Let f: R → R: f(x) = (2x + 1) and g: R → R: g(x) = (x2 – 2).Wri

1.	Let f: R → R: f(x) = (2x + 1) and g: R → R: g(x) = (x2 – 2).Write down the formulae for(i) (g o f) (ii) (f o g) (iii) (f o f) (iv) (g o g).
Answer» 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

Let f: R → R: f(x) = (2x + 1) and g: R → R: g(x) = (x2 – 2).Write down the formulae for(i) (g o f) (ii) (f o g) (iii) (f o f) (iv) (g o g).

Answer»

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