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To determine the correct piecewise representation of the absolute value function [tex]\( f(x) = |x+3| \)[/tex], we need to consider the definition of the absolute value function and where the expression inside the absolute value changes sign.
The absolute value function can be written as a piecewise function based on the sign of the expression inside the absolute value:
[tex]\[ |x+3| = \begin{cases} x + 3 & \text{if } x + 3 \geq 0 \\ -(x + 3) & \text{if } x + 3 < 0 \end{cases} \][/tex]
This simplifies to:
[tex]\[ |x+3| = \begin{cases} x + 3 & \text{if } x \geq -3 \\ -x - 3 & \text{if } x < -3 \end{cases} \][/tex]
Now, let's compare this with the given options:
A. [tex]\( x+3 \)[/tex], [tex]\( x \geq -3 \)[/tex]
This matches the condition [tex]\( |x+3| = x + 3 \)[/tex] for [tex]\( x \geq -3 \)[/tex].
B. [tex]\( -x-3 \)[/tex], [tex]\( x < 3 \)[/tex]
This suggests the piece [tex]\( -x-3 \)[/tex] is valid for [tex]\( x < 3 \)[/tex], but in the definition of [tex]\( |x+3| \)[/tex], [tex]\( -x-3 \)[/tex] should be valid for [tex]\( x < -3 \)[/tex]. This option is incorrect.
C. [tex]\( -x+3 \)[/tex], [tex]\( x < -3 \)[/tex]
This does not match either part of the piecewise function, as [tex]\( -x + 3 \)[/tex] is not a correct transformation of [tex]\( |x+3| \)[/tex]. This option is incorrect.
D. [tex]\( x+3 \)[/tex], [tex]\( x \geq 3 \)[/tex]
This condition [tex]\( x \geq 3 \)[/tex] is valid for [tex]\( |x+3| = x + 3 \)[/tex], but it is not properly capturing the entire range where [tex]\( x + 3 \)[/tex] is non-negative; it should be [tex]\( x \geq -3 \)[/tex]. This option is incorrect.
Thus, the correct piecewise function that [tex]\( f(x) = |x+3| \)[/tex] will include is:
A. [tex]\( x+3 \)[/tex], [tex]\( x \geq -3 \)[/tex]
Hence, the correct answer is Option A.
The absolute value function can be written as a piecewise function based on the sign of the expression inside the absolute value:
[tex]\[ |x+3| = \begin{cases} x + 3 & \text{if } x + 3 \geq 0 \\ -(x + 3) & \text{if } x + 3 < 0 \end{cases} \][/tex]
This simplifies to:
[tex]\[ |x+3| = \begin{cases} x + 3 & \text{if } x \geq -3 \\ -x - 3 & \text{if } x < -3 \end{cases} \][/tex]
Now, let's compare this with the given options:
A. [tex]\( x+3 \)[/tex], [tex]\( x \geq -3 \)[/tex]
This matches the condition [tex]\( |x+3| = x + 3 \)[/tex] for [tex]\( x \geq -3 \)[/tex].
B. [tex]\( -x-3 \)[/tex], [tex]\( x < 3 \)[/tex]
This suggests the piece [tex]\( -x-3 \)[/tex] is valid for [tex]\( x < 3 \)[/tex], but in the definition of [tex]\( |x+3| \)[/tex], [tex]\( -x-3 \)[/tex] should be valid for [tex]\( x < -3 \)[/tex]. This option is incorrect.
C. [tex]\( -x+3 \)[/tex], [tex]\( x < -3 \)[/tex]
This does not match either part of the piecewise function, as [tex]\( -x + 3 \)[/tex] is not a correct transformation of [tex]\( |x+3| \)[/tex]. This option is incorrect.
D. [tex]\( x+3 \)[/tex], [tex]\( x \geq 3 \)[/tex]
This condition [tex]\( x \geq 3 \)[/tex] is valid for [tex]\( |x+3| = x + 3 \)[/tex], but it is not properly capturing the entire range where [tex]\( x + 3 \)[/tex] is non-negative; it should be [tex]\( x \geq -3 \)[/tex]. This option is incorrect.
Thus, the correct piecewise function that [tex]\( f(x) = |x+3| \)[/tex] will include is:
A. [tex]\( x+3 \)[/tex], [tex]\( x \geq -3 \)[/tex]
Hence, the correct answer is Option A.
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