polynomial puzzle ;)

$x^{ 2 }-2x+2≤f(x)≤2x^{ 2 }-4x+3\\ \Rightarrow \left( x-1 \right) ^{ 2 }+1≤f(x)≤2\left( x-1 \right) ^{ 2 }+1\\ \text{both }\left( x-1 \right) ^{ 2 }+1\text{ and } 2\left( x-1 \right) ^{ 2 }+1\text{ have vertex at } \left( 1,1 \right) \\ \text{ so f(x) have vertex on } \left( 1,1 \right) \\ \Rightarrow f(x)=a\left( x-1 \right) ^{ 2 }+1\\ f(11)=a(100)+1=181\Rightarrow a=\frac { 18 }{ 10 } \\ \Rightarrow f(x)=\frac { 18 }{ 10 } \left( x-1 \right) ^{ 2 }+1\\ f(21)=\frac { 18 }{ 10 } \left( 400 \right) +1=721$

Let $f(x)$ be $ax^2 +bx + c$ . Because at $x=1$ $x^2 - 2x + 2 = 2x^2 - 4x + 3 = 1$ , $f(x)$ also has to be equal to $1$ at $x=1$ . We now have that $a + b + c = 1$ . Also, at $x=1$ , the derivative of $x^2 -2x + 2$ and $2x^2 - 4x + 3$ are both $0$ , and thus the derivative of $f(x)$ has to be $0$ at $x=1$ . Now we also know that $f'(1) = 2a + b = 0$ . Lastly, by using the given fact that $f(11) = 181$ , we can derive that $121a + 11b + c = 181$ . By solving this system of equations we obtained - $a + b + c = 1$ - $2a + b = 0$ - $121a + 11b + c = 181$ we get that $f(x) =\frac{18}{10}x^2 - \frac{36}{10}x + \frac{28}{10}$ , and thus $f(21) = 721$