It is given that
f(x) = (2x + 1) and g(x) = (x2 – 2)
(i) g o f = g o f (x) = g{f(x)} = g {2x + 1}
So we get
= (2x + 1)2 – 2
It can be written as
= 4x2 + 4x + 1 – 2
On further calculation
= 4x2 + 4x – 1
(ii) f o g = f o g (x) = f {g(x)} = f{x2 – 2}
So we get
= 2 {x2 – 2} + 1
It can be written as
= 2x2 – 4 + 1
On further calculation
= (2x2 – 3)
(iii) f o f = f o f (x) = f{f(x)} = f(2x + 1)
So we get
= 2 (2x + 1) + 1
It can be written as
= 4x + 2 + 1
On further calculation
= 4x + 3
(iv) g o g = g o g (x) = g{g(x)} = g { x2 – 2}
So we get
= (x2 – 2)2 – 2
On further calculation
= x4 – 4x2 + 4 – 2
We get
= x4 – 4x2 + 2