Q. 103.8( 5 Votes )

# Let f : N → N : f(x) = 2x, g : N → N : g(y) = 3y + 4 and h : N → N : h(z) = sin z. Show that h o (g o f ) = (h o g) o f.

Answer :

To show: h o (g o f ) = (h o g) o f

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

(ii) g o f = g(f(x))

Given: (i) f : N → N : f(x) = 2x

(ii) g : N → N : g(y) = 3y + 4

(iii) h : N → N : h(z) = sin z

Solution: We have,

LHS = h o (g o f )

⇒ h o (g(f(x))

⇒ h(g(2x))

⇒ h(3(2x) + 4)

⇒ h(6x +4)

⇒ sin(6x + 4)

RHS = (h o g) o f

⇒ (h(g(x))) o f

⇒ (h(3x + 4)) o f

⇒ sin(3x+4) o f

Now let sin(3x+4) be a function u

RHS = u o f

⇒ u(f(x))

⇒ u(2x)

⇒ sin(3(2x) + 4)

⇒ sin(6x + 4) = LHS

Hence Proved.

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