Why is $2$ double root of the derivative?

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up vote
2
down vote

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A polynomial function $P(x)$ with degree $5$ increases in the interval
$(-infty, 1)$ and $(3, infty)$ and decreases in the interval
$(1,3)$. Given that $P'(2)= 0$ and $P(0) = 4$, find $P'(6)$.

In this problem, I have recognised that $2$ is an inflection point and the derivative will be of the form: $P'(x)= 5(x-1)(x-3)(x-2)(x-alpha)$ .

But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $alpha =2$?). It's not making sense to me. I need help with that part.

calculus derivatives stationary-point

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edited Jul 15 at 19:09

TheSimpliFire

9,69261951

asked Jul 15 at 18:58

Abcd

2,3831624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $2$ was a single root then $P$ would switch to increasing at $x=2$.
  – copper.hat
  Jul 15 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  why @copper.hat ...
  – Abcd
  Jul 15 at 19:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because $P'(2)=0$ and $P'$ has the form $P'(x) = (...)(x-2)(...)$, where the other parts keep the same sign close to $x=2$.
  – copper.hat
  Jul 15 at 19:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $a$ were not $2$, then $P'$ would change sign at $2$ ... but we know $P'$ is negative on both sides of $2$.
  – GEdgar
  Jul 15 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, we know $P'$ has at most 4 roots, and we know two of them, so the only remaining options are a single or double root at $x=2$.
  – copper.hat
  Jul 15 at 19:14
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

A polynomial function $P(x)$ with degree $5$ increases in the interval
$(-infty, 1)$ and $(3, infty)$ and decreases in the interval
$(1,3)$. Given that $P'(2)= 0$ and $P(0) = 4$, find $P'(6)$.

In this problem, I have recognised that $2$ is an inflection point and the derivative will be of the form: $P'(x)= 5(x-1)(x-3)(x-2)(x-alpha)$ .

But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $alpha =2$?). It's not making sense to me. I need help with that part.

calculus derivatives stationary-point

share|cite|improve this question

edited Jul 15 at 19:09

TheSimpliFire

9,69261951

asked Jul 15 at 18:58

Abcd

2,3831624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $2$ was a single root then $P$ would switch to increasing at $x=2$.
  – copper.hat
  Jul 15 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  why @copper.hat ...
  – Abcd
  Jul 15 at 19:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because $P'(2)=0$ and $P'$ has the form $P'(x) = (...)(x-2)(...)$, where the other parts keep the same sign close to $x=2$.
  – copper.hat
  Jul 15 at 19:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $a$ were not $2$, then $P'$ would change sign at $2$ ... but we know $P'$ is negative on both sides of $2$.
  – GEdgar
  Jul 15 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, we know $P'$ has at most 4 roots, and we know two of them, so the only remaining options are a single or double root at $x=2$.
  – copper.hat
  Jul 15 at 19:14
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

A polynomial function $P(x)$ with degree $5$ increases in the interval
$(-infty, 1)$ and $(3, infty)$ and decreases in the interval
$(1,3)$. Given that $P'(2)= 0$ and $P(0) = 4$, find $P'(6)$.

In this problem, I have recognised that $2$ is an inflection point and the derivative will be of the form: $P'(x)= 5(x-1)(x-3)(x-2)(x-alpha)$ .

But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $alpha =2$?). It's not making sense to me. I need help with that part.

calculus derivatives stationary-point

share|cite|improve this question

edited Jul 15 at 19:09

TheSimpliFire

9,69261951

asked Jul 15 at 18:58

Abcd

2,3831624

A polynomial function $P(x)$ with degree $5$ increases in the interval
$(-infty, 1)$ and $(3, infty)$ and decreases in the interval
$(1,3)$. Given that $P'(2)= 0$ and $P(0) = 4$, find $P'(6)$.

In this problem, I have recognised that $2$ is an inflection point and the derivative will be of the form: $P'(x)= 5(x-1)(x-3)(x-2)(x-alpha)$ .

