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

Ask by Ramos Smith. in the United States
Jan 23,2025

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Answer

The domain is all real numbers: \( (-\infty, \infty) \).

Solution

### Problem (a) Given the function: \[ w(x) = |x + 3| + 6 \] **Domain:** The absolute value function \( |x + 3| \) is defined for all real numbers. Adding 6 does not impose any restrictions on the domain. \[ \boxed{(-\infty, \infty)} \]

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The domain of the function \( w(x) = |x+3| + 6 \) is all real numbers since the absolute value function is defined for all \( x \). In interval notation, this is expressed as \( (-\infty, \infty) \). For \( y(x) = \frac{x}{|x+3| + 6} \), the denominator \( |x+3| + 6 \) is always positive (since the absolute value is non-negative and adding 6 keeps it positive). Therefore, the domain is also all real numbers, written as \( (-\infty, \infty) \). In the case of \( z(x) = \frac{x}{|x+3| - 6} \), we need to consider when the denominator \( |x+3| - 6 \) is equal to zero. Setting \( |x+3| - 6 = 0 \) gives \( |x+3| = 6 \), leading to \( x + 3 = 6 \) or \( x + 3 = -6 \). Solving these gives \( x = 3 \) or \( x = -9 \). Thus, those points are excluded from the domain, resulting in the intervals \( (-\infty, -9) \cup (-9, 3) \cup (3, \infty) \) in interval notation.

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