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The point of inflection on the curve $𝑦=𝑥^3−𝑎𝑥^2−𝑏𝑥+𝑐$ is a stationary point of inflexion. Show that $b=8a^2$. [duplicate]

Show that $b=8a^2$Find the coordinates of any stationary points on the curve $y = {1 over {1 + {x^2}}}$ and state it's natureStrictly monotone real function: stationary point, non-differentiable pointIs $t=0$ a stationary point?The second and third derivitive tests give unexpected $0$How do I show that something only has one stationary point?Sufficient condition for inflection pointThis second derivative is showing a point of inflection rather than a minimum pointa point of inflectionNormal to a curve at the point x=1Show that if the curve $y = f(x)$ has a maximum stationary point at $x = a$

0

1

$begingroup$

This question already has an answer here:

• 
  Show that $b=8a^2$
  
  2 answers
  
  

The point of inflection on the curve $𝑦=𝑥^3−𝑎𝑥^2−𝑏𝑥+𝑐$ is a stationary point of inflection. Show that $b=8a2$.

Thank you for your help.

Edit: The solution to this question drafted in another post is wrong, according to the book. That is why I post it again, in a separate post.

Edit 2: I am not sure, then, why some of you still mark it as a duplicate.

calculus derivatives

share|cite|improve this question

edited Mar 17 at 16:57

Marcin

asked Mar 14 at 12:19

MarcinMarcin

83

$endgroup$

marked as duplicate by Dietrich Burde, RRL, Saad, Parcly Taxel, John Omielan Mar 14 at 14:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The thing is that in that post the solution does not show that this is actually the case - that b = 8a2; and such, according to the solution in the book, is wrong.
  $endgroup$
  – Marcin
  Mar 14 at 12:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Please use MathJax
  $endgroup$
  – J. W. Tanner
  Mar 14 at 13:20
  
  
  

  
  
  
  

add a comment | 

0

1

$begingroup$

This question already has an answer here:

• 
  Show that $b=8a^2$
  
  2 answers
  
  

The point of inflection on the curve $𝑦=𝑥^3−𝑎𝑥^2−𝑏𝑥+𝑐$ is a stationary point of inflection. Show that $b=8a2$.

Thank you for your help.

Edit: The solution to this question drafted in another post is wrong, according to the book. That is why I post it again, in a separate post.

Edit 2: I am not sure, then, why some of you still mark it as a duplicate.

calculus derivatives

share|cite|improve this question

edited Mar 17 at 16:57

Marcin

asked Mar 14 at 12:19

MarcinMarcin

83

$endgroup$

marked as duplicate by Dietrich Burde, RRL, Saad, Parcly Taxel, John Omielan Mar 14 at 14:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The thing is that in that post the solution does not show that this is actually the case - that b = 8a2; and such, according to the solution in the book, is wrong.
  $endgroup$
  – Marcin
  Mar 14 at 12:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Please use MathJax
  $endgroup$
  – J. W. Tanner
  Mar 14 at 13:20
  
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Show that $b=8a^2$
  
  2 answers
  
  

The point of inflection on the curve $𝑦=𝑥^3−𝑎𝑥^2−𝑏𝑥+𝑐$ is a stationary point of inflection. Show that $b=8a2$.

Thank you for your help.

Edit: The solution to this question drafted in another post is wrong, according to the book. That is why I post it again, in a separate post.

Edit 2: I am not sure, then, why some of you still mark it as a duplicate.

calculus derivatives

share|cite|improve this question

edited Mar 17 at 16:57

Marcin

asked Mar 14 at 12:19

MarcinMarcin

83

$endgroup$

share|cite|improve this question

edited Mar 17 at 16:57

Marcin

asked Mar 14 at 12:19

MarcinMarcin

83

share|cite|improve this question

edited Mar 17 at 16:57

Marcin

asked Mar 14 at 12:19

MarcinMarcin

83

asked Mar 14 at 12:19

MarcinMarcin

83

asked Mar 14 at 12:19

MarcinMarcin

83

1 Answer
1

active

oldest

votes

0

$begingroup$

To find stationary points we need $f'(x)=0$ and to find inflection points we need to find where $f''(x)=0$.

$$f(x)=x^3-ax^2-bx+c$$
$$f'(x)=3x^2-2ax-b$$
$$f''(x)=6x-2a$$

So to find the inflection point we need $$6x-2a=0$$ $$6x=2a$$ $$x=frac{a}{3}.$$

In order for this point to be stationary we need $f'(frac{a}{3})=0$.

