Answer
(a) $f \circ g = \frac{sin~2x}{1+sin~2x}$
domain: all real numbers except $(\frac{3\pi}{4}+n\pi)$, where $n$ is an integer
(b) $g \circ f = sin~(\frac{2x}{1+x})$
domain: $(-\infty, -1) \cup (-1, \infty)$
(c) $f \circ f = \frac{x}{1+2x}$
domain: $(-\infty, -1) \cup (-1,-\frac{1}{2}) \cup (-\frac{1}{2}, \infty)$
(d) $g \circ g = sin~(2~sin~2x)$
domain: all real numbers
Work Step by Step
$f(x) = \frac{x}{1+x}$, $g(x) = sin~2x$
(a) We can find $f \circ g$:
$f \circ g = \frac{sin~2x}{1+sin~2x}$
The domain includes all real numbers except:
$1 + sin~2x = 0$
$sin~2x = -1$
$x = \frac{3\pi}{4}+n\pi$, where $n$ is an integer
domain: all real numbers except $(\frac{3\pi}{4}+n\pi)$, where $n$ is an integer
(b) We can find $g \circ f$:
$g \circ f = sin~(\frac{2x}{1+x})$
domain: $(-\infty, -1) \cup (-1, \infty)$
(c) We can find $f \circ f$:
$f \circ f = \frac{(\frac{x}{1+x})}{1+(\frac{x}{1+x})}$
$f \circ f = \frac{\frac{x}{1+x}}{\frac{1+x}{1+x}~+\frac{x}{1+x}}$
$f \circ f = \frac{\frac{x}{1+x}}{\frac{1+2x}{1+x}}$
$f \circ f = \frac{x}{1+x}\cdot \frac{1+x}{1+2x}$
$f \circ f = \frac{x}{1+2x}$
The domain includes all real numbers except $-\frac{1}{2}$ and $-1$
domain: $(-\infty, -1) \cup (-1,-\frac{1}{2}) \cup (-\frac{1}{2}, \infty)$
(d) We can find $g \circ g$:
$g \circ g = sin~(2~sin~2x)$
domain: all real numbers
