Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 33

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.

$f(x) = 1-3x , \qquad g(x) = \cos x$


$
\begin{equation}
\begin{aligned}

\text{(a)} \qquad \quad f \circ g &= f(g(x))\\
f(\cos x) &= 1-3x
&& \text{ Substitute the given function $g(x$) to the value of $x$ of the function $f(x)$}\\

\end{aligned}
\end{equation}
$


$\boxed{f \circ g=1-3 \cos x}$

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



$
\begin{equation}
\begin{aligned}

\text{(b)} \qquad \quad g \circ f &= g(f(x))\\
g(1-3x) &= \cos x && \text{ Substitute the given function $g(x$) to the value of $x$ of the function $f(x)$}\\

\end{aligned}
\end{equation}
$


$\boxed{g \circ f = \cos (1-3x)}$
$\boxed{ \text{ The domain of this function is }(-\infty,\infty)}$



$
\begin{equation}
\begin{aligned}

\text{(c)} \qquad \quad f \circ f &= f(f(x))\\
f(1-3)x &= 1-3x && \text{ Substitute the given function $g(x$) to the value of $x$ of the function $f(x)$}\\
f(1-3)x &= 1-3(1-3x) && \text{ Simplify the equation}\\
f(1-3)x &= 1-3+9x && \text{ Combine like terms}

\end{aligned}
\end{equation}
$


$\boxed{f \circ f=9x-2}$
$\boxed{ \text{ The domain of this function is } (-\infty,\infty)}$





$
\begin{equation}
\begin{aligned}

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

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

\end{aligned}
\end{equation}
$


$\boxed{g \circ g =\cos(\cos x)}$
$\boxed{\text{ The domain of this function is } (-\infty,\infty)}$

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