$f(x)$ is a polynomial of the form ($b,c$ are real numbers) $$f(x) = x^2+bx+c$$ such that $|f(x)| \leq 8 \: \forall \: x \in [1,9]$. Find all $f(x)$ satisfying the given condition.
I found that the problem is data specific since there is a unique solution. Indeed, the following identity gives it away:
$$f(1) - 2f(5) + f(9) = 32$$
it is easy to see that $f(1)=-f(5)=f(9)=8$ and this yields two equations in two variables which can be solved for $b, c$.
However, I cannot see what is so special about this data. Is there a geometric interpretation of the problem? I am basically looking for insight into how this problem might have been conceived and the key idea behind it.
$f(1)-2f(5)+f(9) = (f(1)-f(5)) + (f(9)-f(5))$ Geometrically, this is a sum of two distances. We can consider each piece via a taylor expansion. Apply the mean value theorem to conclude $f(1)=f(5)-(2(3)+b)4 = -4b -24 +f(5)$ and $f(9) = f(5) (2(3)+b)4= 4b+56 + f(5)$, because the second derivative is constant. This means that $f(1) + f(9) = 2f(5) +32$.
Hope this helps with the geometric intuition.