To write the absolute value function [tex]\( f(x) = |x - 5| \)[/tex] as a piecewise function, we need to consider the definition of the absolute value. The absolute value function [tex]\( |u| \)[/tex] is defined as follows:
[tex]\[
|u| = \begin{cases}
u, & \text{if } u \geq 0 \\
-u, & \text{if } u < 0
\end{cases}
\][/tex]
For our specific function [tex]\( f(x) = |x - 5| \)[/tex], we can identify [tex]\( u = x - 5 \)[/tex]. Applying the absolute value definition to [tex]\( u = x - 5 \)[/tex] gives two cases:
1. When [tex]\( x - 5 \geq 0 \)[/tex]:
[tex]\(|x - 5| = x - 5\)[/tex]
This simplifies to:
[tex]\[
\begin{cases}
x - 5, & \text{if } x \geq 5
\end{cases}
\][/tex]
2. When [tex]\( x - 5 < 0 \)[/tex]:
[tex]\(|x - 5| = -(x - 5) = 5 - x\)[/tex]
This simplifies to:
[tex]\[
\begin{cases}
5 - x, & \text{if } x < 5
\end{cases}
\][/tex]
Combining these two parts, we get the piecewise function representation:
[tex]\[
f(x) = \begin{cases}
x - 5, & \text{if } x \geq 5 \\
5 - x, & \text{if } x < 5
\end{cases}
\][/tex]
Given the provided options:
A. [tex]\( f(x) = \left\{\begin{array}{ll}x - 5, & x \geq -5 \\ -x - 5, & x < -5\end{array}\right. \)[/tex]
B. [tex]\( f(x) = \left\{\begin{array}{ll}x - 5, & x \geq -5 \\ -x + 5, & x < -5\end{array}\right. \)[/tex]
C. [tex]\( f(x) = \left\{\begin{array}{ll}x - 5, & x \geq 5 \\ -x + 5, & x < 5\end{array}\right. \)[/tex]
D. [tex]\( f(x) = \left\{\begin{array}{ll}x - 5, & x \geq 5 \\ -x - 5, & x < 5\end{array}\right. \)[/tex]
From our piecewise representation of [tex]\( f(x) = |x - 5| \)[/tex]:
[tex]\[
f(x) = \begin{cases}
x - 5, & \text{if } x \geq 5 \\
5 - x, & \text{if } x < 5
\end{cases}
\][/tex]
The correct answer is:
C. [tex]\( f(x) = \left\{\begin{array}{ll}x - 5, & x \geq 5 \\ -x + 5, & x < 5\end{array}\right. \)[/tex]
Therefore, the correct option is [tex]\( \boxed{3} \)[/tex].
