At what \( x \) - value will the point discontinuities of the given rational function below occur? \( f(x)=\frac{16 x(x+3)(x+6)}{x^{2}-36} \) \( x=0, x=-3 \), and \( x=-6 \) \( x=0 \) and \( x=-3 \) \( x=-6 \) \( x=0 \) only \( x \)

To determine the \( x \)-values where point discontinuities occur for the given rational function: \[ f(x) = \frac{16x(x+3)(x+6)}{x^2 - 36} \] **Step 1: Factor the Denominator** The denominator can be factored using the difference of squares: \[ x^2 - 36 = (x - 6)(x + 6) \] So, the function becomes: \[ f(x) = \frac{16x(x+3)(x+6)}{(x - 6)(x + 6)} \] **Step 2: Identify Potential Discontinuities** Point discontinuities occur where both the numerator and the denominator are zero, leading to an undefined expression that can potentially be removed by canceling common factors. 1. **At \( x = -6 \):** - Numerator: \( 16(-6)(-6+3)(-6+6) = 16(-6)(-3)(0) = 0 \) - Denominator: \( (-6 - 6)(-6 + 6) = (-12)(0) = 0 \) - Both numerator and denominator are zero, indicating a potential point discontinuity. 2. **At \( x = 6 \):** - Numerator: \( 16(6)(6+3)(6+6) \neq 0 \) - Denominator: \( (6 - 6)(6 + 6) = 0 \times 12 = 0 \) - Only the denominator is zero, indicating a vertical asymptote, not a point discontinuity. 3. **At \( x = 0 \) and \( x = -3 \):** - The denominator is not zero, so these points are continuous. **Step 3: Simplify and Confirm** After canceling the common factor \( (x + 6) \): \[ f(x) = \frac{16x(x+3)}{x - 6} \quad \text{for} \quad x \neq -6 \] This confirms that \( x = -6 \) is a removable discontinuity (a point discontinuity). **Conclusion** The function \( f(x) \) has a point discontinuity at: **\( x = -6 \)** Answer: \( x=-6 \)