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

$(i) ( f + g)oh = foh + goh$

$(ii) ( f . g)oh = (foh) . (goh)$

Solution:

To prove:

$ (f+g) \circ h=f o h+g \circ h \\ $

Consider:

$ \begin{aligned} &((f+g) \circ h)(x) \\\\ &=(f+g)(h(x)) \\\\ &=f(h(x))+g(h(x)) \\\\ &=(f \circ h)(x)+(g \circ h)(x) \\\\ &=\{(f \circ h)+(g \circ h)\}(x) \\\\ &\therefore((f+g) \circ h)(x)=\{(f \circ h)+(g \circ h)\}(x) \quad \forall x \in \mathbf{R} \\ \end{aligned} $

Hence,

$ (f+g) \circ h=f \circ h+g \mathrm{o} h \\ $.

To prove:

$ (f \cdot g) \circ h=(f \circ h) \cdot(g \circ h) \\ $

Consider:

$ \begin{aligned} &((f \cdot g) \mathrm{o} h)(x) \\\\ &=(f \cdot g)(h(x)) \\\\ &=f(h(x)) \cdot g(h(x)) \\\\ &=(f \mathrm{o} h)(x) \cdot(g \mathrm{o} h)(x) \\\\ &=\{(f \mathrm{o} h) \cdot(g \circ h)\}(x) \\\\ &\therefore((f \cdot g) \circ h)(x)=\{(f \circ h) \cdot(g \circ h)\}(x) \forall x \in \mathrm{R} \\ \end{aligned} $

Hence,

$ (f \cdot g)$ o $h=(f \circ h) \cdot(g \circ h) \\ $.