Complete the table. $$f ( s ) = \frac { | s - 2 | } { s - 2 }$$
$$\begin{array} { | l | l | l | l | l | l | }
\hline \ s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & & & & & \
\hline
\end{array}$$
A)$$\begin{array} { | l | l | l | l | l | l | }
\hline \ s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & - 1 & 1 & 1 & 1 & 1\
\hline
\end{array}$$
B)$$\begin{array} { | l | l | l | l | l | l | }
\hline \s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & - 1 & 1 & 1 & 3&1 \
\hline
\end{array}$$
C)$$\begin{array} { | l | l | l | l | l | l | }
\hline \s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & - 1 & 1 & 3 & 1&1 \
\hline
\end{array}$$
D)$$\begin{array} { | l | l | l | l | l | l | }
\hline \s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & 1 & 1 & 3 & 1&1 \
\hline
\end{array}$$
E)$$\begin{array} { | l | l | l | l | l | l | }
\hline \s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & - 1 & 3 & 1 & 1&1 \
\hline
\end{array}$$

Question

Complete the table. $$f ( s ) = \frac { | s - 2 | } { s - 2 }$$ 
$$\begin{array} { | l | l | l | l | l | l | } 
\hline \ s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & & & & & \
\hline
\end{array}$$
A)$$\begin{array} { | l | l | l | l | l | l | } 
\hline \ s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & - 1 & 1 & 1 & 1 & 1\
\hline
\end{array}$$
B)$$\begin{array} { | l | l | l | l | l | l | } 
\hline \s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & - 1 & 1 & 1 & 3&1 \
\hline
\end{array}$$
C)$$\begin{array} { | l | l | l | l | l | l | } 
\hline \s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & - 1 & 1 & 3 & 1&1 \
\hline
\end{array}$$
D)$$\begin{array} { | l | l | l | l | l | l | } 
\hline \s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & 1 & 1 & 3 & 1&1 \
\hline
\end{array}$$
E)$$\begin{array} { | l | l | l | l | l | l | } 
\hline \s & 0 & 4 & \frac { 9 } { 2 } & \frac { 13 } { 2 } & 7 \\
\hline f ( s ) & - 1 & 3 & 1 & 1&1 \
\hline
\end{array}$$

Quizplus · Accepted Answer

The function $\frac{|s-2|}{s-2}$ evaluates to 1 when $s > 2$ because the numerator and denominator are the same, resulting in 1. When $s < 2$, the numerator becomes positive due to the absolute value, but the denominator is negative, resulting in -1. For $s = 2$, the function is undefined, but this value is not in the table. For all values in the table greater than 2, the function evaluates to 1, and for $s = 0$, which is less than 2, it evaluates to -1.

Complete the Table f(s)=∣s−2∣s−2f ( s ) = \frac { | s - 2 | } { s - 2 }f(s)=s−2∣s−2∣​

Complete the Table $f ( s ) = \frac { | s - 2 | } { s - 2 }$