Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 34

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

$f(x) = \sqrt{x} , \qquad g(x) = \sqrt[3]{1-x}$



$
\begin{equation}
\begin{aligned}

\text{(a)} \qquad \quad f \circ g &= f(g(x))\\

f(\sqrt[3]{1-x})&= \sqrt{x}
&& \text{ Substitute the given function $g(x)$ to the value of $x $ of the function $ f(x)$}\\
\displaystyle f(\sqrt[3]{1-x}) &= [(1-x)^{\frac{1}{3}}]^{\frac{1}{2}}
&& \text{ Simplify the equation}

\end{aligned}
\end{equation}
$



$\displaystyle \boxed{ f \circ g = (1-x)^{\frac{1}{6}} \text{ or } \sqrt[6]{1-x}} $


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



$
\begin{equation}
\begin{aligned}

\text{(b)} \qquad \quad g \circ f &= g(f(x)) \\

g(\sqrt{x}) &= \sqrt[3]{1-x}
&& \text{ Substitute the given function g(x ) to the value of x of the function f(x) }\\

\end{aligned}
\end{equation}
$


$\boxed{g \circ f = \sqrt[3]{1-\sqrt{x}}} $


$\boxed{\text{ The domain of this function is } [0,1]}$



$
\begin{equation}
\begin{aligned}
\text{(c)} \qquad \quad f \circ f &= f(f(x)) \\
f(\sqrt{x}) &= \sqrt{x}
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\

f(\sqrt{x}) &= [(x)^{\frac{1}{2}}]^{\frac{1}{2}}
&& \text{Simplify the equation}\\

\end{aligned}
\end{equation}
$


$\boxed{f \circ f = \sqrt[4]{x}} $


$\boxed{\text{ The domain of this function is } [0,\infty)} $



$
\begin{equation}
\begin{aligned}
\text{(d)} \qquad \quad g \circ g &= g(g(x)) \\
g(\sqrt[3]{1-x}) &= \sqrt[3]{1-x}
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$ }

\end{aligned}
\end{equation}
$


$ \boxed{g \circ g=\sqrt[3]{1-\sqrt[3]{1-x}}} $


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