But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $alpha =2$?). It's not making sense to me. I need help with that part.

calculus derivatives stationary-point

share|cite|improve this question

edited Jul 15 at 19:09

TheSimpliFire

9,69261951

asked Jul 15 at 18:58

Abcd

2,3831624

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 19:09

TheSimpliFire

9,69261951

edited Jul 15 at 19:09

TheSimpliFire

9,69261951

edited Jul 15 at 19:09

TheSimpliFire

9,69261951

9,69261951

asked Jul 15 at 18:58

Abcd

2,3831624

asked Jul 15 at 18:58

Abcd

2,3831624

asked Jul 15 at 18:58

Abcd

2,3831624

2,3831624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $2$ was a single root then $P$ would switch to increasing at $x=2$.
  – copper.hat
  Jul 15 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  why @copper.hat ...
  – Abcd
  Jul 15 at 19:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because $P'(2)=0$ and $P'$ has the form $P'(x) = (...)(x-2)(...)$, where the other parts keep the same sign close to $x=2$.
  – copper.hat
  Jul 15 at 19:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $a$ were not $2$, then $P'$ would change sign at $2$ ... but we know $P'$ is negative on both sides of $2$.
  – GEdgar
  Jul 15 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, we know $P'$ has at most 4 roots, and we know two of them, so the only remaining options are a single or double root at $x=2$.
  – copper.hat
  Jul 15 at 19:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $2$ was a single root then $P$ would switch to increasing at $x=2$.
  – copper.hat
  Jul 15 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  why @copper.hat ...
  – Abcd
  Jul 15 at 19:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because $P'(2)=0$ and $P'$ has the form $P'(x) = (...)(x-2)(...)$, where the other parts keep the same sign close to $x=2$.
  – copper.hat
  Jul 15 at 19:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $a$ were not $2$, then $P'$ would change sign at $2$ ... but we know $P'$ is negative on both sides of $2$.
  – GEdgar
  Jul 15 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, we know $P'$ has at most 4 roots, and we know two of them, so the only remaining options are a single or double root at $x=2$.
  – copper.hat
  Jul 15 at 19:14
  

  
  
  
  

If $2$ was a single root then $P$ would switch to increasing at $x=2$.
– copper.hat
Jul 15 at 19:10

If $2$ was a single root then $P$ would switch to increasing at $x=2$.
– copper.hat
Jul 15 at 19:10

why @copper.hat ...
– Abcd
Jul 15 at 19:11

why @copper.hat ...
– Abcd
Jul 15 at 19:11

1

1

Because $P'(2)=0$ and $P'$ has the form $P'(x) = (...)(x-2)(...)$, where the other parts keep the same sign close to $x=2$.
– copper.hat
Jul 15 at 19:12

Because $P'(2)=0$ and $P'$ has the form $P'(x) = (...)(x-2)(...)$, where the other parts keep the same sign close to $x=2$.
– copper.hat
Jul 15 at 19:12

1

1

If $a$ were not $2$, then $P'$ would change sign at $2$ ... but we know $P'$ is negative on both sides of $2$.
– GEdgar
Jul 15 at 19:12

If $a$ were not $2$, then $P'$ would change sign at $2$ ... but we know $P'$ is negative on both sides of $2$.
– GEdgar
Jul 15 at 19:12

Also, we know $P'$ has at most 4 roots, and we know two of them, so the only remaining options are a single or double root at $x=2$.
– copper.hat
Jul 15 at 19:14

Also, we know $P'$ has at most 4 roots, and we know two of them, so the only remaining options are a single or double root at $x=2$.
– copper.hat
Jul 15 at 19:14

add a comment | 

4 Answers
4

active

oldest

votes

up vote
3
down vote



accepted

HINT:

If $2$ is an inflexion point then it is necessary that $P''(2)=0$, which means that you have a factor of $(x-2)^2$ in $P'(x)$.

share|cite|improve this answer

answered Jul 15 at 19:07

TheSimpliFire

9,69261951

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does factor of (x-2)^2 ensure that P''(2) = 0?
  – Abcd
  Jul 15 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Differentiate $P'(x)$ and see what you get :)
  – TheSimpliFire
  Jul 15 at 19:09
  

  
  
  
  

add a comment | 

up vote
4
down vote

Since $2$ is an inflection point, we know that $P'(2)=0$ and $P''(2)=0$. (Convince yourself of this. What happens if $P''(2) neq 0$?)

$P''(x)$ is then $-15ax^2 + 60ax - 55a + 20x^3 - 90x^2 + 110x - 30$. Plug in $2$ to get

$$P''(2) = 5 a - 10$$

And so $P''(2)$ is $0$ when $a=2$.

EDIT: mechanodroid gives us a really clever shortcut:

If $P′$ has a double root at $2$ then $P′(x)=(x−2)^2 Q(x)$. This immediately gives $P′(2)=0$. Differentiating $P′$ gives: $P″(x)=2(x−2)Q(x)+(x−2)^2 Q′(x)$
so $P″(2)=0$.

