Question

Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z for all x, y, z ∈ N. Show that ho (gof) = (hog) of.

Answer 1

Given, f : N → N, g : N → N and h : N → R
$\Rightarrow$gof : N → N and hog : N → R
$\Rightarrow$ho (gof) : N → R and (hog) of : N → R
So, both have the same domains.
$$\begin{gathered} \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(2x\right)=3\left(2x\right)+4=6x+4...\left(1\right) \\ \left(hog\right)\left(x\right)=h\left(g\left(x\right)\right)=h\left(3x+4\right)=\sin \left(3x+4\right)...\left(2\right) \\ \text{Now,} \\ \left(ho\left(gof\right)\right)\left(x\right)=h\left(\left(gof\right)\left(x\right)\right)=h\left(6x+4\right)=\sin \left(6x+4\right)[\text{from}\left(1\right)\text{]} \\ \left(\left(hog\right)of\right)\left(x\right)=\left(hog\right)\left(f\left(x\right)\right)=\left(hog\right)\left(2x\right)=\sin \left(6x+4\right)[\text{from}\left(2\right)\text{]} \\ \text{So,}\left(ho\left(gof\right)\right)\left(x\right)=\left(\left(hog\right)of\right)\left(x\right),\forall x\in N \\ \text{Hence,}ho\left(gof\right)=\left(hog\right)of \end{gathered}$$