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Find the Rule That Defines Each Piecewise-Defined Function f(x)={x+1 if 0x312x if 3<x5f ( x ) = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x & \text { if } 3 < x \leq 5 \end{array} \right.

Question 3

Multiple Choice

Find the rule that defines each piecewise-defined function.
- Find the rule that defines each piecewise-defined function. -   A)  f ( x )  = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x & \text { if } 3 < x \leq 5 \end{array} \right.  B)   f ( x )  = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x + 2 & \text { if } 3 < x \leq 5 \end{array} \right.  C)   f ( x )  = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x + \frac { 1 } { 2 } & \text { if } 3 < x \leq 5 \end{array} \right.  D)   f ( x )  = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x - \frac { 1 } { 2 } & \text { if } 3 < x \leq 5 \end{array} \right.


A) f(x) ={x+1 if 0x312x if 3<x5f ( x ) = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x & \text { if } 3 < x \leq 5 \end{array} \right.
B) f(x) ={x+1 if 0x312x+2 if 3<x5f ( x ) = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x + 2 & \text { if } 3 < x \leq 5 \end{array} \right.
C) f(x) ={x+1 if 0x312x+12 if 3<x5f ( x ) = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x + \frac { 1 } { 2 } & \text { if } 3 < x \leq 5 \end{array} \right.
D) f(x) ={x+1 if 0x312x12 if 3<x5f ( x ) = \left\{ \begin{array} { l l } x + 1 & \text { if } 0 \leq x \leq 3 \\ \frac { 1 } { 2 } x - \frac { 1 } { 2 } & \text { if } 3 < x \leq 5 \end{array} \right.

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