That is, when $$ \begin{aligned} 2 x^{2}+11 x-6 & =0 \ (x+6)(2 x-1) & =0 \ x & =\frac{1}{2} \text { and } x=-6 \end{aligned} $$ Part: $$ 2 / 3 $$ Part 3 of 3 The values $$ x=-6 $$ and $$ x=\frac{1}{2} $$ must be excluded from the domain. Therefore, the domain of the function in interval notation is $$ \square $$

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To determine the domain of the function given the roots $$ x = -6 $$ and $$ x = \frac{1}{2} $$, we need to exclude these values from the set of all real numbers.

1. **Identify the set of all real numbers**: The domain of a function is typically all real numbers, which can be represented in interval notation as $$ (-\infty, \infty) $$.

2. **Exclude the roots**: Since the values $$ x = -6 $$ and $$ x = \frac{1}{2} $$ must be excluded, we need to break the interval into parts that do not include these points.

3. **Construct the intervals**:
   - From $$ -\infty $$ to $$ -6 $$: This interval is $$ (-\infty, -6) $$.
   - From $$ -6 $$ to $$ \frac{1}{2} $$: This interval is $$ (-6, \frac{1}{2}) $$.
   - From $$ \frac{1}{2} $$ to $$ \infty $$: This interval is $$ (\frac{1}{2}, \infty) $$.

4. **Combine the intervals**: The domain of the function, excluding the points $$ -6 $$ and $$ \frac{1}{2} $$, can be expressed in interval notation as:
   $$
   (-\infty, -6) \cup (-6, \frac{1}{2}) \cup (\frac{1}{2}, \infty)
   $$

Thus, the domain of the function in interval notation is:
$$
(-\infty, -6) \cup (-6, \frac{1}{2}) \cup (\frac{1}{2}, \infty)
$$
The domain of the function is all real numbers except $$ x = -6 $$ and $$ x = \frac{1}{2} $$. In interval notation, this is: $$ (-\infty, -6) \cup (-6, \frac{1}{2}) \cup (\frac{1}{2}, \infty) $$

That is, when

Part:

Part 3 of 3

The values and must be excluded from the domain.

Therefore, the domain of the function in interval notation is

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That is, when Part: Part 3 of 3 The values and must be excluded from the domain. Therefore, the domain of the function in interval notation is