This means that knowing that $P'(2) = P''(2) = 0$ is sufficient to conclude that $(x-2)^2|P'(x)$, hence $alpha=2$.

share|cite|improve this answer

edited Jul 15 at 19:24

answered Jul 15 at 19:07

Tiwa Aina

2,576319

add a comment | 

up vote
2
down vote

Hint:

$2$ is an inflexion point so necessarily $P''(2) = 0$.

Now $P'(2) = P''(2) = 0$ so $2$ is a double root of $P'$.

share|cite|improve this answer

answered Jul 15 at 19:06

mechanodroid

22.3k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is true that $P'(x)$ must have a root at 2 and that $P''(x)$ must have a root at 2, but how do we know these conditions are both satisfied when $P''(x)$ has a double root at $2$?
  – Tiwa Aina
  Jul 15 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TiwaAina If $P'$ has a double root at $2$ then $P'(x) = (x-2)^2Q(x)$. This immediately gives $P'(2) = 0$. Differentiating $P'$ gives: $$P''(x) = 2(x-2)Q(x) + (x-2)^2Q'(x)$$ so $P''(2) = 0$.
  – mechanodroid
  Jul 15 at 19:19
  

  
  
  
  

add a comment | 

up vote
1
down vote

The fact that $x=2$ is an inflection point implies that the second derivative $P''(2)=0$ as well. That means $2$ should not just be a root of $P'$, but should still be a root if you take the derivative again. That is the case if it is a double root.

share|cite|improve this answer

answered Jul 15 at 19:16

molarmass

1,325513

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do we know that a root is preserved (after differentiation) when it has multiplicity 2 or more? Especially when the polynomial has several roots?
  – Tiwa Aina
  Jul 15 at 19:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By the power rule, the $(x-2)^2$ term will not disappear if you take the derivative, whereas $(x-2)$ (without the square) would.
  – molarmass
  Jul 15 at 19:21
  

  
  
  
  

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

HINT:

If $2$ is an inflexion point then it is necessary that $P''(2)=0$, which means that you have a factor of $(x-2)^2$ in $P'(x)$.

share|cite|improve this answer

answered Jul 15 at 19:07

TheSimpliFire

9,69261951

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does factor of (x-2)^2 ensure that P''(2) = 0?
  – Abcd
  Jul 15 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Differentiate $P'(x)$ and see what you get :)
  – TheSimpliFire
  Jul 15 at 19:09
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

HINT:

If $2$ is an inflexion point then it is necessary that $P''(2)=0$, which means that you have a factor of $(x-2)^2$ in $P'(x)$.

share|cite|improve this answer

answered Jul 15 at 19:07

TheSimpliFire

9,69261951

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does factor of (x-2)^2 ensure that P''(2) = 0?
  – Abcd
  Jul 15 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Differentiate $P'(x)$ and see what you get :)
  – TheSimpliFire
  Jul 15 at 19:09
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

HINT:

If $2$ is an inflexion point then it is necessary that $P''(2)=0$, which means that you have a factor of $(x-2)^2$ in $P'(x)$.

share|cite|improve this answer

answered Jul 15 at 19:07

TheSimpliFire

9,69261951

HINT:

If $2$ is an inflexion point then it is necessary that $P''(2)=0$, which means that you have a factor of $(x-2)^2$ in $P'(x)$.

share|cite|improve this answer

answered Jul 15 at 19:07

TheSimpliFire

9,69261951

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 19:07

TheSimpliFire

9,69261951

answered Jul 15 at 19:07

TheSimpliFire

9,69261951

answered Jul 15 at 19:07

TheSimpliFire

9,69261951

9,69261951

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does factor of (x-2)^2 ensure that P''(2) = 0?
  – Abcd
  Jul 15 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Differentiate $P'(x)$ and see what you get :)
  – TheSimpliFire
  Jul 15 at 19:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does factor of (x-2)^2 ensure that P''(2) = 0?
  – Abcd
  Jul 15 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Differentiate $P'(x)$ and see what you get :)
  – TheSimpliFire
  Jul 15 at 19:09
  

  
  
  
  

How does factor of (x-2)^2 ensure that P''(2) = 0?
– Abcd
Jul 15 at 19:08

How does factor of (x-2)^2 ensure that P''(2) = 0?
– Abcd
Jul 15 at 19:08

Differentiate $P'(x)$ and see what you get :)
– TheSimpliFire
Jul 15 at 19:09

Differentiate $P'(x)$ and see what you get :)
– TheSimpliFire
Jul 15 at 19:09

add a comment | 

up vote
4
down vote

Since $2$ is an inflection point, we know that $P'(2)=0$ and $P''(2)=0$. (Convince yourself of this. What happens if $P''(2) neq 0$?)

