Q. 25

# Functions f, g : R → R are defined, respectively, by f (x) = x^{2} + 3x + 1, g (x) = 2x – 3, find

(i) f o g (ii) g o f (iii) f o f (iv) g o g

Answer :

Given that, f (x) = x^{2} + 3x + 1, g (x) = 2x – 3

(i) f o g

fog = f(g(x)) = f(2x-3)

= (2x-3)^{2} + 3(2x-3) + 1

= (4x^{2}-12x+9) + 6x – 9 +1

= 4x^{2} - 6x + 1

∴ fog = 4x^{2} - 6x + 1

(ii) g o f

gof = g(f(x)) = g(x^{2} + 3x + 1)

= 2(x^{2} + 3x + 1) – 3

= 2x^{2} + 6x + 2 – 3

= 2x^{2} + 6x – 1

∴ gof = 2x^{2} + 6x – 1

(iii) f o f

fof = f(f(x)) = f(x^{2} + 3x + 1)

= (x^{2} + 3x + 1)^{2} + 3(x^{2} + 3x + 1) + 1

= x^{4}+9x^{2}+1+6x^{3}+6x+2x^{2}+3x^{2}+9x+3+1

= x^{4}+6x^{3}+14x^{2}+15x+5

∴ fof = x^{4}+6x^{3}+14x^{2}+15x+5

(iv) g o g

gog = g(g(x)) = g(2x-3)

= 2(2x-3) – 3

= 4x-6-3

= 4x-9

∴ gog = 4x-9

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