Answer
a) $\dfrac{3}{\sqrt[3]4+1}$; b) $1$; c) $\dfrac{6}{5}$; d) $0$.
Work Step by Step
We are given the functions:
$f(x)=\dfrac{3}{x+1}$
$g(x)=\sqrt[3] x$
a) $(f\circ g)(4)=f(g(4))=f\left(\sqrt[3] 4\right)=\dfrac{3}{\sqrt[3]4+1}$
b) $(g\circ f)(2)=g(f(2))=g\left(\dfrac{3}{2+1}\right)=g(1)=\sqrt[3] 1=1$
c) $(f\circ f)(1)=f(f(1))=f\left(\dfrac{3}{1+1}\right)=f\left(\dfrac{3}{2}\right)=\dfrac{3}{\dfrac{3}{2}+1}=\dfrac{6}{5}$
d) $(g\circ g)(0)=g(g(0))=g\left(\sqrt[3]0\right)=g\left(0\right)=\sqrt[3]0=0$
