(i) g o f

To find: g o f

Formula used: g o f = g(f(x))

Given: (i) f : R → R : f(x) = (x² + 3x + 1)

(ii) g: R → R : g(x) = (2x - 3)

Solution: We have,

g o f = g(f(x)) = g(x² + 3x + 1) = [ 2(x² + 3x + 1) – 3 ]

⇒ 2x² + 6x + 2 – 3

⇒ 2x² + 6x – 1

Ans). g o f (x) = 2x² + 6x – 1

(ii) f o g

To find: f o g

Formula used: f o g = f(g(x))

Given: (i) f : R → R : f(x) = (x² + 3x + 1)

(ii) g: R → R : g(x) = (2x - 3)

Solution: We have,

f o g = f(g(x)) = f(2x - 3) = [ (2x - 3)² + 3(2x – 3) + 1 ]

⇒ 4x² - 12x + 9 + 6x – 9 + 1

⇒ 4x² - 6x + 1

Ans). f o g (x) = 4x² - 6x + 1

(iii) g o g

To find: g o g

Formula used: g o g = g(g(x))

Given: (i) g: R → R : g(x) = (2x - 3)

Solution: We have,

g o g = g(g(x)) = g(2x - 3) = [ 2(2x – 3) - 3 ]

⇒ 4x – 6 - 3

⇒ 4x - 9

Ans). g o g (x) = 4x – 9