Write the domain in interval notation. \( \begin{array}{l}\text { (a) } k(x)=\frac{x+7}{x-6} \\ \text { (b) } j(x)=\frac{x+7}{x^{2}+6} \\ \text { (c) } p(x)=\frac{x+7}{x^{2}-6}\end{array} \)

Let's determine the domain for each function by identifying the values of \( x \) that make the denominator zero (since division by zero is undefined). We'll express the domain in interval notation. ### (a) \( k(x) = \frac{x+7}{x-6} \) - **Denominator:** \( x - 6 \) - **Undefined when:** \( x - 6 = 0 \Rightarrow x = 6 \) **Domain:** All real numbers except \( x = 6 \). **Interval Notation:** \[ (-\infty, 6) \cup (6, \infty) \] --- ### (b) \( j(x) = \frac{x+7}{x^2 + 6} \) - **Denominator:** \( x^2 + 6 \) - **Undefined when:** \( x^2 + 6 = 0 \Rightarrow x^2 = -6 \) Since \( x^2 = -6 \) has no real solutions, the denominator is never zero. **Domain:** All real numbers. **Interval Notation:** \[ (-\infty, \infty) \] --- ### (c) \( p(x) = \frac{x+7}{x^2 - 6} \) - **Denominator:** \( x^2 - 6 \) - **Undefined when:** \( x^2 - 6 = 0 \Rightarrow x^2 = 6 \Rightarrow x = \sqrt{6} \) or \( x = -\sqrt{6} \) **Domain:** All real numbers except \( x = \sqrt{6} \) and \( x = -\sqrt{6} \). **Interval Notation:** \[ (-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty) \] --- **Summary:** - **(a)** \( (-\infty, 6) \cup (6, \infty) \) - **(b)** \( (-\infty, \infty) \) - **(c)** \( (-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty) \)