Answer
a) $f\circ g = \dfrac{4}{4+x}$
Domain: $\{x|x\ne-4,x\ne0,x\ne1 \}$
b) $g\circ f =-\dfrac{4x-4}{x}$
Domain: $\{x|x\ne0,x\ne1 \}$
c) $f\circ f =x^2-x$
Domain: $\{x|x\ne1 \}$
d) $g\circ g =x$
Domain: $\{x|x\ne0\}$
Work Step by Step
a) $f\circ g = f(g(x)) = $
$\dfrac{-\frac{4}{x}}{-\frac{4}{x}-1}=$
$\dfrac{-\frac{4}{x}}{\frac{-4}{x}-\frac{x}{x}}=$
$\dfrac{-\frac{4}{x}}{\frac{-4-x}{x}}=$
$\dfrac{-4(x)}{x(-4-x)}$
$\dfrac{-4}{-4-x}$
$\dfrac{4}{4+x}$
Domain: $\{x|x\ne-4,x\ne0,x\ne1 \}$
b) $g\circ f = g(f(x)) =$
$-\dfrac{4}{\frac{x}{x-1}}=$
$-\dfrac{4(x-1)}{x}=$
$-\dfrac{4x-4}{x}$
Domain: $\{x|x\ne0,x\ne1 \}$
c) $f\circ f = f(f(x))=$
$\dfrac{x}{\frac{x}{x-1}-1}=$
$\dfrac{x}{\frac{x}{x-1}-\frac{x-1}{x-1}}=$
$\dfrac{x}{\frac{x-(x-1)}{x-1}}=$
$\dfrac{x(x-1)}{x-x+1}=$
$x^2-x$
Domain: $\{x|x\ne1 \}$
d) $g\circ g = g(g(x))=$
$-\dfrac{4}{-\frac{4}{x}}=$
$\dfrac{4x}{4}=$
$x$
Domain: $\{x|x\ne0\}$
