Suppose that $f(x) = \sqrt{x}$ and $g(x) = \sin x $. Find the functions a) $f \circ g$, $\quad$ b) $g \circ f$, $\quad$ c) $f \circ f$ $\quad$ d) $g \circ g$ and their domains.
$
\begin{equation}
\begin{aligned}
f(x) =& \sqrt{x} \qquad \quad g(x) = \sin x\\
a) f \circ g =& f(g(x))\\
f(\sin x) =& \sqrt{x} \\
f \circ g =& \sqrt{\sin x} \\
\end{aligned}
\end{equation}
$
The square root function is defined only for non negative values of $x$, as we refer to the graph of $\sin x$, the values where the funcvtion is defined is from $0 \leq x \leq \pi$ and $2\pi$ which equals to 0.
Therefore, the domain of the function is $[2n\pi, \pi + 2n\pi]$ for its multiple cycles where $n$ is an integer.
$
\begin{equation}
\begin{aligned}
\text{ b)} g \circ f =& g(f(x))\\
g(\sqrt{x}) =& \sin x \\
g \circ f =& \sin (\sqrt{x})
\end{aligned}
\end{equation}
$
domain: $[0, \infty]$
$
\begin{equation}
\begin{aligned}
\text{ c)} f \circ f =& f(f(x)) \\
f(\sqrt{x}) =& \sqrt{x}\\
f(\sqrt{x}) =& \sqrt{\sqrt{x}} \\
f \circ f =& \sqrt[4]{x}
\end{aligned}
\end{equation}
$
domain: $[0, \infty]$
$
\begin{equation}
\begin{aligned}
\text{ d)} g \circ g =& g(g(x))\\
g(\sin x ) =& \sin x \\
g \circ g =& \sin (\sin x)
\end{aligned}
\end{equation}
$
domain: $(-\infty, \infty)$
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