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![\bf \begin{cases} f(x)=x-3\\ g(x)=x^2+1\\ h(x)=-4x+1\\ ----------\\ (g\circ h)(x)=g(~~h(x)~~) \end{cases} \\\\\\ g(~~h(x)~~)=[h(x)]^2+1\implies g(~~h(x)~~)=[-4x+1]^2+1 \\\\\\ g(~~h(x)~~)=[-4x+1][-4x+1]+1 \\\\\\ g(~~h(x)~~)=\stackrel{FOIL}{[16x^2-8x+1]}+1\implies g(~~h(x)~~)=16x^2-8x+2](/tpl/images/0085/4064/66ad1.png)
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