Answer
(a) $ \sqrt {3+3x+3x^2}$, domain $(-\infty,\infty)$.
(b) $ 1+\sqrt {3x}+3x$, domain $[0,\infty)$.
(c) $ \sqrt {3\sqrt {3x}}$, domain $[0,\infty)$.
(d) $ 3+3x+4x^2+2x^3+x^4$, domain $(-\infty,\infty)$.
Work Step by Step
Given $f(x)=\sqrt {3x}$ and $g(x)=1+x+x^2$, we have:
(a) $(f\circ g)(x)=\sqrt {3(1+x+x^2)}=\sqrt {3+3x+3x^2}$, domain $(-\infty,\infty)$.
(b) $(g\circ f)(x)=1+(\sqrt {3x})+(\sqrt {3x})^2=1+\sqrt {3x}+3x$, domain $[0,\infty)$.
(c) $(f\circ f)(x)=\sqrt {3(\sqrt {3x})}=\sqrt {3\sqrt {3x}}$, domain $[0,\infty)$.
(d) $(g\circ g)(x)=1+(1+x+x^2)+(1+x+x^2)^2=3+3x+4x^2+2x^3+x^4$, domain $(-\infty,\infty)$.
