![\bf \begin{cases} f(x)=x-3\\ g(x)=x^2+1\\ h(x)=-4x+1\\ -------&--------\\ (h\circ g\circ f)(x)=&(h\circ g(~f(x)~)\\ &h(~~g(~~f(x)~~)~~) \end{cases} \\\\\\ g(~~f(x)~~)=[f(x)]^2+1\implies g(~~f(x)~~)=[x-3]^2+1 \\\\\\ g(~~f(x)~~)=[x^2-6x+9]+1\implies \boxed{g(~~f(x)~~)=x^2-6x+10} \\\\\\ h(~~g(~~f(x)~~)~~)=-4[g(~f(x)~)]+1 \\\\\\ h(~~g(~~f(x)~~)~~)=-4[x^2-6x+10]+1 \\\\\\ h(~~g(~~f(x)~~)~~)=-4x^2+24x-40+1 \\\\\\ h(~~g(~~f(x)~~)~~)=-4x^2+24x-39](/tpl/images/0085/4051/6a4bc.png)
Can anyone me with number 6 & 8?
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Mathematics, 13.07.2019 16:30 synite
Can anyone me with number 6 & 8?
Answers: 2
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