If $f ( x ) = \frac { 2 } { x - 5 }$ find $f ( - 5 ) , f ( 0 ) , f ( 5 ) , f ( - 3 )$ .
A) $f ( - 5 ) = - \frac { 1 } { 5 } , f ( 0 ) = - \frac { 2 } { 5 } , f ( 5 ) \text { does not exist, } f ( - 3 ) = - \frac { 1 } { 4 }$
B) $f ( - 5 ) \text { does not exist, } f ( 0 ) = - \frac { 2 } { 5 } , f ( 5 ) = \frac { 1 } { 5 } , f ( - 3 ) = - \frac { 1 } { 4 }$
C) $f ( - 5 ) = \frac { 1 } { 5 } , f ( 0 ) = - \frac { 2 } { 5 } , f ( 5 ) \text { does not exist, } f ( - 3 ) = - \frac { 1 } { 4 }$
D) $f ( - 5 ) = - \frac { 1 } { 5 } , f ( 0 ) = - \frac { 2 } { 5 } , f ( 5 ) = \frac { 1 } { 5 } , f ( - 3 ) = \frac { 1 } { 4 }$

Question

Quizplus · Accepted Answer

- To find f(-5), substitute -5 for x in the equation: 
f(-5) = 2/(-5-5) = -2/10 = -1/5
- To find f(0), substitute 0 for x in the equation:
f(0) = 2/(0-5) = -2/5
- To find f(5), substitute 5 for x in the equation:
f(5) = 2/(5-5) -> the denominator is 0, which makes the fraction undefined, meaning f(5) does not exist
- To find f(-3), substitute -3 for x in the equation:
f(-3) = 2/(-3-5) = -2/8 = -1/4 
Therefore, the best choice is A.

If f(x)=2x−5f ( x ) = \frac { 2 } { x - 5 }f(x)=x−52​

If $f ( x ) = \frac { 2 } { x - 5 }$