Respuesta :

[tex]\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[/tex]