Respuesta :

[tex]\bf \begin{cases} h(x)=(f\circ g)(x)\\ h(x)=\sqrt{x+5}\\ f(x)=\sqrt{x+2} \end{cases}\\\\ -------------------------------\\\\ (f\circ g)(x)\implies f[\ g(x)\ ]\implies f[\ g(x)\ ]=\sqrt{g(x)+2}\qquad thus \\\\\\ \sqrt{g(x)+2}=(f\circ g)(x)=h(x)=\sqrt{x+5}\qquad therefore \\\\\\ \sqrt{g(x)+2}=\sqrt{x+5}\impliedby \textit{squaring both sides}\\\\\\ g(x)+2=x+5\implies g(x)=x+5-2\implies \boxed{g(x)=x+3}[/tex]