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I want to show that the inflection point of a degree 3 polynomial $f(x)$ is at $$ x = \frac{x_1 + x_2 + x_3}{3}, $$ if $f(x_1) = f(x_2) = f(x_3) = 0$. I was trying to show that the sign of $f''(x)$ changes at $x$, but how can I show this for an arbitrary polynomial?

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up vote 2 down vote accepted

Let $f(x)=ax^3+bx^2+cx+d$. That means the sign of $f''(x)$ changes when $6ax+2b=0$, so when $x=\frac{-b}{3a}$.

Since the sum of the roots of $f$ equals $\frac{-b}{a}$, we have that $x=\frac{-b}{3a}=\frac{x_1+x_2+x_3}{3}$.

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why is the sum $x_1+x_2+x_3$ equal to $-b/a$? – somebody Oct 30 '12 at 11:57
@somebody Because $$\begin{eqnarray*} f(x) &=&ax^{3}+bx^{2}+cx+d \\ &=&a(x-x_{1})(x-x_{2})(x-x_{3}) \\ &=&ax^{3}-a\left( x_{1}+x_{2}+x_{3}\right) x^{2}+a(x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})x-ax_{1}x_{2}x_{3}\text{,} \\ &\Rightarrow &-a\left( x_{1}+x_{2}+x_{3}\right) =b \end{eqnarray*}$$ – Américo Tavares Oct 30 '12 at 12:02
See the example with the cubic function here: – barto Oct 30 '12 at 12:04

You know that $f(x)=ax^3+bx^2+cx+d$, so $f''(x)=6ax+2b=0$ implies that $x=$ $___$. What do you know about the relationship between the coefficients $a,b,c,d$ and the roots $x_1,x_2,x_3$?

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