Give a rule for the piecewise-defined function. Then give the domain and range.
- $Give a rule for the piecewise-defined function. Then give the domain and range. - A) f ( x ) = \left\{ \begin{array} { l l } \sqrt [ 3 ] { x } & \text { if } x < 1 ; \text { Domain: } ( \infty , \infty ) , \text { Range: } ( \infty , 1 ) \cup [ 3 , \infty ) \\ x + 2 & \text { if } x \geq 1 \end{array} \right. B) f ( x ) = \left\{ \begin{array} { l l } - x ^ { 3 } & \text { if } x < 1 \\ x - 2 & \text { if } x \geq 1 \end{array} \right. Domain: ( \infty , 1 ) \cup [ 3 , \infty ) , Range: ( \infty , \infty ) C) f ( x ) = \left\{ \begin{array} { l l } x ^ { 3 } & \text { if } x < 1 \\ x + 2 & \text { if } x \geq 1 \end{array} \right. Domain: ( \infty , \infty ) , Range: ( \infty , 1 ) \cup [ 3 , \infty ) D) f ( x ) = \left\{ \begin{array} { l l } x ^ { 3 } & \text { if } x < 1 \\ x - 2 & \text { if } x \geq 1 \end{array} \right. ; Domain: ( \infty , 1 ) \cup [ 3 , \infty ) , Range: ( \infty , \infty )$
A)

f ( x ) = \left\{ \begin{array} { l l } \sqrt [ 3 ] { x } & \text { if } x < 1 ; \text { Domain: } ( \infty , \infty ) , \text { Range: } ( \infty , 1 ) \cup [ 3 , \infty ) \\ x + 2 & \text { if } x \geq 1 \end{array} \right.

f ( x ) = \left\{ \begin{array} { l l } - x ^ { 3 } & \text { if } x < 1 \\ x - 2 & \text { if } x \geq 1 \end{array} \right.

Domain:

( \infty , 1 ) \cup [ 3 , \infty )

, Range:

( \infty , \infty )

f ( x ) = \left\{ \begin{array} { l l } x ^ { 3 } & \text { if } x < 1 \\ x + 2 & \text { if } x \geq 1 \end{array} \right.

Domain:

( \infty , \infty )

, Range:

( \infty , 1 ) \cup [ 3 , \infty )

f ( x ) = \left\{ \begin{array} { l l } x ^ { 3 } & \text { if } x < 1 \\ x - 2 & \text { if } x \geq 1 \end{array} \right.

; Domain:

( \infty , 1 ) \cup [ 3 , \infty )

, Range:

( \infty , \infty )

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