Question

Let $f=\{(3,1),(9,3),(12,4)\}$ and $g=\{(1,3),(3,3),(4,9),(5,9)\}$. Show that $g\circ f$ and $f\circ g$ are both defined. Also, find $f\circ g$ and $g\circ f$.

Answer 1

$f\circ g$ means $g\left(x\right)$ function is in $f\left(x\right)$ function.

$g\circ f$ means $f\left(x\right)$ function is in $g\left(x\right)$ function.

$f=\left\{\right(3,1),(9,3),(12,4\left)\right\}$ and $g=\left\{\right(1,3),(3,3),(4,9),(5,9\left)\right\}$

$f=\{3,9,12\}\to \{1,3,4\}$ and $g:\{1,3,4,5\}\to \{3,9\}$

Co-domain of $f$ is a subset of the domain of $g.$

So, $g\circ f$ exists and $g\circ f:\{3,9,12\}\to \{3,9\}$

$\Rightarrow$ $(g\circ f)\left(3\right)=g\left[f\right(3\left)\right]=g\left(1\right)=3$

$\Rightarrow$ $(g\circ f)\left(9\right)=g\left[f\right(9\left)\right]=g\left(9\right)=3$

$\Rightarrow$ $(g\circ f)\left(12\right)=g\left[f\right(12\left)\right]=g\left(4\right)=9$

$\Rightarrow$ $g\circ f=\left\{\right(3,3),(9,3),(12,9\left)\right\}$

Co-domain of $g$ is a subset of the domain of $f$.

So, $f\circ g$ exists and $f\circ g:\{1,3,4,5\}\to \{3,9,12\}$

$\Rightarrow$ $(f\circ g)\left(1\right)=f\left[g\right(3\left)\right]=f\left(3\right)=1$

$\Rightarrow$ $(f\circ g)\left(3\right)=f\left[g\right(3\left)\right]=f\left(3\right)=1$

$\Rightarrow$ $(f\circ g)\left(4\right)=f\left[g\right(4\left)\right]=f\left(9\right)=3$

$\Rightarrow$ $(f\circ g)\left(5\right)=f\left[g\right(5\left)\right]=f\left(9\right)=3$

$\Rightarrow$ $f\circ g=\left\{\right(1,1),(3,1),(4,3),(5,3\left)\right\}$