Finding the value of $a$ which minimizes the absolute maximum of $f(x)$ I know that this is an elementary problem in calculus and so it has a routine way of proof. I faced  it and brought it up here just because it is one of R.A.Silverman's interesting problem. Let me learn your approaches like an student. Thanks.

What value of $a$ minimizes the absolute maximum of the function $$f(x)=|x^2+a|$$ in the interval $[-1,1]$?

 A: Note the symmetry of $x^2+a$ about the $y$-axis. Given the interval, the absolute maximum is $\max\{-a,1+a\}$. (Why?) Minimization, then, will be achieved when $-a=1+a$--that is, when $a=-\frac12$.
A: In my opinion, a calculus student would likely start by looking at the graph of this function for a few values of $a$. Then, they would notice that the absolute maxima occur on the endpoints or at $x=0$. So, looking at $f(-1)=f(1)=|1+a|$ and $f(0)=|a|$, I think it would be fairly easy for a calculus student to see that $a=-\frac{1}{2} \hspace{1 mm} $ minimizes the absolute maximum.
A: Denote $f(x)=x^{2}$ then $f([-1,0])=f(]0,1])$ and so we need to
minimize $f$ over $[0,1]$.
It is clear that $f([0,1])=[0,1]$ and thus $(f+a)([0,1])=[a,a+1]$.
So we can move $[0,1]$ by $a$ and we want to minimize the absolute
value of numbers in $[a,a+1]$.
It is easy to see (by symmetry) that $a=-0.5$ (and there is a formal
way to get to it by examining which numbers in $[a,a+1]$ are positive)
is the desired $a$. 
After we see this choice, it can be easily proved
it is optimal (any change of $a$ to $a+a'$ you will either increase
the interval to include numbers greater then $0.5$ or smaller then
$-0.5$)
A: Let $y=x^2$. We want to minimize the maximum of $|y+a|$ as $y$ ranges over $[0,1]$. This happens at $a=-1/2$. 
A: We have $0\leq x^2\leq1$ for the values given. We want to shift this function so that its minimum negative value is equal to its maximum positive value - cutting this range of values in half. So we have to shift it down by 0.5.
