Solve the following rational inequality \( \frac{x-3}{x^{2}-36}>0 \). Give your answer using interval notation.

To solve the rational inequality \( \frac{x-3}{x^{2}-36}>0 \), we first need to factor the denominator. The expression \( x^{2}-36 \) can be factored as \( (x-6)(x+6) \). Thus, our inequality becomes: \[ \frac{x-3}{(x-6)(x+6)} > 0 \] Next, we identify the critical points where the numerator and denominator are zero: 1. Numerator: \( x - 3 = 0 \) gives \( x = 3 \) 2. Denominator: \( (x-6)(x+6) = 0 \) gives \( x = 6 \) and \( x = -6 \) The critical points are \( x = -6, 3, 6 \). We will now test the intervals formed by these points: \( (-\infty, -6) \), \( (-6, 3) \), \( (3, 6) \), and \( (6, \infty) \). 1. **Interval \( (-\infty, -6) \)**: Choose \( x = -7 \): \[ \frac{-7-3}{(-7-6)(-7+6)} = \frac{-10}{(-13)(-1)} = \frac{-10}{13} < 0 \] 2. **Interval \( (-6, 3) \)**: Choose \( x = 0 \): \[ \frac{0-3}{(0-6)(0+6)} = \frac{-3}{(-6)(6)} = \frac{-3}{-36} = \frac{1}{12} > 0 \] 3. **Interval \( (3, 6) \)**: Choose \( x = 5 \): \[ \frac{5-3}{(5-6)(5+6)} = \frac{2}{(-1)(11)} = \frac{2}{-11} < 0 \] 4. **Interval \( (6, \infty) \)**: Choose \( x = 7 \): \[ \frac{7-3}{(7-6)(7+6)} = \frac{4}{(1)(13)} = \frac{4}{13} > 0 \] Now, we summarize the signs for each interval: - \( (-\infty, -6) \): negative - \( (-6, 3) \): positive - \( (3, 6) \): negative - \( (6, \infty) \): positive The inequality \( \frac{x-3}{(x-6)(x+6)} > 0 \) holds true in the intervals \( (-6, 3) \) and \( (6, \infty) \). Lastly, we check the critical points: - At \( x = -6 \) and \( x = 6 \), the expression is undefined. - At \( x = 3 \), the expression is zero (not included). Thus, the solution in interval notation is: \[ (-6, 3) \cup (6, \infty) \]