Quadratic Inequality

Note that $f(x) - 1$ is a polynomial that satisfies

$(x - 1)^2 \le f(x)-1 \le 2(x - 1)^2$

The above inequality shows that only zero of $f(x)-1$ is at $x = 1$ ; and furthermore, this zero has multiplicity $2$ .

Thus, $f(x)-1 = a(x-1)^2$ where $a>0$ .

$=> f(x)=a(x-1)^2+1$ ,

From given data:

$\begin{aligned} f(11)&=181\\ 100a+1&=181\\ a&=\frac{9}{5} \end{aligned}$

Thus

$f(16)=\frac{9}{5}(16-1)^2+1=406$