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Show that a cubic function (a third-degree polynomial) always has exactly one point of inflection. If its graph has three $ x $-intercepts $ x_1 $, $ x_2 $, and $ x_3 $, show that the $ x $-coordinate of the inflection point is $ (x_1 + x_2 + x_3) /3 $.

Let the cubic function be $f(x)=a x^{3}+b x^{2}+c x+d \Rightarrow f^{\prime}(x)=3 a x^{2}+2 b x+c \Rightarrow f^{\prime \prime}(x)=6 a x+2 b$

So $f$ is $\mathrm{CU}$ when $6 a x+2 b>0 \Leftrightarrow x>-b /(3 a), \mathrm{CD}$ when $x<-b /(3 a),$ and so the only point of inflection occurs

when $x=-b /(3 a) .$ If the graph has three $x$ -intercepts $x_{1}, x_{2}$ and $x_{3}$, then the expression for $f(x)$ must factor as $f(x)=a\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right) .$ Multiplying these factors together gives us

\[

f(x)=a\left[x^{3}-\left(x_{1}+x_{2}+x_{3}\right) x^{2}+\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right) x-x_{1} x_{2} x_{3}\right]

\]

Equating the coefficients of the $x^{2}$ -terms for the two forms of $f$ gives us $b=-a\left(x_{1}+x_{2}+x_{3}\right) .$ Hence, the $x$ -coordinate of

the point of inflection is $-\frac{b}{3 a}=-\frac{-a\left(x_{1}+x_{2}+x_{3}\right)}{3 a}=\frac{x_{1}+x_{2}+x_{3}}{3}$

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so we have to show that a cubic function of the degree point has exactly one point of inflection. And if it's grant has to be. Accidents have Expo next to an ex tree so that the extraordinary inflection point X one protects two four, six three, divided by three. So remember that the order to find the inflection point we have to take the second derivative and the second derivative a cubic function is a linear function, because to kill, it went against return to some form of to some form of a linear function that have a one degree power. So we know that there will be exactly one point of infection could That's what it means to be linear function. And so, with that in mind, we can go ahead and write a general function for the cubic function tro con's say some constant K times x minus x one times X minus X two and times X Mine is extra and the reason I'm ready like this because we're told that there are three accidents. Havoc wanted to an extreme. So these are the three in the steppe. Now I take the derivatives that this is going to be a lot of product looked. So this is going to come out to be a K Time's tags. My sex two times x minus X three clothes K times X minus X one times X on his X tree and then again, plus que and then x minus x one on X minus two and then you can simplify this further. And this work, you can rewrite that prime of access. Uh, what this what this is doing is foiling it out. Oh, I pulled out of contractors. Cases would be to explore money too. Times x one plus X two plus x three and this is our multiplied by axe and in Class X one x two class act two three on the plus X three x one. Now we take the second derivative and this will give us six X K minus two. Okay, times x one project too three and the inflection point chords on the second rivers. He called zero. So when he saw for ex actually you will get accident for the Act one Protect too those extra three all over three and that will prove well, we were looking for