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

Ask by Fitzgerald Knight. in the United States
Feb 25,2025

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(a) Domain: (-∞, 2) ∪ (2, ∞) (b) Domain: (-∞, ∞) (c) Domain: (-∞, -√2) ∪ (-√2, √2) ∪ (√2, ∞)

Solution

(a) For k(x) = (x + 6)/(x - 2), the denominator cannot be zero. Setting x - 2 = 0 gives x = 2. Thus, the domain is all real numbers except 2.  Domain: (-∞, 2) ∪ (2, ∞) (b) For j(x) = (x + 6)/(x² + 2), note that x² is always nonnegative, so x² + 2 is always at least 2 (and never zero). Therefore, j(x) is defined for all real numbers.  Domain: (-∞, ∞) (c) For p(x) = (x + 6)/(x² - 2), set the denominator equal to zero: x² - 2 = 0. Solving gives x² = 2, or x = ±√2. These are the values where the function is undefined.  Domain: (-∞, -√2) ∪ (-√2, √2) ∪ (√2, ∞)

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For the function \( k(x)=\frac{x+6}{x-2} \), the domain is all real numbers except where the denominator is zero. Thus, \( x-2 \neq 0 \) which gives us \( x \neq 2 \). In interval notation, the domain is \( (-\infty, 2) \cup (2, \infty) \). For the function \( j(x)=\frac{x+6}{x^{2}+2} \), the denominator \( x^2 + 2 \) is never zero for any real number, since \( x^2 \) is always non-negative and adding 2 keeps it positive. Therefore, the domain in interval notation is \( (-\infty, \infty) \). For the function \( p(x)=\frac{x+6}{x^{2}-2} \), the denominator needs to be considered, \( x^2 - 2 \neq 0 \). This gives us \( x^2 \neq 2 \) or \( x \neq \pm \sqrt{2} \). Thus, the domain in interval notation is \( (-\infty, -\sqrt{2}) \cup (-\sqrt{2}, \sqrt{2}) \cup (\sqrt{2}, \infty) \).

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