1.

Consider the functions f(x), g(x), h(x) as given below. Show that (fog)oh = fo(goh) in each case. (i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x2 (ii) f(x) = x2, g(x) = 2x and h(x) = x + 4 (iii) f(x) = x – 4, g(x) = x2 and h(x) = 3x – 5

Answer»

(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x2 

f(x) = x – 1 

g(x) = 3x + 1 

f(x) = x2 

(fog)oh = fo(goh) 

LHS = (fog)oh 

fog = f(g(x)) = f(3x + 1) = 3x + 1 – 1 = 3x 

(fog)oh = (fog)(h(x)) = (fog)(x2) = 32 … (1) 

RHS = fo(goh) 

goh = g(h(x)) = g(x2) = 3x2 + 1

fo(goh) = f(3x2 + 1) = 3x2 + 1 – 1= 3x2 … (2) 

LHS = RHS 

Hence it is verified.

(ii) f(x) = x2, g(x) = 2x, h(x) = x + 4 

(fog)oh = fo(goh) 

LHS = (fog)oh 

fog = f(g(x)) = f(2x) = (2x)2 = 4x2

(fog)oh = (fog) h(x) = (fog) (x + 4) 

= 4(x + 4)2 = 4(x2 + 8x + 16) 

= 4x2 + 32x + 64 … (1) 

RHS = fo(goh) goh = g(h(x)) = g(x + 4) 

= 2(x + 4) = (2x + 8) 

fo(goh) = f(goh) = f(2x + 8) = (2x + 8)2 

= 4x2 + 32x + 64 … (2) 

(1) = (2) 

LHS = RHS 

∴ (fog)oh = fo(goh) 

It is proved.

(iii) f(x) = x – 4, g(x) = x2, h(x) = 3x – 5 

(fog)oh = fo(goh)

LHS = (fog)oh 

fog = f(g(x)) = f(x2) = x2 – 4 

(fog)oh = (fog)(3x – 5) = (3x – 5)2 – 4 

= 9x2 – 30x + 25 - 4 

= 9x2 – 30x + 21 … (1) 

∴ RHS = fo(goh) 

(goh) = g(h(x)) = g(3x – 5) = (3x – 5)2 

= 9x2 – 30x + 25 

fo(goh) = f(9x2 – 30 x + 25) 

= 9x2 – 30x + 25 – 4 

= 9x2 – 30x + 21 … (2) 

(1) = (2) 

LHS = RHS 

∴ (fog)oh = fo(goh) 

It is proved.


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