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Point of Inflection


Date: 05/17/97 at 12:47:07
From: Andrew Reniers
Subject: A Calculus Question

Hi Dr. Math,

I have this question that I have tried and don't understand.  I have 
no clue what to do.  I hope that you can help me.  Here it is:

Suppose that the cubic polynomial h(x) = x^3 - 3bx^2 + 3cx + d has a 
local maximum at A: (x1, y1) and a local minimum at B: (x2, y2).  
Prove that the point of inflection of h is at the midpoint of the line 
segment AB.

I hope you can help me with this problem.  

Thank you,
Andrew Reniers

P.S.: This web site is a great idea!


Date: 05/19/97 at 05:45:37
From: Doctor Anthony
Subject: Re: A Calculus Question

It is easy to show that the x-coordinate of the point of inflection is 
midway between the maximum and minimum turning points.

h'(x) = 3x^2 - 6bx + 3c = 0 for max and min turning points.

          x^2 - 2bx + c = 0

The sum of the roots is 2b,  so we have x1+x2 = 2b

                                    (x1+x2)/2 = b

Also the product of the roots is equal to c, so x1*x2 = c

h''(x) = 6x - 6b = 0 for point of inflection
           x - b = 0   
               x = b at point of inflection = (x1+x2)/2

So the x-coordinate of point of inflection is midway between the 
maximum and minimum turning points.

We must now consider the y-coordinate of the point of inflection.

We have y1 = x1^3 - 3b.x1^2 + 3c.x1 + d
        y2 = x2^3 - 3b.x2^2 + 3c.x2 + d

  (y1+y2)/2 = (x1^3+x2^3)/2 - 3b(x1^2+x2^2)/2 + 3c(x1+x2)/2 + d

Now put b = (x1+x2)/2 and c = x1*x2   

This simplifies to:

(y1+y2)/2 = -x1^3/4 -x2^3/4 + 3x1^2.x2/4 + 3x1.x2^2/4 + d   (1)

Finally put x = (x1+x2)/2 into the equation of the cubic, and simplify 
again with b = (x1+x2)/2  and c = x1*x2 and we will get the same 
expression as (1) above. It follows that the point 
[(x1+x2)/2, (y1+y2)/2] is a point on the curve. We already know that 
the point of inflection is at x = (x1+x2)/2 and so the y-value of the 
point of inflection is (y1+y2)/2.

So the point of inflection is at the midpoint of the line joining 
(x1,y1) and (x2,y2).

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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