Point of InflectionDate: 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/ |
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