$P''(x)$ is then $-15ax^2 + 60ax - 55a + 20x^3 - 90x^2 + 110x - 30$. Plug in $2$ to get

$$P''(2) = 5 a - 10$$

And so $P''(2)$ is $0$ when $a=2$.

EDIT: mechanodroid gives us a really clever shortcut:

If $P′$ has a double root at $2$ then $P′(x)=(x−2)^2 Q(x)$. This immediately gives $P′(2)=0$. Differentiating $P′$ gives: $P″(x)=2(x−2)Q(x)+(x−2)^2 Q′(x)$
so $P″(2)=0$.

This means that knowing that $P'(2) = P''(2) = 0$ is sufficient to conclude that $(x-2)^2|P'(x)$, hence $alpha=2$.

share|cite|improve this answer

edited Jul 15 at 19:24

answered Jul 15 at 19:07

Tiwa Aina

2,576319

add a comment | 

up vote
4
down vote

Since $2$ is an inflection point, we know that $P'(2)=0$ and $P''(2)=0$. (Convince yourself of this. What happens if $P''(2) neq 0$?)

$P''(x)$ is then $-15ax^2 + 60ax - 55a + 20x^3 - 90x^2 + 110x - 30$. Plug in $2$ to get

$$P''(2) = 5 a - 10$$

And so $P''(2)$ is $0$ when $a=2$.

EDIT: mechanodroid gives us a really clever shortcut:

If $P′$ has a double root at $2$ then $P′(x)=(x−2)^2 Q(x)$. This immediately gives $P′(2)=0$. Differentiating $P′$ gives: $P″(x)=2(x−2)Q(x)+(x−2)^2 Q′(x)$
so $P″(2)=0$.

This means that knowing that $P'(2) = P''(2) = 0$ is sufficient to conclude that $(x-2)^2|P'(x)$, hence $alpha=2$.

share|cite|improve this answer

edited Jul 15 at 19:24

answered Jul 15 at 19:07

Tiwa Aina

2,576319

add a comment | 

up vote
4
down vote

up vote
4
down vote

Since $2$ is an inflection point, we know that $P'(2)=0$ and $P''(2)=0$. (Convince yourself of this. What happens if $P''(2) neq 0$?)

$P''(x)$ is then $-15ax^2 + 60ax - 55a + 20x^3 - 90x^2 + 110x - 30$. Plug in $2$ to get

$$P''(2) = 5 a - 10$$

And so $P''(2)$ is $0$ when $a=2$.

EDIT: mechanodroid gives us a really clever shortcut:

If $P′$ has a double root at $2$ then $P′(x)=(x−2)^2 Q(x)$. This immediately gives $P′(2)=0$. Differentiating $P′$ gives: $P″(x)=2(x−2)Q(x)+(x−2)^2 Q′(x)$
so $P″(2)=0$.

This means that knowing that $P'(2) = P''(2) = 0$ is sufficient to conclude that $(x-2)^2|P'(x)$, hence $alpha=2$.

share|cite|improve this answer

edited Jul 15 at 19:24

answered Jul 15 at 19:07

Tiwa Aina

2,576319

Since $2$ is an inflection point, we know that $P'(2)=0$ and $P''(2)=0$. (Convince yourself of this. What happens if $P''(2) neq 0$?)

$P''(x)$ is then $-15ax^2 + 60ax - 55a + 20x^3 - 90x^2 + 110x - 30$. Plug in $2$ to get

$$P''(2) = 5 a - 10$$

And so $P''(2)$ is $0$ when $a=2$.

EDIT: mechanodroid gives us a really clever shortcut:

If $P′$ has a double root at $2$ then $P′(x)=(x−2)^2 Q(x)$. This immediately gives $P′(2)=0$. Differentiating $P′$ gives: $P″(x)=2(x−2)Q(x)+(x−2)^2 Q′(x)$
so $P″(2)=0$.

