Answer :
To write the absolute value function [tex]\( f(x) = |x - 5| \)[/tex] as a piecewise function, we need to consider the definition of the absolute value function. The absolute value function [tex]\( |y| \)[/tex] is defined as:
[tex]\[ |y| = \begin{cases} y & \text{if } y \geq 0 \\ -y & \text{if } y < 0 \end{cases} \][/tex]
For the given function [tex]\( f(x) = |x - 5| \)[/tex]:
1. When [tex]\( x - 5 \geq 0 \)[/tex] (i.e., [tex]\( x \geq 5 \)[/tex]):
[tex]\[ |x - 5| = x - 5 \][/tex]
So, [tex]\( f(x) = x - 5 \)[/tex] when [tex]\( x \geq 5 \)[/tex].
2. When [tex]\( x - 5 < 0 \)[/tex] (i.e., [tex]\( x < 5 \)[/tex]):
[tex]\[ |x - 5| = -(x - 5) = -x + 5 \][/tex]
So, [tex]\( f(x) = -x + 5 \)[/tex] when [tex]\( x < 5 \)[/tex].
Putting these two cases together, we can write the piecewise function as:
[tex]\[ f(x) = \begin{cases} x - 5 & \text{if } x \geq 5 \\ -x + 5 & \text{if } x < 5 \end{cases} \][/tex]
Now we compare this with the given options:
A. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq 5 \\ -x+5, & x<5\end{array}\right. \)[/tex]
B. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq-5 \\ -x-5, & x<-5\end{array}\right. \)[/tex]
C. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq-5 \\ -x+5, & x<-5\end{array}\right. \)[/tex]
D. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq 5 \\ -x-5, & x<5\end{array}\right. \)[/tex]
From our piecewise function, we see that the correct form matches option A:
[tex]\[ f(x) = \left\{\begin{array}{ll}x-5, & x \geq 5 \\ -x+5, & x < 5\end{array}\right. \][/tex]
Thus, the correct answer is:
A. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq 5 \\ -x+5, & x<5\end{array}\right. \)[/tex]
[tex]\[ |y| = \begin{cases} y & \text{if } y \geq 0 \\ -y & \text{if } y < 0 \end{cases} \][/tex]
For the given function [tex]\( f(x) = |x - 5| \)[/tex]:
1. When [tex]\( x - 5 \geq 0 \)[/tex] (i.e., [tex]\( x \geq 5 \)[/tex]):
[tex]\[ |x - 5| = x - 5 \][/tex]
So, [tex]\( f(x) = x - 5 \)[/tex] when [tex]\( x \geq 5 \)[/tex].
2. When [tex]\( x - 5 < 0 \)[/tex] (i.e., [tex]\( x < 5 \)[/tex]):
[tex]\[ |x - 5| = -(x - 5) = -x + 5 \][/tex]
So, [tex]\( f(x) = -x + 5 \)[/tex] when [tex]\( x < 5 \)[/tex].
Putting these two cases together, we can write the piecewise function as:
[tex]\[ f(x) = \begin{cases} x - 5 & \text{if } x \geq 5 \\ -x + 5 & \text{if } x < 5 \end{cases} \][/tex]
Now we compare this with the given options:
A. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq 5 \\ -x+5, & x<5\end{array}\right. \)[/tex]
B. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq-5 \\ -x-5, & x<-5\end{array}\right. \)[/tex]
C. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq-5 \\ -x+5, & x<-5\end{array}\right. \)[/tex]
D. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq 5 \\ -x-5, & x<5\end{array}\right. \)[/tex]
From our piecewise function, we see that the correct form matches option A:
[tex]\[ f(x) = \left\{\begin{array}{ll}x-5, & x \geq 5 \\ -x+5, & x < 5\end{array}\right. \][/tex]
Thus, the correct answer is:
A. [tex]\( f(x)=\left\{\begin{array}{ll}x-5, & x \geq 5 \\ -x+5, & x<5\end{array}\right. \)[/tex]
