Short Answer TypeConsider f : N → N, g : N → N and h : N → R defined as f (x) = 2 x, g(y) = 3 y + 4 and h(z) = sin z ∀ x, y and z in N. Show that h o (g o f) = (h o g) o f.
Let f, g and h be function from R to R. Show that (f + g) o h = f o h + g o h
(f . g) o h = (f o h) . (g o h)
i) Since f, g and h are functions from R to R.
∴ (f + g) o h : R ∴ R and f o h + g o h : R ∴ R Now, ((f + g) oh) (x) = ((f + g) h(x))
⇒ ((f + g) o h) (x) = f (h (x)) + g(h(x))
⇒ ((f + g) o h) (x) = (f o h) (x) + (g o h) (x) for all x ∈ R ∴ ( f + g) o h = f o h + g o h (ii) Again (f . g) o h : R → R and (f o h) . (g o h) : R → R Now {(f . g) o h } (x) = ( f . g) (h(x))
⇒ { (f . g) o h } (x) = f(h(x)) . g(h(x))
⇒ {(f . g) o h }(x) = (f o h) (x) . (g o h) (x)
⇒ { (f . g) o h }(x) = {(f o h) . (g o h)} (x) for all x ∈ R ∴ (f . g) o h = (f o h) . (g o h).
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