Let f: R → R an g: R → R be defined by f(x) = x + 1 and g(x) = x – 1. Show that fog = gof = IR.
Let f: R → R and g: R → R are defined as
f (x) = x + 1 and g (x) = x – 1
Now,
fog(x) = f(g(x)) = f(x – 1) = x – 1 + 1
= x =IR ……(i)
Again,
fog(x ) = f(g(x)) = g(x + 1) = x + 1 – 1
= x = IR ……(ii)
from (i)& (ii)
fog = gof = IR