b+c}{3}$
answered Jan 20 at 0:48
user289143user289143
903313
$endgroup$
$begingroup$
Thank you very much. Real quick: Can you remind me why we’re able to write the cubic function like that? What's the reason?
$endgroup$
– Scoofjeer
Jan 20 at 1:36
add a comment |
$begingroup$
If the coefficient of $x^3$ is not equal to one, we may divide through by that without changing the problem. Then, for
$$ x^3 - A x^2 + B x - C, $$
the sum of the three (real, by hypothesis) roots is $A$ because
$$ (x-r_1)(x-r_2) (x-r_3) = x^3 - A x^2 + B x - C.$$
The first derivative is $3x^2 - 2Ax,$ the second derivative is $6x - 2A,$ with point of inflection located at $x = A/3.$
answered Jan 20 at 0:46
Will JagyWill Jagy
103k5102200
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Let's call the three different real roots of our cubic $a,b,c$. We can write our cubic as $(x-a)(x-b)(x-c)=(x^2-(a+b)x+ab)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$
Now the derivative of this cubic is $3x^2-2(a+b+c)x+ab+ac+bc$ and the second derivative is $6x-2(a+b+c)$. We have that in a point of inflection the second derivative is $=0$ thus this point has $x$ coordinate equal to $frac{a+b+c}{3}$
answered Jan 20 at 0:48
user289143user289143
903313
$endgroup$
$begingroup$
Thank you very much. Real quick: Can you remind me why we’re able to write the cubic function like that? What's the reason?
$endgroup$
– Scoofjeer
Jan 20 at 1:36
add a comment |
$begingroup$
Let's call the three different real roots of our cubic $a,b,c$. We can write our cubic as $(x-a)(x-b)(x-c)=(x^2-(a+b)x+ab)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$
Now the derivative of this cubic is $3x^2-2(a+b+c)x+ab+ac+bc$ and the second derivative is $6x-2(a+b+c)$. We have that in a point of inflection the second derivative is $=0$ thus this point has $x$ coordinate equal to $frac{a+b+c}{3}$
answered Jan 20 at 0:48
user289143user289143
903313
$endgroup$
$begingroup$
Thank you very much. Real quick: Can you remind me why we’re able to write the cubic function like that? What's the reason?
$endgroup$
– Scoofjeer
Jan 20 at 1:36
add a comment |
$begingroup$
Let's call the three different real roots of our cubic $a,b,c$. We can write our cubic as $(x-a)(x-b)(x-c)=(x^2-(a+b)x+ab)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$
Now the derivative of this cubic is $3x^2-2(a+b+c)x+ab+ac+bc$ and the second derivative is $6x-2(a+b+c)$. We have that in a point of inflection the second derivative is $=0$ thus this point has $x$ coordinate equal to $frac{a+b+c}{3}$
answered Jan 20 at 0:48
user289143user289143
903313
$endgroup$
Let's call the three different real roots of our cubic $a,b,c$. We can write our cubic as $(x-a)(x-b)(x-c)=(x^2-(a+b)x+ab)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$
Now the derivative of this cubic is $3x^2-2(a+b+c)x+ab+ac+bc$ and the second derivative is $6x-2(a+b+c)$. We have that in a point of inflection the second derivative is $=0$ thus this point has $x$ coordinate equal to $frac{a+b+c}{3}$
answered Jan 20 at 0:48
user289143user289143
903313
answered Jan 20 at 0:48
user289143user289143
903313
answered Jan 20 at 0:48
user289143user289143
903313
answered Jan 20 at 0:48
user289143user289143
903313
903313
$begingroup$
Thank you very much. Real quick: Can you remind me why we’re able to write the cubic function like that? What's the reason?
$endgroup$
– Scoofjeer
Jan 20 at 1:36
add a comment |
$begingroup$
Thank you very much. Real quick: Can you remind me why we’re able to write the cubic function like that? What's the reason?
$endgroup$
– Scoofjeer
Jan 20 at 1:36
$begingroup$
Thank you very much. Real quick: Can you remind me why we’re able to write the cubic function like that? What's the reason?
$endgroup$
– Scoofjeer
Jan 20 at 1:36
$begingroup$
Thank you very much. Real quick: Can you remind me why we’re able to write the cubic function like that? What's the reason?
$endgroup$
– Scoofjeer
Jan 20 at 1:36
add a comment |
$begingroup$
If the coefficient of $x^3$ is not equal to one, we may divide through by that without changing the problem. Then, for
$$ x^3 - A x^2 + B x - C, $$
the sum of the three (real, by hypothesis) roots is $A$ because
$$ (x-r_1)(x-r_2) (x-r_3) = x^3 - A x^2 + B x - C.$$
The first derivative is $3x^2 - 2Ax,$ the second derivative is $6x - 2A,$ with point of inflection located at $x = A/3.$
answered Jan 20 at 0:46
Will JagyWill Jagy
103k5102200
$endgroup$
add a comment |
$begingroup$
If the coefficient of $x^3$ is not equal to one, we may divide through by that without changing the problem. Then, for
$$ x^3 - A x^2 + B x - C, $$
the sum of the three (real, by hypothesis) roots is $A$ because
$$ (x-r_1)(x-r_2) (x-r_3) = x^3 - A x^2 + B x - C.$$
The first derivative is $3x^2 - 2Ax,$ the second derivative is $6x - 2A,$ with point of inflection located at $x = A/3.$
answered Jan 20 at 0:46
Will JagyWill Jagy
103k5102200
$endgroup$
add a comment |
$begingroup$
If the coefficient of $x^3$ is not equal to one, we may divide through by that without changing the problem. Then, for
$$ x^3 - A x^2 + B x - C, $$
the sum of the three (real, by hypothesis) roots is $A$ because
$$ (x-r_1)(x-r_2) (x-r_3) = x^3 - A x^2 + B x - C.$$
The first derivative is $3x^2 - 2Ax,$ the second derivative is $6x - 2A,$ with point of inflection located at $x = A/3.$
answered Jan 20 at 0:46
Will JagyWill Jagy
103k5102200
$endgroup$
If the coefficient of $x^3$ is not equal to one, we may divide through by that without changing the problem. Then, for
$$ x^3 - A x^2 + B x - C, $$
the sum of the three (real, by hypothesis) roots is $A$ because
$$ (x-r_1)(x-r_2) (x-r_3) = x^3 - A x^2 + B x - C.$$
The first derivative is $3x^2 - 2Ax,$ the second derivative is $6x - 2A,$ with point of inflection located at $x = A/3.$
answered Jan 20 at 0:46
Will JagyWill Jagy
103k5102200
answered Jan 20 at 0:46
Will JagyWill Jagy
103k5102200
answered Jan 20 at 0:46
Will JagyWill Jagy
103k5102200
answered Jan 20 at 0:46
Will JagyWill Jagy
103k5102200
103k5102200
add a comment |
add a comment |
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