Respuesta :

When a problem asks [tex] g\circ f [/tex], it really means: [tex] g(f(x)) [/tex]

So, whenever you have a function say f(x), and you want to find f(1), you plug 1 wherever you see an x. This problem is no differen, except we're plugging in an f(x) wherever we see an x in g(x) to get:

[tex] g(f(x))=f(x)^2+16 \implies\\
g(f(x))=(\sqrt{x})^2+16 \implies\\
g(f(x))=x+16 [/tex]