Find the x-values (if any) at which the function $f ( x ) = \frac { x } { x ^ { 2 } + 16 }$ is not continuous. Which of the discontinuities are removable?
A)4 and -4, not removable
B)continuous everywhere
C)4 and -4, removable
D)discontinuous everywhere
E)none of the above

Question

Quizplus · Accepted Answer

The function $f(x) = \frac{x}{x^2 + 16}$ is continuous everywhere because the denominator $x^2 + 16$ is never zero for any real number x. The denominator is always positive, so there are no points of discontinuity.

Find the X-Values (If Any) at Which the Function f(x)=xx2+16f ( x ) = \frac { x } { x ^ { 2 } + 16 }f(x)=x2+16x​

Find the X-Values (If Any) at Which the Function $f ( x ) = \frac { x } { x ^ { 2 } + 16 }$