To transform the given absolute value function [tex]\( f(x) = |x + 3| \)[/tex] into a piecewise function, we need to consider how the absolute value affects the expression within it.
The absolute value function [tex]\( |a| \)[/tex] is defined as:
[tex]\[
|a| =
\begin{cases}
a & \text{if } a \geq 0, \\
-a & \text{if } a < 0.
\end{cases}
\][/tex]
Applying this definition to [tex]\( f(x) = |x + 3| \)[/tex]:
[tex]\[
f(x) = |x + 3| =
\begin{cases}
x + 3 & \text{if } x + 3 \geq 0, \\
-(x + 3) & \text{if } x + 3 < 0.
\end{cases}
\][/tex]
Now we can solve the conditions for [tex]\( x \)[/tex]:
1. For [tex]\( x + 3 \geq 0 \)[/tex]:
[tex]\[
x + 3 \geq 0 \implies x \geq -3
\][/tex]
Therefore:
[tex]\[
f(x) = x + 3 \text{ for } x \geq -3
\][/tex]
2. For [tex]\( x + 3 < 0 \)[/tex]:
[tex]\[
x + 3 < 0 \implies x < -3
\][/tex]
Therefore:
[tex]\[
f(x) = -(x + 3) = -x - 3 \text{ for } x < -3
\][/tex]
So the piecewise function can be written as:
[tex]\[
f(x) =
\begin{cases}
x + 3 & \text{if } x \geq -3, \\
-x - 3 & \text{if } x < -3.
\end{cases}
\][/tex]
From the given multiple-choice options, we can identify the correct piece:
A. [tex]\( x+3, x \geq 3 \)[/tex] – This is incorrect because it does not cover the correct domain for when [tex]\( x \geq -3 \)[/tex].
B. [tex]\( x + 3, x \geq -3 \)[/tex] – This is correct because it matches our piecewise condition for [tex]\( x \geq -3 \)[/tex].
C. [tex]\( -x + 3, x < -3 \)[/tex] – This is incorrect because the correct expression for [tex]\( x < -3 \)[/tex] is [tex]\( -x - 3\)[/tex].
D. [tex]\( -x - 3, x < 3 \)[/tex] – This is incorrect because it does not match the condition [tex]\( x < -3 \)[/tex] and misrepresents the function [tex]\( -x - 3 \)[/tex].
Therefore, the correct answer is:
B. [tex]\( x + 3, x \geq -3 \)[/tex]
