Activity 11

prepared by Sherry Nolan

•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

*    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) *    The third line translates f (x) so that the point of inflection (-b/(3 a), f (-b/(3 a))), is at the origin and calls this new function f1 (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:  Two questions:
      (a) The coordinates of the point of inflection are:

f[-b/(3 a)]

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

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

and

Expand[f1[x]]

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

(b) Students who can produce the above know what they are doing.  Do we really need to make them perform the procedure by hand to be convinced that they understand the mathematical concept?

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 are some examples of the cubic polynomial being symmetric about its point of inflection.
Here is how to plot an output.  First, shut off the message that comes up when a complex solution occurs in the plotting.

Off[Plot :: plnr]

Next, give values to the coefficients so that there is something to plot.

a = 1 ;

b = 2 ;

c = 1 ;

d = 1 ;

Plot[{f[x], f ''[x]}, {x, -5, 5}, PlotStyle -> {{RGBColor[1, 0, 0]}, {RGBColor[0, 1, 0]}}, Epilog -> {Point[{-b/(3 a) , (2 b^3)/(27 a^2) - (b c)/(3 a) + d}]}]

[Graphics:HTMLFiles/a11_23.gif]

To help see the point of inflection I will remove the x-axis.

Plot[{f[x], f ''[x]}, {x, -5, 5}, PlotStyle -> {{RGBColor[1, 0, 0]}, {RGBColor[0, 1, 0]}}, Epilog -> {Point[{-b/(3 a) , (2 b^3)/(27 a^2) - (b c)/(3 a) + d}]}, Axes -> {False, True}]

[Graphics:HTMLFiles/a11_25.gif]

-Graphics -

a = 8 ;

b = 12 ;

c = 5 ;

d = -6 ;

Plot[{f[x], f ''[x]}, {x, -5, 5}, PlotStyle -> {{RGBColor[1, 0, 0]}, {RGBColor[0, 1, 0]}}, Epilog -> {Point[{-b/(3 a) , (2 b^3)/(27 a^2) - (b c)/(3 a) + d}]}]

[Graphics:HTMLFiles/a11_32.gif]

To help see the point of inflection I will remove the x-axis.

Plot[{f[x], f ''[x]}, {x, -5, 5}, PlotStyle -> {{RGBColor[1, 0, 0]}, {RGBColor[0, 1, 0]}}, Epilog -> {Point[{-b/(3 a) , (2 b^3)/(27 a^2) - (b c)/(3 a) + d}]}, Axes -> {False, True}]

[Graphics:HTMLFiles/a11_34.gif]

-Graphics -

Converted by Mathematica  (April 22, 2003)