Let $f, g$ and $h$ be functions from $R$ to $R$ . Show that
(i) $(f + g) o h = f o h + g o h$
(ii) $(f . g) o h = (f o h) . (g o h)$

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Solution

(i)
To prove:
$(f + g) o h = f o h + g o h$
Consider:
$((f + g) o h) (x)$
$= (f + g) (h (x))$
$= f (h (x)) + g (h (x))$
$= (f o h) (x) + (g o h) (x)$
$= (f o h) + (g o h) (x)$
$∴ ((f + g) o h) (x) = (f o h) + (g o h) (x)$ for all $x \in R$
Hence, $(f + g) o h = f o h + g o h$

(ii)
To prove: $(f . g) o h = (f o h) . (g o h)$
Consider:
$((f . g) o h) (x)$
$= (f . g) (h (x))$
$= f (h (x)) . g (h (x))$
$= (f o h) (x) . (g o h) (x)$
$= (f o h) . (g o h) (x)$
$∴ ((f . g) o h) (x) = (f o h) . (g o h) (x)$ for all $x \in R$
Hence, $(f . g) o h = (f o h) . (g o h)$

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