Question

If $g:N\to N$ is given by $g\left(n\right)=2n+3$, and $f:N\to N$ is given by $f\left(n\right)=n+1$, then find $g\circ f$, $f\circ g$ and $g\circ g$

Answer 1

Given : $g\left(n\right)=2n+3$ and $f\left(n\right)=n+1$

$gof=g\left(f\right(n\left)\right)$

$=g(n+1)$

$=2(n+1)+3$

$=2n+2+3$

$\therefore$ $\left(gof\right)\left(n\right)=2n+5$

$\left(fog\right)\left(n\right)=f\left(g\right(n\left)\right)$

$=f(2n+3)$

$=2n+3+1$

$=2n+4$

$\therefore$ $\left(fog\right)\left(n\right)=2n+4$

$\left(gog\right)\left(n\right)=g\left(g\right(n\left)\right)$

$=g(2n+3)$

$=2(2n+3)+3$

$=4n+6+3=4n+9$

$\therefore$ $\left(gog\right)\left(n\right)=4n+9$