To write the absolute value function [tex]\( f(x) = |x + 3| \)[/tex] as a piecewise function, we need to consider the definition of the absolute value function. The absolute value of a number is always non-negative, and it is defined as follows:
[tex]\[
|A| =
\begin{cases}
A & \text{if } A \geq 0 \\
-A & \text{if } A < 0
\end{cases}
\][/tex]
In our function [tex]\( f(x) = |x + 3| \)[/tex], we can identify [tex]\( A \)[/tex] as [tex]\( x + 3 \)[/tex].
Thus, we can rewrite [tex]\( f(x) \)[/tex] in piecewise form by determining the conditions where [tex]\( x + 3 \)[/tex] is non-negative and where it is negative.
1. When [tex]\( x + 3 \geq 0 \)[/tex]:
[tex]\[
x + 3 \geq 0 \implies x \geq -3
\][/tex]
In this case, since [tex]\( x + 3 \)[/tex] is non-negative, the absolute value function simplifies to:
[tex]\[
f(x) = x + 3 \text{ when } x \geq -3
\][/tex]
2. When [tex]\( x + 3 < 0 \)[/tex]:
[tex]\[
x + 3 < 0 \implies x < -3
\][/tex]
In this case, since [tex]\( x + 3 \)[/tex] is negative, the absolute value function changes sign:
[tex]\[
f(x) = -(x + 3) = -x - 3 \text{ when } x < -3
\][/tex]
Putting this together, the piecewise function for [tex]\( f(x) = |x + 3| \)[/tex] is:
[tex]\[
f(x) =
\begin{cases}
x + 3 & \text{if } x \geq -3 \\
-x - 3 & \text{if } x < -3
\end{cases}
\][/tex]
Given the multiple choices provided:
A. [tex]\(x + 3, x \geq 3\)[/tex] - This statement is incorrect because the condition should be [tex]\(x \geq -3\)[/tex], not [tex]\(x \geq 3\)[/tex].
B. [tex]\(-x - 3, x < 3\)[/tex] - This statement is incorrect because the correct condition should be [tex]\(x < -3\)[/tex], not [tex]\(x < 3\)[/tex].
C. [tex]\(-x + 3, x < -3\)[/tex] - This statement is incorrect. The piece should be [tex]\(-x - 3\)[/tex], not [tex]\(-x + 3\)[/tex].
D. [tex]\(x + 3, x \geq -3\)[/tex] - This statement is correct as per our piecewise analysis.
Thus, the correct answer is:
[tex]\[
\boxed{D}
\][/tex]
