Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z for all x, y, z ∈ N. Show that ho (gof) = (hog) of.

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Solution

Given, f : N → N, g : N → N and h : N → R
$\Rightarrow$ gof : N → N and hog : N → R
$\Rightarrow$ ho (gof) : N → R and (hog) of : N → R
So, both have the same domains.
$(g o f) (x) = g (f (x)) = g (2 x) = 3 (2 x) + 4 = 6 x + 4 . . . (1) (h o g) (x) = h (g (x)) = h (3 x + 4) = \sin (3 x + 4) . . . (2) Now, (h o (g o f)) (x) = h ((g o f) (x)) = h (6 x + 4) = \sin (6 x + 4) [from (1)] ((h o g) o f) (x) = (h o g) (f (x)) = (h o g) (2 x) = \sin (6 x + 4) [from (2)] So, (h o (g o f)) (x) = ((h o g) o f) (x), \forall x \in N Hence, h o (g o f) = (h o g) o f$

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