This means that knowing that $P'(2) = P''(2) = 0$ is sufficient to conclude that $(x-2)^2|P'(x)$, hence $alpha=2$.

share|cite|improve this answer

edited Jul 15 at 19:24

answered Jul 15 at 19:07

Tiwa Aina

2,576319

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 19:24

edited Jul 15 at 19:24

edited Jul 15 at 19:24

answered Jul 15 at 19:07

Tiwa Aina

2,576319

answered Jul 15 at 19:07

Tiwa Aina

2,576319

answered Jul 15 at 19:07

Tiwa Aina

2,576319

2,576319

add a comment | 

add a comment | 

up vote
2
down vote

Hint:

$2$ is an inflexion point so necessarily $P''(2) = 0$.

Now $P'(2) = P''(2) = 0$ so $2$ is a double root of $P'$.

share|cite|improve this answer

answered Jul 15 at 19:06

mechanodroid

22.3k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is true that $P'(x)$ must have a root at 2 and that $P''(x)$ must have a root at 2, but how do we know these conditions are both satisfied when $P''(x)$ has a double root at $2$?
  – Tiwa Aina
  Jul 15 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TiwaAina If $P'$ has a double root at $2$ then $P'(x) = (x-2)^2Q(x)$. This immediately gives $P'(2) = 0$. Differentiating $P'$ gives: $$P''(x) = 2(x-2)Q(x) + (x-2)^2Q'(x)$$ so $P''(2) = 0$.
  – mechanodroid
  Jul 15 at 19:19
  

  
  
  
  

add a comment | 

up vote
2
down vote

Hint:

$2$ is an inflexion point so necessarily $P''(2) = 0$.

Now $P'(2) = P''(2) = 0$ so $2$ is a double root of $P'$.

share|cite|improve this answer

answered Jul 15 at 19:06

mechanodroid

22.3k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is true that $P'(x)$ must have a root at 2 and that $P''(x)$ must have a root at 2, but how do we know these conditions are both satisfied when $P''(x)$ has a double root at $2$?
  – Tiwa Aina
  Jul 15 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TiwaAina If $P'$ has a double root at $2$ then $P'(x) = (x-2)^2Q(x)$. This immediately gives $P'(2) = 0$. Differentiating $P'$ gives: $$P''(x) = 2(x-2)Q(x) + (x-2)^2Q'(x)$$ so $P''(2) = 0$.
  – mechanodroid
  Jul 15 at 19:19
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Hint:

$2$ is an inflexion point so necessarily $P''(2) = 0$.

Now $P'(2) = P''(2) = 0$ so $2$ is a double root of $P'$.

share|cite|improve this answer

answered Jul 15 at 19:06

mechanodroid

22.3k52041

Hint:

$2$ is an inflexion point so necessarily $P''(2) = 0$.

Now $P'(2) = P''(2) = 0$ so $2$ is a double root of $P'$.

share|cite|improve this answer

answered Jul 15 at 19:06

mechanodroid

22.3k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 19:06

mechanodroid

22.3k52041

answered Jul 15 at 19:06

mechanodroid

22.3k52041

answered Jul 15 at 19:06

mechanodroid

22.3k52041

22.3k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is true that $P'(x)$ must have a root at 2 and that $P''(x)$ must have a root at 2, but how do we know these conditions are both satisfied when $P''(x)$ has a double root at $2$?
  – Tiwa Aina
  Jul 15 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TiwaAina If $P'$ has a double root at $2$ then $P'(x) = (x-2)^2Q(x)$. This immediately gives $P'(2) = 0$. Differentiating $P'$ gives: $$P''(x) = 2(x-2)Q(x) + (x-2)^2Q'(x)$$ so $P''(2) = 0$.
  – mechanodroid
  Jul 15 at 19:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is true that $P'(x)$ must have a root at 2 and that $P''(x)$ must have a root at 2, but how do we know these conditions are both satisfied when $P''(x)$ has a double root at $2$?
  – Tiwa Aina
  Jul 15 at 19:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @TiwaAina If $P'$ has a double root at $2$ then $P'(x) = (x-2)^2Q(x)$. This immediately gives $P'(2) = 0$. Differentiating $P'$ gives: $$P''(x) = 2(x-2)Q(x) + (x-2)^2Q'(x)$$ so $P''(2) = 0$.
  – mechanodroid
  Jul 15 at 19:19
  

  
  
  
  

It is true that $P'(x)$ must have a root at 2 and that $P''(x)$ must have a root at 2, but how do we know these conditions are both satisfied when $P''(x)$ has a double root at $2$?
– Tiwa Aina
Jul 15 at 19:16

It is true that $P'(x)$ must have a root at 2 and that $P''(x)$ must have a root at 2, but how do we know these conditions are both satisfied when $P''(x)$ has a double root at $2$?
– Tiwa Aina
Jul 15 at 19:16

