determine where the following function is continuous.\nm(x)=\frac{x - 1}{4x^{2}-81}\nthe function is…

determine where the following function is continuous.\nm(x)=\frac{x - 1}{4x^{2}-81}\nthe function is continuous on (square).\n(type your answer in interval notation. use integers or fractions for any numbers in the expression.)
Answer
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:
$(-\infty,-\frac{9}{2})\cup(-\frac{9}{2},\frac{9}{2})\cup(\frac{9}{2},\infty)$