1. ## Interval

Hello.

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

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, $x \in ]-\infty, -a-2] \cup [ a-2, \infty[$

or

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

I am not sure about it.

Rapha

2. Remember g(x-2) is a translation to the right by 2 so

g(x-2) = 0, ??

is correct. For eg if x= -a+2 g(-a+2-2) = g(-a) = 0

if x = a+2 g(a+2-2) = g(a) = 0 as it should

On the other hand Consider
g(x-2) = 0,

Then if x = a-2 you would have g(a-4)= 0 which would not be true for
a = 4

3. Alright, thank you, mate