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 $$ x $$

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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: $$ x=-6 $$
The function has a point discontinuity at $$ x = -6 $$.

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 \( x \)

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