Determine where the following function is continuous. \( M(x)=\frac{x-3}{9 x^{2}-64} \) The function is continuous on \( \square \). (Type your answer in interval notation, Use integers or fractions for any numbers in the expression.)

To determine where the function \( M(x) = \frac{x-3}{9x^2 - 64} \) is continuous, we first need to identify where the denominator is equal to zero since the function will be discontinuous at those points. So, we set: \( 9x^2 - 64 = 0 \) Solving for \( x \), we get: \( 9x^2 = 64 \) \( x^2 = \frac{64}{9} \) \( x = \pm \frac{8}{3} \) Thus, the function is discontinuous at \( x = \frac{8}{3} \) and \( x = -\frac{8}{3} \). The intervals where the function is continuous are: \( (-\infty, -\frac{8}{3}) \), \( (-\frac{8}{3}, \frac{8}{3}) \), and \( (\frac{8}{3}, \infty) \). So, the function is continuous on \( (-\infty, -\frac{8}{3}) \cup (-\frac{8}{3}, \frac{8}{3}) \cup (\frac{8}{3}, \infty) \).