Describe the interval $($ s $)$ on which the function $f ( x ) = \frac { x - 4 } { x ^ { 2 } - 16 }$ is continuous.
A) $( - \infty , - 4 ] , ( - 4,4 ] \& ( 4 , \infty )$
B) $( - \infty , 4 ) , ( 4,4 ) \& ( 4 , \infty )$
C) $( - \infty , - 4 ) , ( - 4,4 ) \& ( 4 , \infty )$
D) $( - \infty , - 4 ] , ( - 4,4 ) \& ( 4 , \infty )$
E) $( - \infty , - 4 ] , [ - 4,4 ] \& [ 4 , \infty )$

Question

Quizplus · Accepted Answer

The function $f(x) = \frac{x - 4}{x^2 - 16}$ is continuous everywhere except at the points where the denominator is zero. The denominator $x^2 - 16 = (x - 4)(x + 4)$ is zero at $x = 4$ and $x = -4$. Therefore, the function is continuous on the intervals $(-\infty, -4), (-4, 4),$ and $(4, \infty)$.

Describe the Interval ((( S ))) On Which the Function f(x)=x−4x2−16f ( x ) = \frac { x - 4 } { x ^ { 2 } - 16 }f(x)=x2−16x−4​

Describe the Interval $($ S $)$ On Which the Function $f ( x ) = \frac { x - 4 } { x ^ { 2 } - 16 }$