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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.

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Solution

Given, f : N → N, g : N → N and h : N → R $⇒$gof : N → N and hog : N → R $⇒$ho (gof) : N → R and (hog) of : N → R So, both have the same domains. $\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(2x\right)=3\left(2x\right)+4=6x+4...\left(1\right)\phantom{\rule{0ex}{0ex}}\left(hog\right)\left(x\right)=h\left(g\left(x\right)\right)=h\left(3x+4\right)=\mathrm{sin}\left(3x+4\right)...\left(2\right)\phantom{\rule{0ex}{0ex}}\text{Now,}\phantom{\rule{0ex}{0ex}}\left(ho\left(gof\right)\right)\left(x\right)=h\left(\left(gof\right)\left(x\right)\right)=h\left(6x+4\right)=\mathrm{sin}\left(6x+4\right)\left[\text{from}\left(1\right)\text{]}\phantom{\rule{0ex}{0ex}}\left(\left(hog\right)of\right)\left(x\right)=\left(hog\right)\left(f\left(x\right)\right)=\left(hog\right)\left(2x\right)=\mathrm{sin}\left(6x+4\right)\left[\text{from}\left(2\right)\text{]}\phantom{\rule{0ex}{0ex}}\text{So,}\left(ho\left(gof\right)\right)\left(x\right)=\left(\left(hog\right)of\right)\left(x\right),\forall x\in N\phantom{\rule{0ex}{0ex}}\text{Hence,}ho\left(gof\right)=\left(hog\right)of\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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