State the domain, using interval notation, of the piecewise-defined function x + 2 if 6 13.

Algebra and Trigonometry (6th Edition)
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Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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How do I find the domain of this piecewise function?
**Problem Statement:**

State the domain, using interval notation, of the piecewise-defined function:

\[ 
f(x) = 
\begin{cases} 
\dfrac{x + 2}{x - 8} & \text{if } 6 \leq x < 10 \\
x^2 - 4x + 7 & \text{if } x \geq 13 
\end{cases} 
\]

**Explanation:**

This piecewise-defined function has two expressions, each applicable over specific intervals of \( x \).

1. **First Case:**

   - Expression: \(\dfrac{x + 2}{x - 8}\)
   - Interval: \(6 \leq x < 10\)

   Here, the function is defined for \( x \) values between 6 and 10, including 6 but not including 10. However, we must ensure the denominator is not zero, i.e., \( x \neq 8 \).

2. **Second Case:**

   - Expression: \(x^2 - 4x + 7\)
   - Interval: \(x \geq 13\)

   This polynomial is defined for all \( x \) values greater than or equal to 13.

**Domain in Interval Notation:**

To find the domain of \( f(x) \), consider the defined parts and remove any excluded values:

- From the first case: \([6, 8) \cup (8, 10)\)
- From the second case: \([13, \infty)\)

Thus, the domain of \( f(x) \) in interval notation is:
\[ [6, 8) \cup (8, 10) \cup [13, \infty) \]
Transcribed Image Text:**Problem Statement:** State the domain, using interval notation, of the piecewise-defined function: \[ f(x) = \begin{cases} \dfrac{x + 2}{x - 8} & \text{if } 6 \leq x < 10 \\ x^2 - 4x + 7 & \text{if } x \geq 13 \end{cases} \] **Explanation:** This piecewise-defined function has two expressions, each applicable over specific intervals of \( x \). 1. **First Case:** - Expression: \(\dfrac{x + 2}{x - 8}\) - Interval: \(6 \leq x < 10\) Here, the function is defined for \( x \) values between 6 and 10, including 6 but not including 10. However, we must ensure the denominator is not zero, i.e., \( x \neq 8 \). 2. **Second Case:** - Expression: \(x^2 - 4x + 7\) - Interval: \(x \geq 13\) This polynomial is defined for all \( x \) values greater than or equal to 13. **Domain in Interval Notation:** To find the domain of \( f(x) \), consider the defined parts and remove any excluded values: - From the first case: \([6, 8) \cup (8, 10)\) - From the second case: \([13, \infty)\) Thus, the domain of \( f(x) \) in interval notation is: \[ [6, 8) \cup (8, 10) \cup [13, \infty) \]
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