Question

Consider all monic polynomials f(x)=x^2+bx+c, where b and c are real numbers. What is the minimum value of N, where

N=max x∈[−10,10]|f(x)|?

Answer 1

$$f(x) = x^2 + bx + c$$ $$N = \max |f(x)|, \, x \in [-10, 10]$$

The maximum $|f(x)|$ is reached at boundary points $-10, 10$ or at inner local extrema point.

Let's consider $N$ as function of $b, c$: $N(b, c)$. Since the interval $[-10; 10]$ is symmetric about 0 and $x^2 + bx + c = x(x + b) + c$ N is even function of b:

$N(b, c) = N(-b, c)$. Since N reaches minimum the minima point of b is $b = 0$. So

$$f(x) = x^2 + c.$$ $$f(-10) = f(10) = 100 + c; \, f(0) = c$$

Thus

$$N = \min_c \max_x |f(x, c)|$$

Minimum of N is reached if

$$|100 + c| = |c|$$

So

$$c = -50$$ $$N = 50$$