Question 11 (5 points) Determine where the function \( f(x)=\frac{x^{2}+7}{x^{2}+x-6} \) is continuous. \( (-3, \infty) \) \( (-\infty,-3) \cup(2, \infty) \) \( (-\infty, 2) \) \( (-3,2) \cup(2, \infty) \) \( (-\infty,-3) \cup(-3,2) \) \( (-\infty,-3) \cup(-3,2) \cup(2, \infty) \)

To determine where the function \( f(x)=\frac{x^{2}+7}{x^{2}+x-6} \) is continuous, we first need to identify any discontinuities. The function is continuous wherever the denominator is not zero. The denominator can be factored as: \[ x^{2}+x-6 = (x-2)(x+3) \] Setting the denominator equal to zero gives us the critical points \( x=2 \) and \( x=-3 \). Therefore, the function is undefined at these points. The function \( f(x) \) is continuous in the intervals where the denominator does not result in zero. This results in the following intervals of continuity: 1. \( (-\infty, -3) \) 2. \( (-3, 2) \) 3. \( (2, \infty) \) Combining these intervals, the function is continuous on: \[ (-\infty, -3) \cup (-3, 2) \cup (2, \infty) \] So, the correct answer is: \[ (-\infty,-3) \cup(-3,2) \cup(2, \infty) \] Now for some extra relevant tidbits! Did you know that the rational function like \( f(x) \) can often show interesting behavior when you graph them? They typically have vertical asymptotes at points where the denominator is zero, like here at \( x = -3 \) and \( x = 2 \). As a result, the function will spike up or down infinitely around those vertical lines! For the math enthusiasts out there, if you're ever stuck on identifying discontinuities, it helps to remember that factoring is your best buddy! Always break down polynomial expressions to find the critical points, and don't forget to check what happens at those points for insight into the behavior of the function!