Why put $x=1,x=-1,x=0$ here?

Question

Problem

Assume $f(x)=ax^2+bx+c$ and $|f(x)|\leq 1(x \in [-1,1])$. Prove $|cx^2-bx+a|\leq 2(x \in [-1,1])$.

Solution

One kind of solution is like below:

Put $x=1,x=-1,x=0$ into the functional expression. We have $$a+b+c=f(1),~~a-b+c=f(-1),~~~c=f(0).$$ Thus, we may solve the equations system to obtain $$a=\frac{f(1)+f(-1)}{2},~~b=\frac{f(1)-f(-1)}{2},~~c=f(0).$$ According to these, we may write $|cx^2-bx+a|$ as a expression with respect to $f(1),f(-1)$ and $f(1)$, and can make further transformation to the inequality.

Question

Why put $x=1,x=-1,x=0$? The three values are choosen randomly or deliberately? Can we choose other three values? For example, $x=1/2,1/3,1$?

Answer 1

You can use any three points you want. You will get a set of three equations in the three unknowns $a,b,c$. Using $\pm 1,0$ makes the equations very simple because $0$ isolates $c$ and $\pm 1$ are easy to square and the opposite signs give the nice $a+b$ and $a-b$.