Question

select the correct answer.

consider this absolute value function.

\\f(x) = |x - 5|\\

how can function \\(f\\) be written as a piecewise function?

\\(\bigcirc f(x) = \

$$\begin{cases} x - 5, & x \\ge 5 \\\\ -x - 5, & x < 5 \\end{cases}$$

\\)

\\(\bigcirc f(x) = \

$$\begin{cases} x - 5, & x \\ge 5 \\\\ -x + 5, & x < 5 \\end{cases}$$

\\)

\\(\bigcirc f(x) = \

$$\begin{cases} x - 5, & x \\ge -5 \\\\ -x - 5, & x < -5 \\end{cases}$$

\\)

\\(\bigcirc f(x) = \

$$\begin{cases} x - 5, & x \\ge -5 \\\\ -x + 5, & x < -5 \\end{cases}$$

\\)

Explanation:

Define the absolute value piecewise conditions

$$ |u| =
LATEXBLOCK0
$$

Substitute the inner expression

$$
LATEXBLOCK1
$$

Simplify the inequalities and expressions

$$ f(x) =
LATEXBLOCK2
$$

Answer:

• \(f(x) =

$$\begin{cases} x - 5, & x \ge 5 \\ -x - 5, & x < 5 \end{cases}$$

\)

• **\(f(x) =

$$\begin{cases} x - 5, & x \ge 5 \\ -x + 5, & x < 5 \end{cases}$$

\) (Correct answer)**

• \(f(x) =

$$\begin{cases} x - 5, & x \ge -5 \\ -x - 5, & x < -5 \end{cases}$$

\)

• \(f(x) =

$$\begin{cases} x - 5, & x \ge -5 \\ -x + 5, & x < -5 \end{cases}$$

\)