$$3bigg(frac{a}{3}bigg)^2-2abigg(frac{a}{3}bigg)-b=0$$
$$frac{a^2}{3}-frac{2a^2}{3}-b=0$$

Now you should be able to plug in $a$ and $b$ and show that this equation is correct. Is your question saying that $a=2$ and $b=8$? Or $b=8a^2$? In either case, this inflection point would not be a stationary point in this function.

share|cite|improve this answer

answered Mar 14 at 14:16

Jake OJake O

1665

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The latter one. It looks like there is a mistake in the answers, then. Thank you very much for solving it.
  $endgroup$
  – Marcin
  Mar 14 at 14:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You’re welcome, glad I could help.
  $endgroup$
  – Jake O
  Mar 14 at 14:42
  

  
  
  
  

add a comment | 

0

$begingroup$

To find stationary points we need $f'(x)=0$ and to find inflection points we need to find where $f''(x)=0$.

$$f(x)=x^3-ax^2-bx+c$$
$$f'(x)=3x^2-2ax-b$$
$$f''(x)=6x-2a$$

So to find the inflection point we need $$6x-2a=0$$ $$6x=2a$$ $$x=frac{a}{3}.$$

In order for this point to be stationary we need $f'(frac{a}{3})=0$.

$$3bigg(frac{a}{3}bigg)^2-2abigg(frac{a}{3}bigg)-b=0$$
$$frac{a^2}{3}-frac{2a^2}{3}-b=0$$

Now you should be able to plug in $a$ and $b$ and show that this equation is correct. Is your question saying that $a=2$ and $b=8$? Or $b=8a^2$? In either case, this inflection point would not be a stationary point in this function.

share|cite|improve this answer

answered Mar 14 at 14:16

Jake OJake O

1665

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The latter one. It looks like there is a mistake in the answers, then. Thank you very much for solving it.
  $endgroup$
  – Marcin
  Mar 14 at 14:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You’re welcome, glad I could help.
  $endgroup$
  – Jake O
  Mar 14 at 14:42
  

  
  
  
  

add a comment | 

0

$begingroup$

To find stationary points we need $f'(x)=0$ and to find inflection points we need to find where $f''(x)=0$.

$$f(x)=x^3-ax^2-bx+c$$
$$f'(x)=3x^2-2ax-b$$
$$f''(x)=6x-2a$$

So to find the inflection point we need $$6x-2a=0$$ $$6x=2a$$ $$x=frac{a}{3}.$$

In order for this point to be stationary we need $f'(frac{a}{3})=0$.

$$3bigg(frac{a}{3}bigg)^2-2abigg(frac{a}{3}bigg)-b=0$$
$$frac{a^2}{3}-frac{2a^2}{3}-b=0$$

Now you should be able to plug in $a$ and $b$ and show that this equation is correct. Is your question saying that $a=2$ and $b=8$? Or $b=8a^2$? In either case, this inflection point would not be a stationary point in this function.

share|cite|improve this answer

answered Mar 14 at 14:16

Jake OJake O

1665

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The latter one. It looks like there is a mistake in the answers, then. Thank you very much for solving it.
  $endgroup$
  – Marcin
  Mar 14 at 14:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You’re welcome, glad I could help.
  $endgroup$
  – Jake O
  Mar 14 at 14:42
  

  
  
  
  

add a comment | 

$begingroup$

To find stationary points we need $f'(x)=0$ and to find inflection points we need to find where $f''(x)=0$.

$$f(x)=x^3-ax^2-bx+c$$
$$f'(x)=3x^2-2ax-b$$
$$f''(x)=6x-2a$$

So to find the inflection point we need $$6x-2a=0$$ $$6x=2a$$ $$x=frac{a}{3}.$$

In order for this point to be stationary we need $f'(frac{a}{3})=0$.

$$3bigg(frac{a}{3}bigg)^2-2abigg(frac{a}{3}bigg)-b=0$$
$$frac{a^2}{3}-frac{2a^2}{3}-b=0$$

Now you should be able to plug in $a$ and $b$ and show that this equation is correct. Is your question saying that $a=2$ and $b=8$? Or $b=8a^2$? In either case, this inflection point would not be a stationary point in this function.

share|cite|improve this answer

answered Mar 14 at 14:16

Jake OJake O

1665

$endgroup$

share|cite|improve this answer

answered Mar 14 at 14:16

Jake OJake O

1665

answered Mar 14 at 14:16

Jake OJake O

1665

answered Mar 14 at 14:16

Jake OJake O

1665