$\text { Determine the domain and range of the function } f ( x ) = \left\{ \begin{array} { l }
3 x + 2 , x

Question

$\text { Determine the domain and range of the function } f ( x ) = \left\{ \begin{array} { l } 
3 x + 2 , x < 0 \
3 x + 6 , x \geq 0
\end{array} \right. \text {. }$
A) domain: $( - \infty , 2 )$
range: $( - \infty , 2 ) \cap [ 6 , \infty ]$

B) domain: $( - \infty , \infty )$
range: $( - \infty , 2 ) \cup [ 6 , \infty )$

C) domain: $( - \infty , \infty )$
range: $( - \infty , 2 ) \cup ( \infty , 6 ]$

D) domain: $( - \infty , \infty )$
range: $( \infty , 2 ) \cup ( 6 , - \infty )$

E) domain: $( - \infty , 3 )$
range: $( - \infty , 2 ) \cap [ 6 , \infty )$

Quizplus · Accepted Answer

The function is defined for all real numbers, hence the domain is $(- \infty, \infty)$. The range is continuous for all values less than 0 and all values greater than or equal to 0, but there is no overlap in the output values, hence the range is a union of two intervals.

A) Domain (−∞,2)( - \infty , 2 )(−∞,2) Range (−∞,2)∩[6,∞]( - \infty , 2 ) \cap [ 6 , \infty ](−∞,2)∩[6,∞]

A) Domain $( - \infty , 2 )$ Range $( - \infty , 2 ) \cap [ 6 , \infty ]$