Certainly! To determine the correct piecewise representation of the function [tex]\( f(x)=|x+3| \)[/tex], we need to consider the definition of the absolute value.
The absolute value function [tex]\( |a| \)[/tex] can be defined piecewise as:
[tex]\[ |a| = \begin{cases}
a & \text{if } a \geq 0, \\
-a & \text{if } a < 0.
\end{cases} \][/tex]
In our case, [tex]\( a = x + 3 \)[/tex]. Hence, the function [tex]\( f(x) = |x + 3| \)[/tex] can be rewritten as:
[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 solve the inequalities in these conditions:
- For [tex]\( x + 3 \geq 0 \)[/tex]:
[tex]\[ x \geq -3 \][/tex]
So, [tex]\( f(x) = x + 3 \)[/tex] when [tex]\( x \geq -3 \)[/tex].
- For [tex]\( x + 3 < 0 \)[/tex]:
[tex]\[ x < -3 \][/tex]
So, [tex]\( f(x) = -(x + 3) = -x - 3 \)[/tex] when [tex]\( x < -3 \)[/tex].
Thus, the piecewise function is:
[tex]\[ f(x) = \begin{cases}
x + 3, & x \geq -3, \\
-x - 3, & x < -3.
\end{cases} \][/tex]
Given the choices:
A. [tex]\( x + 3, x \geq 3 \)[/tex] - This is incorrect because the condition should be [tex]\( x \geq -3 \)[/tex], not [tex]\( x \geq 3 \)[/tex].
B. [tex]\( x + 3, x \geq -3 \)[/tex] - This is correct.
C. [tex]\( -x + 3, x < -3 \)[/tex] - This is incorrect because the expression should be [tex]\( -x - 3 \)[/tex], not [tex]\( -x + 3 \)[/tex].
D. [tex]\( -x - 3, x < 3 \)[/tex] - This is incorrect because the condition should be [tex]\( x < -3 \)[/tex], not [tex]\( x < 3 \)[/tex].
Thus, the correct answer is:
B. [tex]\( x + 3, x \geq -3 \)[/tex]
