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 Cubic Quandary (Posted on 2014-09-13)
Prove that graphs of all cubic functions are rotationally symmetrical about their point of inflection, or provide a counter-example.

 No Solution Yet Submitted by Dustin No Rating

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 Solution | Comment 1 of 2
I was hoping to find a solution that avoids calculus.  This solution, however, makes use of derivatives to find the point of inflection.

f(x) = ax³+ bx² + cx + d
f"(x)= 6ax + 2b
0 = 6ax + 2b
x = -b/(3a)
The coordinates of the point of inflection are
(-b/(3a) , 2b³/(27a²) - bc/(3a) + d)
If we translate f(x) so this point is on the origin we get
g(x) = f(x - b/(3a)) - (2b³/(27a²) - bc/(3a) + d)

simplifying this we get the odd cubic
g(x) = ax³ + (c - b²/3a)x
which is clearly origin symmetric.
So the original f(x) is symmetric about its point of inflection.

 Posted by Jer on 2014-09-15 14:02:41

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