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Question

# Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f : A → B, g : B → C be defined as f(x) = 2x + 1 and g(x) = x2 − 2. Express (gof)−1 and f−1 og−1 as the sets of ordered pairs and verify that (gof)−1 = f−1 og−1.

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Solution

## $f\left(x\right)=2x+1\phantom{\rule{0ex}{0ex}}⇒f=\left\{\left(1,2\left(1\right)+1\right),\left(2,2\left(2\right)+1\right),\left(3,2\left(3\right)+1\right),\left(4,2\left(4\right)+1\right)\right\}=\left\{\left(1,3\right),\left(2,5\right),\left(3,7\right),\left(4,9\right)\right\}\phantom{\rule{0ex}{0ex}}g\left(x\right)={x}^{2}-2\phantom{\rule{0ex}{0ex}}⇒g=\left\{\left(3,{3}^{2}-2\right),\left(5,{5}^{2}-2\right),\left(7,{7}^{2}-2\right),\left(9,{9}^{2}-2\right)\right\}=\left\{\left(3,7\right),\left(5,23\right),\left(7,47\right),\left(9,79\right)\right\}\phantom{\rule{0ex}{0ex}}\text{Clearly}\mathit{\text{f}}\text{and}\mathit{\text{g}}\text{are bijections and, hence,}{f}^{-1}:B\to A\text{and}{g}^{-1}:C\to B\text{exist.}\phantom{\rule{0ex}{0ex}}\mathrm{So},{f}^{-1}=\left\{\left(3,1\right),\left(5,2\right),\left(7,3\right),\left(9,4\right)\right\}\phantom{\rule{0ex}{0ex}}\mathrm{and}{g}^{-1}=\left\{\left(7,3\right),\left(23,5\right),\left(47,7\right),\left(79,9\right)\right\}\phantom{\rule{0ex}{0ex}}\text{Now,}\left({f}^{-1}o{g}^{-1}\right)\text{:}C\to A\phantom{\rule{0ex}{0ex}}{f}^{-1}o{g}^{-1}=\left\{\left(7,1\right),\left(23,2\right),\left(47,3\right),\left(79,4\right)\right\}...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Also,}f:A\to B\text{and}g:B\to C,\phantom{\rule{0ex}{0ex}}\text{⇒}gof:A\to C,{\left(gof\right)}^{-1}:C\to A\phantom{\rule{0ex}{0ex}}\text{So,}{f}^{-1}o{g}^{-1}\text{and}{\left(gof\right)}^{-1}\text{have same domains.}\phantom{\rule{0ex}{0ex}}\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(2x+1\right)={\left(2x+1\right)}^{2}-2\phantom{\rule{0ex}{0ex}}⇒\left(gof\right)\left(x\right)=4{x}^{2}+4x+1-2\phantom{\rule{0ex}{0ex}}⇒\left(gof\right)\left(x\right)=4{x}^{2}+4x-1\phantom{\rule{0ex}{0ex}}\mathrm{Then},\left(gof\right)\left(1\right)=g\left(f\left(1\right)\right)=4+4-1=7,\phantom{\rule{0ex}{0ex}}\left(gof\right)\left(2\right)=g\left(f\left(2\right)\right)=4+4-1=23,\phantom{\rule{0ex}{0ex}}\left(gof\right)\left(3\right)=g\left(f\left(3\right)\right)=4+4-1=47\mathrm{and}\phantom{\rule{0ex}{0ex}}\left(gof\right)\left(4\right)=g\left(f\left(4\right)\right)=4+4-1=79\phantom{\rule{0ex}{0ex}}\mathrm{So},gof=\left\{\left(1,7\right),\left(2,23\right),\left(3,47\right),\left(4,79\right)\right\}\phantom{\rule{0ex}{0ex}}⇒{\left(gof\right)}^{-1}=\left\{\left(7,1\right),\left(23,2\right),\left(47,3\right),\left(79,4\right)\right\}...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{From}\left(1\right)\text{and}\left(2\right)\text{, we get:}\phantom{\rule{0ex}{0ex}}\text{}{\left(gof\right)}^{-1}={f}^{-1}o{g}^{-1}\phantom{\rule{0ex}{0ex}}$

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