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.)
Roots of $f_n(x) = 1 + (1-x)^2 - (x+3)(1-x)^{n+1}$ in the interval $[0,1]$
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1@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 $n$. – 2011-10-23