Question

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

Solution

## $$f\circ g$$ means $$g(x)$$ function is in $$f(x)$$ function. $$g\circ f$$ means $$f(x)$$ function is in $$g(x)$$ function.$$f=\{(3,1),(9,3),(12,4)\}$$ and $$g=\{(1,3),(3,3),(4,9),(5,9)\}$$$$f=\{3,9,12\}\rightarrow \{1,3,4\}$$ and $$g:\{1,3,4,5\}\rightarrow \{3,9\}$$Co-domain of $$f$$ is a subset of the domain of $$g.$$So, $$g\circ f$$ exists and $$g\circ f:\{3,9,12\}\rightarrow \{3,9\}$$$$\Rightarrow$$  $$(g\circ f)(3)=g[f(3)]=g(1)=3$$$$\Rightarrow$$  $$(g\circ f)(9)=g[f(9)]=g(9)=3$$$$\Rightarrow$$  $$(g\circ f)(12)=g[f(12)]=g(4)=9$$$$\Rightarrow$$  $$g\circ f=\{(3,3),(9,3),(12,9)\}$$Co-domain of $$g$$ is a subset of the domain of $$f$$.So, $$f\circ g$$ exists and $$f\circ g:\{1,3,4,5\}\rightarrow \{3,9,12\}$$$$\Rightarrow$$  $$(f\circ g)(1)=f[g(3)]=f(3)=1$$$$\Rightarrow$$  $$(f\circ g)(3)=f[g(3)]=f(3)=1$$$$\Rightarrow$$  $$(f\circ g)(4)=f[g(4)]=f(9)=3$$$$\Rightarrow$$  $$(f\circ g)(5)=f[g(5)]=f(9)=3$$$$\Rightarrow$$  $$f\circ g=\{(1,1),(3,1),(4,3),(5,3)\}$$Mathematics

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