It is given that
f(x) = (x2 + 3x + 1) and g(x) = (2x – 3)
(i) g o f = g o f (x) = g{f(x)} = g(x2 + 3x + 1)
So we get
= 2 (x2 + 3x + 1) – 3
It can be written as
= 2x2 + 6x + 2 – 3
On further calculation
= 2x2 + 6x – 1
(ii) f o g = f o g (x) = f{g(x)} = f {2x – 3}
So we get
= (2x – 3)2 + 3 (2x – 3) + 1
It can be written as
= 4x2 – 12x + 9 + 6x – 9 + 1
On further calculation
= 4x2 – 6x + 1
(iii) g o g = g o g (x) = g {g(x)} = g{2x – 3}
So we get
= 2 (2x – 3) – 3
It can be written as
= 4x – 6 – 3
On further calculation
= 4x – 9