Question

determine where the following function is continuous.
m(x)=\frac{x - 1}{4x^{2}-81}
the function is continuous on (square).
(type your answer in interval notation. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find the denominator's zeros

Set $4x^{2}-81 = 0$. Then $4x^{2}=81$, so $x^{2}=\frac{81}{4}$, and $x=\pm\frac{9}{2}$.

Step2: Determine the continuity intervals

A rational - function is continuous everywhere except where the denominator is zero. So the function $M(x)=\frac{x - 1}{4x^{2}-81}$ is continuous on $(-\infty,-\frac{9}{2})\cup(-\frac{9}{2},\frac{9}{2})\cup(\frac{9}{2},\infty)$.

Answer:

$(-\infty,-\frac{9}{2})\cup(-\frac{9}{2},\frac{9}{2})\cup(\frac{9}{2},\infty)$