Answer
$a.\quad(f\circ g)(1) = 2$
$b.\displaystyle \quad(g\circ f)(1) = \frac{1}{2}$
$c.\displaystyle \quad (f\circ g)(x) = \frac{2}{\sqrt{x}} $
$d.\displaystyle \quad (g\circ f)(x) = \frac{1}{2\sqrt{x}}$
Work Step by Step
$(f\circ g)(x)=f[g(x)]=\sqrt{4g(x)}$
$= \sqrt{4\cdot\frac{1}{x}} $
$= \displaystyle \frac{2}{\sqrt{x}} \qquad ... \quad(c)$
$(f\displaystyle \circ g)(1)= \frac{2}{\sqrt{1}} =2 \qquad ... \quad(a)$
$(g\displaystyle \circ f)(x)=g[f(x)]=\frac{1}{f(x)}$
$= \displaystyle \frac{1}{\sqrt{4x}} $
$= \displaystyle \frac{1}{2\sqrt{x}} \qquad ... \quad(d)$
$(g\displaystyle \circ f)(1)=\frac{1}{2\sqrt{1}}=\frac{1}{2} \qquad ... \quad(b)$
