# Roots of $f_n(x) = 1 + (1-x)^2 - (x+3)(1-x)^{n+1}$ in the interval $[0,1]$

Does the polynomial $f_n(x) = 1 + (1-x)^2 - (x+3)(1-x)^{n+1}$ have exactly one root in the interval $[0,1]$ for all non-negative integers $n$? It has at least one root because $f_n(0) = -1$ and $f_n(1) = 1$, but I am not sure if the root is unique. (The plots for different values of $n$ suggest so.)

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@SL2:The derivative is not always positive. The derivative is $[ (n+2)x + 3n + 2 ](1-x)^n - 2(1-x)$ and its sign is determined by whether $a_n(x) = [(n+2)x + 3n + 2 ](1-x)^{n-1}$ is greater than or less than $2$. As $n \to \infty$, $a_n(x) \to 0$, for $x \in [0,1]$. Thus, the derivative, $a_n(x) - 2$, is negative for large $n$. – Aditya Oct 23 '11 at 4:14

I can see for some values of n that a real root exists, but the root is not the same for all values of n as the plots below suggest.

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Yes, it is easy to see that the roots form a decreasing sequence. $f_{n+1}(x) - f_n(x) > 0$. If $a_n$ is a root of $f_n$ then, $f_{n+1}(a_n) > 0$. Thus, there is a root of $f_{n+1}$ between $0$ (where $f_{n+1}$ is negative) and $a_n$ (where $f_{n+1}$ is positive). – Aditya Oct 23 '11 at 6:12