Recall that:
[tex](f\circ f)(x)=f(f(x)).[/tex]
Therefore:
[tex](f\circ f)(x)=f(\sqrt[]{x+2})=\sqrt[]{\sqrt[]{x+2}+2}.[/tex]
Now, the above function is well defined as long as x+2 remains positive, therefore, it is well defined as long as x is greater or equal to -2.
Answer: The domain is:
[tex]\lbrack-2,\infty).[/tex]
The composition is:
[tex](f\circ f)(x)=f(\sqrt[]{x+2})=\sqrt[]{\sqrt[]{x+2}+2}.[/tex]
