Answer
$a.\displaystyle \qquad \frac{2\sqrt{10}}{25}$
$b.\displaystyle \qquad\frac{4\sqrt{2}-2}{7}$
$c.\qquad 1$
$d.\displaystyle \qquad\frac{2}{3}$
Work Step by Step
$a.$
$(f\circ g)(4)=f[g(4)]$
$g(4)=\displaystyle \frac{2}{4+1}=\frac{2}{5}$
$f[g(4)]=f(\displaystyle \frac{2}{5})=(\frac{2}{5})^{3/2}=\frac{\sqrt{8}}{\sqrt{125}}$
$=\displaystyle \frac{2\sqrt{2}}{5\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{10}}{25}$
$b.$
$(g\circ f)(2)=g[f(2)] $
$f(2)=2^{3/2}=2\sqrt{2}$
$g[f(2)] =g(2\displaystyle \sqrt{2})=\frac{2}{2\sqrt{2}+1}$
$=\displaystyle \frac{2}{2\sqrt{2}+1}\cdot\frac{2\sqrt{2}-1}{2\sqrt{2}-1}=\frac{4\sqrt{2}-2}{8-1}$
$=\displaystyle \frac{4\sqrt{2}-2}{7}$
$c.$
$(f\circ f)(1) =f[f(1)]$
$f(1)=1^{3/2}=1$
$f[f(1)]=f(1)=1$
$d.$
$(g\circ g)(0)=g[g(0)]$
$g(0) =\displaystyle \frac{2}{0+1}=2$
$g[g(0)]=g(2)=\displaystyle \frac{2}{2+1}=\frac{2}{3}$
