Minimum of absolute value of quadratic

Note that $\begin{aligned}f(10)&=100+10b+c\\f(-10)&=100-10b+c\\f(0)&=c\end{aligned}$ Hence, $\begin{aligned}200&=f(10)+f(-10)-2f(0)\\&\leq|f(10)|+|f(-10)|+2|f(0)|\end{aligned}$ Therefore, it is not possible for each of $|f(10)|$ , $|f(-10)|$ and $|f(0)|$ to be less than 50. Hence $N\geq50$ .

On the other hand, by choosing $f(x)=x^2-50$ it is clear that $-50\leq f(x)\leq50$ when $x\in[-10,10]$ . Hence $N\leq50$ .

Therefore $N=50$ , as desired.

It is easy to see that $\max \limits_{x \in [-10, 10]} |f(x)|$ occurs at either $x=-10$ , $x=10$ , or $x=0$ , so $\max \limits_{x \in [-10, 10]} |f(x)| \in \{f(10), f(-10), f(0) \}$ . Note that $f(10)= 100 + 10b + c$ $f(-10)= 100-10b+c$ $f(0)= c$ Then, note that $f(10) + f(-10) - 2f(0)= 200$ From the triangle inequality, we have $f(10) + f(-10) - 2f(0) \leq |f(10)| + |f(-10)| + 2|f(0)|$ $\implies |f(10)| + |f(-10)| + 2|f(0)| \geq 200$ Let $a= \max (|f(10)|, |f(-10)|, |f(0)|$ . Then, $|f(10)| + |f(-10)| + 2|f(0)| \leq 4a$ $\implies 4a \geq 200 \implies a \geq 200$ This shows that $N \geq 50$ . It can be checked that equality holds for $f(x)= x^2 - 50$ . Thus $\boxed{N=50}$ is the smallest possible value of $N$ .

Note that if b is fixed, then the minimum value is obtained when

c = - \frac 12 (min {x \in [-10,10]} |x^2 + bx| + max {x \in [-10,10]} |x^2 + bx|)

= - \frac 12 (max_{x \in [-10,10]} |x^2 + bx|), since |x^2 + bx| = 0 when x = 0

The resulting minimum value will then be

\frac 12 (max_{x \in [-10,10]} |x^2 + bx|)

The problem is therefore equivalent to minimising (max_{x \in [-10,10]} |x^2 + bx|).

Furthermore, |x^2 + bx| is maximum when x = 10 if b \geq 0, or x = -10 if b \leq 0

WLOG let x, b \geq 0

Then

max_{x \in [0, 10]} |x^2 + bx| = |10^2 + 10b|

Since b \geq 0, the minimum possible value is when b = 0.

Then c = - 50

And so the minimum value of max_{x \in [-10,10]} |f(x)| is obtained when f(x) = x^2 - 50, and the value is 50.

A monic polynomial is just a translation of $x^2$ which will have it's smallest range between -10 and 10 when the vertex is on the the x-axis. The max would be 100 but we can move the vertex to (0, -50) and then the max is 50.

We begin by considering $b$ . Since $x$ ranges from -10 to 10, for all $x$ in the range, there is also $-x$ . Thus the equation can be written as

$\displaystyle \max_{x \in [0,10]}{(|x^2+bx+c|, |x^2-bx+c|)}$ .

As $|b|$ increases, either $|x^2+bx+c|$ or $|x^2-bx+c|$ increases. Thus to decrease $\displaystyle \max_{x \in [0,10]}{(|x^2+bx+c|, |x^2-bx+c|)}$ , we must decrease $|b|$ , and hence $b=0$ .

Now we consider $c$ . Since $b=0$ , $f(x)=x^2+c$ is symmetric about the y-axis and since $x$ ranges from -10 to 10, $|f(x)|$ has critical values at $|f(0)|$ and $|f(\pm10)|$ . The original equation can thus be written as $\displaystyle \max{(|100+c|, |c|)}$ . This is minimized at $c=-50$ . Thus $f(x)=x^2-50$ and $\displaystyle \max_{x \in [-10,10]}{|f(x)|} = 50$ .

This polinamial represents a parabola. If the polinomial is desired to have minimum values in interval -10 < x < 10, it is obvious that the bottom of the curve represented by the polinomial must be at x=0 (mid point of the interval).

df/dx = 2x + b

df/dx|(x=0) = 0 --> b = 0

Therefore;

f(x) = x^2+c

Let's assume |f(10)| = |f(0)| = max|f(x)| and c < 0 since |f(x)| have minimum values in given interval.

100+c = -c --> c = -50 --> f(x) = x^2 - 50

Finally, we can get;

max|f(x)| = |f(10)| = |f(0)| = 50

The horizontal shift of $f(x)$ is not relevant for this problem, so we can say that $f_0(x)=x^2+c$ . In the range $[-10,10]$ , $f_0(x)$ has a maximum value at $(±10,f_0(±10))$ . The minimum maximum of the function occurs when the minimum and maximum of $f_0(x)$ have equal absolute values. Thus, $c=-50$ , and the minimum maximum $|N|$ is $\boxed{50}$ .

We can easily prove by logical thinking that for minimum, coefficint of x must be zero for any value of c.now to find c we have to apply a little graphical analysis and we can easily prove that f(0)=-f(10) for minimum. Interesting question sir!

Use symmetry!!

Any other way would simply get one side, f(-10) or f(10), greater than the other.

We know we have an extremum at x = 0.

$\Rightarrow (\frac {d f(x)}{dx})_{x=0} = 0$

$\Rightarrow 2x +b = 0$

$\Rightarrow b = -2x = 0$

Now we have $f(x) = x^2 + c$

$\Rightarrow f(10) = f(-10) = 100+c ; f(0) = c$

Equating absolutes of these, | f(10) | = | f(0) | , we get the minimum as $\boxed{50}$ .