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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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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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    See the example with the cubic function here: http://en.wikipedia.org/wiki/$V$ieta%27s_formulas#$E$xample2012-10-30
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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 $x_1,x_2,x_3$?