Which answer includes the intervals that contain the solution to the inequality? (x^2-1)/(3x+9)leqslant 0 [-3,-1)cup (1,infty ) (-3,-1]cup [1,infty ) C [-infty ,-3)cup (-1,1) (-infty ,-3)cup [-1,1]

To solve the inequality $\frac{x^{2}-1}{3x+9}\leqslant 0$, we need to find the values of $x$ for which the expression is less than or equal to zero.<br /><br />First, let's find the critical points by setting the numerator and denominator equal to zero:<br /><br />$x^{2}-1=0$ gives $x=\pm 1$<br /><br />$3x+9=0$ gives $x=-3$<br /><br />Now, we can test the intervals between the critical points to determine where the expression is less than or equal to zero.<br /><br />Interval $(-\infty,-3)$:<br />Choose $x=-4$ as a test point.<br />$\frac{(-4)^{2}-1}{3(-4)+9}=\frac{15}{-3}=-5<0$<br /><br />Interval $(-3,-1)$:<br />Choose $x=-2$ as a test point.<br />$\frac{(-2)^{2}-1}{3(-2)+9}=\frac{3}{3}=1>0$<br /><br />Interval $(-1,1)$:<br />Choose $x=0$ as a test point.<br />$\frac{0^{2}-1}{3(0)+9}=\frac{-1}{9}<0$<br /><br />Interval $(1,\infty)$:<br />Choose $x=2$ as a test point.<br />$\frac{2^{2}-1}{3(2)+9}=\frac{3}{15}>0$<br /><br />Therefore, the intervals that contain the solution to the inequality are $(-\infty,-3)\cup (-1,1)$.<br /><br />So, the correct answer is C $[-\infty,-3)\cup (-1,1)$.