1

1

@TiwaAina If $P'$ has a double root at $2$ then $P'(x) = (x-2)^2Q(x)$. This immediately gives $P'(2) = 0$. Differentiating $P'$ gives: $$P''(x) = 2(x-2)Q(x) + (x-2)^2Q'(x)$$ so $P''(2) = 0$.
– mechanodroid
Jul 15 at 19:19

@TiwaAina If $P'$ has a double root at $2$ then $P'(x) = (x-2)^2Q(x)$. This immediately gives $P'(2) = 0$. Differentiating $P'$ gives: $$P''(x) = 2(x-2)Q(x) + (x-2)^2Q'(x)$$ so $P''(2) = 0$.
– mechanodroid
Jul 15 at 19:19

add a comment | 

up vote
1
down vote

The fact that $x=2$ is an inflection point implies that the second derivative $P''(2)=0$ as well. That means $2$ should not just be a root of $P'$, but should still be a root if you take the derivative again. That is the case if it is a double root.

share|cite|improve this answer

answered Jul 15 at 19:16

molarmass

1,325513

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do we know that a root is preserved (after differentiation) when it has multiplicity 2 or more? Especially when the polynomial has several roots?
  – Tiwa Aina
  Jul 15 at 19:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By the power rule, the $(x-2)^2$ term will not disappear if you take the derivative, whereas $(x-2)$ (without the square) would.
  – molarmass
  Jul 15 at 19:21
  

  
  
  
  

add a comment | 

up vote
1
down vote

The fact that $x=2$ is an inflection point implies that the second derivative $P''(2)=0$ as well. That means $2$ should not just be a root of $P'$, but should still be a root if you take the derivative again. That is the case if it is a double root.

share|cite|improve this answer

answered Jul 15 at 19:16

molarmass

1,325513

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do we know that a root is preserved (after differentiation) when it has multiplicity 2 or more? Especially when the polynomial has several roots?
  – Tiwa Aina
  Jul 15 at 19:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By the power rule, the $(x-2)^2$ term will not disappear if you take the derivative, whereas $(x-2)$ (without the square) would.
  – molarmass
  Jul 15 at 19:21
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

The fact that $x=2$ is an inflection point implies that the second derivative $P''(2)=0$ as well. That means $2$ should not just be a root of $P'$, but should still be a root if you take the derivative again. That is the case if it is a double root.

share|cite|improve this answer

answered Jul 15 at 19:16

molarmass

1,325513

The fact that $x=2$ is an inflection point implies that the second derivative $P''(2)=0$ as well. That means $2$ should not just be a root of $P'$, but should still be a root if you take the derivative again. That is the case if it is a double root.

share|cite|improve this answer

answered Jul 15 at 19:16

molarmass

1,325513

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 19:16

molarmass

1,325513

answered Jul 15 at 19:16

molarmass

1,325513

answered Jul 15 at 19:16

molarmass

1,325513

1,325513

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do we know that a root is preserved (after differentiation) when it has multiplicity 2 or more? Especially when the polynomial has several roots?
  – Tiwa Aina
  Jul 15 at 19:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By the power rule, the $(x-2)^2$ term will not disappear if you take the derivative, whereas $(x-2)$ (without the square) would.
  – molarmass
  Jul 15 at 19:21
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do we know that a root is preserved (after differentiation) when it has multiplicity 2 or more? Especially when the polynomial has several roots?
  – Tiwa Aina
  Jul 15 at 19:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  By the power rule, the $(x-2)^2$ term will not disappear if you take the derivative, whereas $(x-2)$ (without the square) would.
  – molarmass
  Jul 15 at 19:21
  

  
  
  
  

How do we know that a root is preserved (after differentiation) when it has multiplicity 2 or more? Especially when the polynomial has several roots?
– Tiwa Aina
Jul 15 at 19:17

How do we know that a root is preserved (after differentiation) when it has multiplicity 2 or more? Especially when the polynomial has several roots?
– Tiwa Aina
Jul 15 at 19:17

By the power rule, the $(x-2)^2$ term will not disappear if you take the derivative, whereas $(x-2)$ (without the square) would.
– molarmass
Jul 15 at 19:21

By the power rule, the $(x-2)^2$ term will not disappear if you take the derivative, whereas $(x-2)$ (without the square) would.
– molarmass
Jul 15 at 19:21

add a comment | 

 

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