How does g(x)= 4/8^x change over a unit interval in x?

Answer 1

(4/8)^x

x

0 1

1 1/2

2 16/64=1/4

3 64/512=8/256=4/128=2/64=1/32

etc

Don't forget the negatives.

Answer 2

g(x) = 4 / 8^x and g(x+1) = 4 / 8^(x+1) = 4 / [8(8^x)] = (1/8) [4 / 8^x]

So, g(x+1) is 1/8 of g(x).