Write the domain in interval notation. (a) \( w(x)=|x+3|+6 \) (b) \( y(x)=\frac{x}{|x+3|+6} \) (c) \( z(x)=\frac{x}{|x+3|-6} \)

Let's determine the domain for each of the given functions by identifying all real numbers \( x \) for which the function is defined. ### (a) \( w(x) = |x + 3| + 6 \) **Analysis:** - The absolute value function \( |x + 3| \) is defined for all real numbers \( x \). - Adding 6 doesn't impose any additional restrictions. **Domain:** All real numbers. **Interval Notation:** \[ (-\infty, \infty) \] --- ### (b) \( y(x) = \frac{x}{|x + 3| + 6} \) **Analysis:** - The denominator \( |x + 3| + 6 \) is always greater than or equal to 6 since \( |x + 3| \geq 0 \). - Therefore, the denominator never equals zero, and the function is defined for all real numbers \( x \). **Domain:** All real numbers. **Interval Notation:** \[ (-\infty, \infty) \] --- ### (c) \( z(x) = \frac{x}{|x + 3| - 6} \) **Analysis:** - The denominator \( |x + 3| - 6 \) must not be zero to avoid division by zero. - Set \( |x + 3| - 6 \neq 0 \): \[ |x + 3| \neq 6 \] - Solving \( |x + 3| = 6 \): \[ x + 3 = 6 \quad \text{or} \quad x + 3 = -6 \] \[ x = 3 \quad \text{or} \quad x = -9 \] - Thus, \( x \neq 3 \) and \( x \neq -9 \). **Domain:** All real numbers except \( x = 3 \) and \( x = -9 \). **Interval Notation:** \[ (-\infty, -9) \cup (-9, 3) \cup (3, \infty) \] --- ### **Summary of Domains** - **(a)** \( w(x) = |x + 3| + 6 \): \[ (-\infty, \infty) \] - **(b)** \( y(x) = \frac{x}{|x + 3| + 6} \): \[ (-\infty, \infty) \] - **(c)** \( z(x) = \frac{x}{|x + 3| - 6} \): \[ (-\infty, -9) \cup (-9, 3) \cup (3, \infty) \]