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.
f : N → N defined by f(x) = 2 x ∀ n ∈ N
g : N → N defined by g(y) = 3 y + 4 ∀ y ∈ N
h : N → R defined by h(z) = sin z ∀ z ∈ N.
Now (h o (g o f))(x) = (h o (g o f) (x))
= h(g(f(x)) = h(g(2 x)) = h (3(2 x) + 4) = h(6 x + 4) = sin (6 x + 4) ∀ x ∈ N.
A Iso, ((h o g) o f) (x) = ((h o g) (f (x)) = (h o g) (2 x) = h(g (2 x))
= h(3(2 x) + 4) = h(6 x + 4)
= sin (6 x + 4), ∀ x ∈ N.
∴ 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)
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