Solve the inequality \( \frac{9}{x+1} \geq-5 \). Use exact values when entering the answer.

To solve the inequality \( \frac{9}{x+1} \geq -5 \), we start by isolating the fraction. We can rewrite the inequality as: \[ \frac{9}{x+1} + 5 \geq 0 \] To combine the terms, we need a common denominator. The common denominator is \( x + 1 \): \[ \frac{9}{x+1} + \frac{5(x+1)}{x+1} \geq 0 \] This simplifies to: \[ \frac{9 + 5(x + 1)}{x + 1} \geq 0 \] Expanding the numerator: \[ \frac{9 + 5x + 5}{x + 1} \geq 0 \] This simplifies to: \[ \frac{5x + 14}{x + 1} \geq 0 \] Next, we need to find the critical points where the expression is equal to zero or undefined. The numerator \( 5x + 14 = 0 \) gives: \[ 5x + 14 = 0 \implies 5x = -14 \implies x = -\frac{14}{5} \] The denominator \( x + 1 = 0 \) gives: \[ x + 1 = 0 \implies x = -1 \] Now we have two critical points: \( x = -\frac{14}{5} \) and \( x = -1 \). We will test the intervals determined by these points: 1. \( (-\infty, -\frac{14}{5}) \) 2. \( (-\frac{14}{5}, -1) \) 3. \( (-1, \infty) \) **Testing the intervals:** 1. **Interval \( (-\infty, -\frac{14}{5}) \)**: Choose \( x = -3 \): \[ \frac{5(-3) + 14}{-3 + 1} = \frac{-15 + 14}{-2} = \frac{-1}{-2} = \frac{1}{2} > 0 \] This interval satisfies the inequality. 2. **Interval \( (-\frac{14}{5}, -1) \)**: Choose \( x = -2 \): \[ \frac{5(-2) + 14}{-2 + 1} = \frac{-10 + 14}{-1} = \frac{4}{-1} = -4 < 0 \] This interval does not satisfy the inequality. 3. **Interval \( (-1, \infty) \)**: Choose \( x = 0 \): \[ \frac{5(0) + 14}{0 + 1} = \frac{14}{1} = 14 > 0 \] This interval satisfies the inequality. **Including the critical points:** - At \( x = -\frac{14}{5} \): \[ \frac{5(-\frac{14}{5}) + 14}{-\frac{14}{5} + 1} = \frac{-14 + 14}{-\frac{14}{5} + 1} = \frac{0}{-\frac{9}{5}} = 0 \] This point is included since the inequality is non-strict (≥). - At \( x = -1 \), the expression is undefined, so this point is not included. **Final solution:** The solution to the inequality is: \[ (-\infty, -1) \cup \left[-\frac{14}{5}, \infty\right) \] Thus, the final answer is: \[ \boxed{(-\infty, -1) \cup \left[-\frac{14}{5}, \infty\right)} \]