Activity 11

prepared by Sherry Nolan, UMass Lowell

•Mathematica Versions of the code

The object of this activity is to prove that the graph of every cubic polynomial is symmetric about its point of inflection.
This is the first set of equations:

f[x_] := a x^3 + b x^2 + c x + d ;

Solve[f ''[x] == 0, x]

{{x -> -b/(3 a)}}

f1[x_] := f[x + -b/(3 a)] - f[-b/(3 a)] ;

SameQ[Simplify[f1[x]], Simplify[-f1[-x]]]

True

•Comments

*    The first line defines the general cubic and names it f (x)

*    The second line finds the x - coordinate of the point of inflection :    the zero of the second derivative of f (x)

RowBox[{*,   , The, , third, , line, , translates, , Cell[TextData[{f, (x)}]], ... nd, , calls, , this, , new, , function,   , RowBox[{Cell[TextData[{f, 1(x)}]], .}]}]

*    The fourth line shows that the equation f1 (x) = -f1 (-x) is an identity (always true) .

We conclude that the translated function is an odd function and symmetric to the origin.  There fore the original f(x) is symmetric to its point of inflection. QED.

Note:
The coordinates of the point of inflection are:

{-b/(3 a), f[-b/(3 a)]}

{-b/(3 a), (2 b^3)/(27 a^2) - (c b)/(3 a) + d}

and f1(x) is

Factor[f1[x]]

(x (-b^2 + 3 a^2 x^2 + 3 a c))/(3 a)

This is the second set of equations which represent an alternaive solution to the posted question.  They are a little more complicated by perhaps more elegant:

f2[x_] := f[-b/(3 a)] - f[-b/(3 a) - k] ;

f3[x_] := f[-b/(3 a) + k] - f[-b/(3 a)] ;

SameQ[Simplify[f2[x]] === Simplify[f3[x]]]

True

*  We would use the same first two lines to define f(x) and take the second derivative.
*  The third line compares the y-coordinates of points k units to the right and left of the inflection point.   It shows that these points are always the same distance below and above the inflection point, thus establishing the symmetry.

•Comments/Extensions

Here is a graphical display of the symmetry about the point of inflection.  You can change the values of the coefficients as you like, but you may need to adjust the plotting range.

a = 1 ;

b = 2 ;

c = 1 ;

d = 5 ;

Plot[{f[x], f ''[x]}, {x, -5, 5}, PlotStyle -> {{RGBColor[1, 0, 0]}, {RGBColor[0, 1, 0]}}, Epilog -> {PointSize[0.02], Point[{-b/(3 a) , f[-b/(3 a)]}]}]

[Graphics:HTMLFiles/a11_25.gif]

-Graphics -

Converted by Mathematica  (May 14, 2003)