Let $f, g, h$ be three functions from $R \rightarrow R$ as: $f(x)=2 x^{3}+5, g(x)=\cos x, h(x)=x^{3}-1 .$ Find $h o(g \circ f)$ and $(h o g) o f$. Are they equal?

Solution:

$ \begin{array}{ll} F: R \rightarrow R & f(x)=2 x^{3}+5 \\ g: R \rightarrow R & g(x)=\cos x \\ h \in R \rightarrow R & h(x)=x^{3}-1 \end{array} $

$ \begin{aligned} h o(g o f) &=h(g(f(x))) \\ &=h\left(g\left(2 x^{3}+5\right)\right) \\ &=h\left(\cos \left(2 x^{3}+5\right)\right) \\ &=\left(\cos \left(2 x^{3}+5\right)\right)^{3}-1 \\ (h o g) o f &=(h g(x)) \text { of } \\ &=((h g(x)) f(x)) \\ &=\left(h g(x)\left(2 x^{3}+5\right)\right) \\ &=h\left[\cos \left(2 x^{3}+5\right)\right] \\ &=\left(\cos \left(2 x^{3}+5\right)\right)^{3}-1 \end{aligned} $

So,

$ L .H .S=R. H. S $

ho(gof) = (hog)of