Respuesta :

Both are D

First one:

f(g(x)) =f(x^2+3) then you x^2+3 is the x you are going to introduce in the function f

f(x^2+3)= sqrroot(x^2+3-3)= sqrroot (x^2) =x

Second:

you need to leave x alone

y = 3x^2 +1

y-1 = 3x^2

(y-1)/3 = x^2

sqrroot ((y-1)/3) = x

sqrroot ((x-1)/3) = h^-1(x)