Hello.

i know g(0) = 1 and g(x) = 0 , iff $\displaystyle x \in ]-\infty, -a] \cup [ a, \infty[$

What do we know about g(x-2) for example?

It is g(2-2) = 0, but what about g=0?

Is it g(x-2) = 0, $\displaystyle x \in ]-\infty, -a-2] \cup [ a-2, \infty[ $

or

g(x-2) = 0, $\displaystyle x \in ]-\infty, -a+2] \cup [ a+2, \infty[ $ ??

I am not sure about it.

Rapha