Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 36

We need to find (a) $f \circ g$, (b) $g \circ f$, (c) $f \circ f$, and (d) $g \circ g$ and state their domains



$f(x) = \frac{x}{1+x} , \qquad \quad g(x) = \sin 2x$


$
\begin{equation}
\begin{aligned}
\text{(a)} \qquad \quad f \circ g =& f(g(x))\\


\displaystyle f(\sin 2x)=& \frac{x}{1+x}
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\


\end{aligned}
\end{equation}
$


$\boxed{\displaystyle f \circ g = \frac{\sin 2x}{1+ \sin 2x}}$

$\boxed{ \text{The domain of this function is } \displaystyle (-\infty, \frac{-\pi}{4}) \bigcup\displaystyle(\frac{-\pi}{4}, \infty)}$


$
\begin{equation}
\begin{aligned}
\text{(b)} \qquad \quad f \circ g =& g(f(x))\\


\displaystyle g \left(\frac{x}{1+x}\right) =& \sin 2x
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\

\displaystyle g \left(\frac{x}{1+x}\right) =& \sin 2\left(\frac{x}{1+x}\right)
&& \text{ Simplify the equation }
\end{aligned}
\end{equation}
$



$\boxed{\displaystyle f \circ g= \sin \left(\frac{2x}{1+x}\right)}$


$\boxed{ \text{The domain of this function is } (-\infty, -1) \bigcup(-1,\infty)}$




$
\begin{equation}
\begin{aligned}
\text{(c)} \qquad \quad f \circ f =& f(f(x))\\


\displaystyle f\left(\frac{x}{1+x}\right) =& \frac{x}{1+x}
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\

\displaystyle f\left(\frac{x}{1+x}\right)=& \frac{\frac{x}{1+x}}{1+\frac{x}{1+x}}
&& \text{ Get the LCD of the numerator and the denominator}\\

\displaystyle f\left(\frac{x}{1+x}\right)=& \frac{\frac{x}{\cancel{1+x}}}{\frac{1+x+x}{\cancel{1+x}}}
&& \text{ Cancel out and combine like terms}

\end{aligned}
\end{equation}
$


$\boxed{\displaystyle f \circ f = \frac{x}{2x+1}}$

$\boxed{\text{ The domain of this function is } \displaystyle(-\infty, \frac{-1}{2}) \bigcup \displaystyle(\frac{-1}{2},\infty)}$


$
\begin{equation}
\begin{aligned}
\text{(d)} \qquad \quad g \circ g =& g(g(x))\\

g(\sin 2x)=& \sin 2x
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\

g(\sin 2x)=& \sin 2(\sin 2x)
&& \text{ Simplify equation}

\end{aligned}
\end{equation}
$


$\boxed{g \circ g= \sin(2 \sin 2x)}$


$\boxed{\text{ The domain of this function is }(-\infty,\infty)}$