$$\text { Find the domain and range of the function } h ( x ) = \frac { 11 } { x + 6 } \text {. }$$
A) domain: $( - \infty , - 6 ) \cup ( - 6 , \infty )$
range: $( - \infty , \infty )$

B) domain: $( - \infty , - 6 ) \cup ( - 6 , \infty )$
range: $( - \infty , 0 ) \cup ( 0 , \infty )$

C) domain: $( - \infty , - 6 ] \cup [ - 6 , \infty )$
range: $( - \infty , 0 ) \cup ( 0 , \infty )$

D) domain: $( - \infty , - 6 )$
range: $( 0 , \infty )$

E) domain: $( - 6 , \infty )$
range: $( 0 , \infty )$

Question

$$\text { Find the domain and range of the function } h ( x ) = \frac { 11 } { x + 6 } \text {. }$$ 
A) domain: $( - \infty , - 6 ) \cup ( - 6 , \infty )$
 range: $( - \infty , \infty )$

B) domain: $( - \infty , - 6 ) \cup ( - 6 , \infty )$ 
range: $( - \infty , 0 ) \cup ( 0 , \infty )$

C) domain: $( - \infty , - 6 ] \cup [ - 6 , \infty )$ 
range: $( - \infty , 0 ) \cup ( 0 , \infty )$

D) domain: $( - \infty , - 6 )$
range: $( 0 , \infty )$

E) domain: $( - 6 , \infty )$
range: $( 0 , \infty )$

Quizplus · Accepted Answer

The function is defined for all values of $x$ except $x=-6$ because the denominator becomes zero. Thus, the domain is $( - \infty , - 6 ) \cup ( - 6 , \infty )$. For the range, note that as $x$ approaches $-6$ from the left, $h(x)$ approaches negative infinity and as $x$ approaches $-6$ from the right, $h(x)$ approaches positive infinity. Thus, the range is $( - \infty , 0 ) \cup ( 0 , \infty )$. The only choice that matches these values is B.

Find the domain and range of the function h(x)=11x+6. \text { Find the domain and range of the function } h ( x ) = \frac { 11 } { x + 6 } \text {. } Find the domain and range of the function h(x)=x+611​.

$\text { Find the domain and range of the function } h ( x ) = \frac { 11 } { x + 6 } \text {. }$