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may I know how to find the point of symmetry of the function $y=2x^3-bx^2+cx$. I have checked some resources, it said that the point of symmetry is $(\frac{b}{6},f(\frac{b}{6})$, due to the fact that $\frac{b}{6}=\frac{p+q}{2}$, where p is the x-coordinate for maximum point, and q is the x-coordinate for minimum point. May I know why does it work?

Thank you very much for you help.

Henry Cai
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1 Answers1

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A cubic has a point of symmetry at the inflection.

Taking the second derivative,

$$12x-2b=0,$$

so $\left(\dfrac b6,-\dfrac{b^3}{54}+\dfrac{bc}6\right)$.

  • Thank you very much, could you explain more on why the cubic equation has a point of symmetry at the inflection? Sorry for any inconvenience caused – Henry Cai Dec 19 '19 at 13:06
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    @HenryCai: Consider Taylor's formula, which is a exact formula for polynomials:$f(x)-f(x_0)=f'(x_0)(x-x_0)+\dfrac{f''(x_0) }2 (x-x_0)^2+\dfrac{f'''(x_0)}6 (x-x_0)^3$. If $f''(x_0)=0$, $f(x)-f(x_0)$ is an odd function of $x-x_0$. – Bernard Dec 19 '19 at 13:47
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    @HenryCai: the first derivative is a parabola, which is symmetric. Can you tell where ? –  Dec 19 '19 at 14:13
  • @YvesDaoust: I could see that the symmetric point for the quadratic equation is $\frac{p+q}{2}$, which is the same point as the point of inflection. So, this implies that the symmetric point for first derivative is the symmetric point for the original equation? – Henry Cai Dec 19 '19 at 14:21
  • @HenryCai: symmetry cannot disappear. –  Dec 19 '19 at 14:22
  • @Bernard: Thank you very much, I can see that if it is an odd function, the point of inflection would be the symmetric point; however, I don't quite understand why $f(x)-f(x_0) =f'(x_0)(x-x_0) $ would be an odd function. Would you mind to explain this for me? – Henry Cai Dec 19 '19 at 14:27
  • @YvesDaoust: Thank you very much. Does this also work for any other functions? for example, if I want to find a symmetric point of a polynomial with degree of 4. So, can I use the second derivative, which is a quadratic polynomial, to find the symmetric point?. Thank you very much for your help. – Henry Cai Dec 19 '19 at 14:28
  • @HenryCai: It an odd function $\color{red}{\text{of }:x-x_0}$, as are all monomials of odd degree. – Bernard Dec 19 '19 at 15:02
  • @bernard: Got it, thank you very much, Sir. – Henry Cai Dec 19 '19 at 15:11