Respuesta :

[tex]\bf \begin{cases} f(x)=\sqrt{x+7}\\ g(x)=4x\\ h(x)=x^2\\ (f\circ g\circ h)(x)=f(~~g(~~h(x)~~)~~) \end{cases}\\\\ -------------------------------\\\\ g(~~h(x)~~)=4\boxed{h(x)}\implies g(~~h(x)~~)=4\boxed{x^2} \\\\\\ f(~~g(~~h(x)~~)~~)=\sqrt{\boxed{g(~~h(x)~~)}+7} \\\\\\ f(~~g(~~h(x)~~)~~)=\sqrt{\boxed{4x^2}+7}[/tex]