Task B-2.4. Let $f(x)$ be a quadratic function such that

$$
x^{2}-2 x+2 \leq f(x) \leq 2 x^{2}-4 x+3
$$

for every $x \in \mathbb{R}$. If $f(11)=181$, what is $f(16)$?
## Solution.

Let's write the given inequality in the following form

$$
(x-1)^{2}+1 \leq f(x) \leq 2(x-1)^{2}+1
$$

For the given inequality to hold for all real numbers $x$, the function $f(x)$ must be of the form

$$
f(x)=a(x-1)^{2}+1, \quad a \in[1,2]
$$

From $f(11)=181$ it follows that

$$
\begin{aligned}
& a(11-1)^{2}+1=181 \\
& a=\frac{9}{5}
\end{aligned}
$$

Then we have

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