Answer
a) $1$
b) $-1$
c) $-\frac{\left| 6+x \right|}{\left( 6+x \right)}$
Work Step by Step
(a)
Let us consider the function $f\left( x \right)=\frac{\left| x+3 \right|}{x+3}$.
By putting the value of $x$ as $5$ in the equation of $f\left( x \right)$ we obtain the value of $f\left( 5 \right)$ as:
$\begin{align}
& f\left( x \right)=\frac{\left| x+3 \right|}{x+3} \\
& f\left( 5 \right)=\frac{\left| 5+3 \right|}{5+3} \\
& =\frac{\left| 8 \right|}{8} \\
& =1
\end{align}$
Hence, the value of $f\left( 5 \right)$ is $1$.
(b)
Let us consider the function $f\left( x \right)=\frac{\left| x+3 \right|}{x+3}$.
By putting the value of $x$ as $-5$ in the equation of $f\left( x \right)$ we obtain the value of $f\left( -5 \right)$ as:
$\begin{align}
& f\left( x \right)=\frac{\left| x+3 \right|}{x+3} \\
& f\left( -5 \right)=\frac{\left| -5+3 \right|}{-5+3} \\
& =\frac{\left| -2 \right|}{-2} \\
& =-1
\end{align}$
Hence, the value of $f\left( -5 \right)$ is $-1$.
(c)
Let us consider the function $f\left( x \right)=\frac{\left| x+3 \right|}{x+3}$.
By putting the value of $x$ as $-9-x$ in the equation of $f\left( x \right)$ we obtain the value of $f\left( -9-x \right)$ as:
$\begin{align}
& f\left( x \right)=\frac{\left| x+3 \right|}{x+3} \\
& f\left( -9-x \right)=\frac{\left| -9-x+3 \right|}{-9-x+3} \\
& =\frac{\left| -6-x \right|}{-6-x} \\
& =-\frac{\left| 6+x \right|}{\left( 6+x \right)}
\end{align}$
Hence, the value of $f\left( -9-x \right)$ is $-\frac{\left| 6+x \right|}{\left( 6+x \right)}$.
