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Question
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
⇒ gof : N → N and hog : N → R
⇒ ho (gof) : N → R and (hog) of : N → R
So, both have the same domains.
(gof) (x) = g (f (x)) = g (2x) = 3 (2x)+4 = 6x+4 ...(1)
(hog) (x) = h(g (x)) = h (3x+4) = sin (3x+4) ... (2)
Now,
( h o (gof)) (x) = h ((gof) (x)) = h (6x+4) = sin (6x+4) [from (1)
((hog) o f) (x) = (hog) (f (x)) = (hog) (2x) = sin (6x+4) [from (2)
So, ( h o (gof)) (x )= ((hog) o f) (x), ∀x ∈ N
Hence, h o (gof) = (hog) o f
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