# Second derivative of a third degree polynomial function

Let $f(x) = ax^3 + bx^2 + cx +d$ be a third degree polynomial function.

$$f'' = 6ax + 2b = 0 \Longrightarrow x = \frac{-b}{3a}$$ This is equal to $\frac{1}{3}$ of the sum of the roots of $f(x)$. So my question is: can we say that the root of the second derivative is equal to $\frac{1}{3}$ of the sum (of roots) for all third degree polynomial functions? If not, why?

• Yeah it can be said. You just proved it. May 10, 2018 at 16:09
• Yes it is true. You are correct May 10, 2018 at 16:32

## 3 Answers

Yes.

There is an inflection point (of the cubic polynomial) at the root of its second derivative. This relation essentially tells you how the inflection point shifts as you change the three roots of the original polynomial.

If the zero of $f''(x)$ is $p$. then the polynomial$f(x)$ has root $p$ with multiplicity 3. i.e,$(x-p)^3|f(x)$ and $(x-p)^4$ does not divides$f(x)$

$(x-p)^3|f(x)$ and $f(x)$ is cubic polynomial $\implies$ $f(x)=k(x-p)^3$ , where $k$ is a real number. therefore roots of $f(x)$ are $p,p,p$.........................(1)

Sum of the roots$=\frac{-b}{a} = p+p+p=3p$ $\implies$ $\frac{1}{3}$( Sum of the roots)$=p$...................(2)

from (1) and (2) , $\frac{1}{3}$( Sum of the roots) is also a root of $f(x)$.

this property is not available for all cubic equations.

HINT: If the average of roots of the cubic equation is also a root of the same equation , then their roots are arranged in ascending or descending order, then it's average is middle term.