Respuesta :

[tex]\bf \begin{cases} (f\circ g)(x)=x^2-6x+8\\\\ (f\circ g)(x)=f(~~g(x)~~)\\\\ g(x)=x-3\\ \end{cases} \\\\\\ \textit{now, one probability is }x^2-1 \\\\\\ f(x)=x^2-1\implies f(~~g(x)~~)=(x-3)^2-1 \\\\\\ f(~~g(x)~~)=(x^2-6x+9)-1\implies f(~~g(x)~~)=x^2-6x+8[/tex]

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