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![\bf \begin{cases} f(x)=x^2\\ g(x)=x+1\\ (f\circ g)(x)=f(~~g(x)~~) \end{cases} \\\\\\ f(~~g(x)~~)\implies [g(x)]^2\implies [x+1]^2\implies x^2+2x+1](/tpl/images/0043/6304/24ded.png)
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