Question

Let f, g and h be functions from R to R. Show that

(f + g) oh = foh + goh
(f . g) oh = (foh) . (goh)

Answer 1

(i) (f + g)oh = foh + goh
$\textbf{Let us consider ((f + g)oh)(x) = (f + g)(h(x))}$
$= f(h(x)) + g(h(x))\\= (foh)(x) + (goh)(x)\\= \{(fog) + (goh)\}(x)$
Then, ((f + g)oh)(x) = {(foh) +(goh)}(x) ∀ x ϵ R
$\textbf{Therefore, (f + g)oh = foh + goh.}$

(ii) (f.g)oh = (foh).(goh)
$\textbf{Let us consider ((f.g)oh)(x) = (f.g)(h(x))}$
$= f(h(x)).g(h(x))\\= f (h(x)).g(h(x))\\= (fog)(x).(goh)(x)\\= \{(fog).(goh)\}(x)$
Then, ((f.g)oh)(x) = {(fog).(goh)}(x) ∀ x ϵ R
$\textbf{Therefore, (f.g)oh = (fog).(goh)}$