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Find the domain of the following function. \( g(x)=\frac{1}{\sqrt{13+x}} \) The domain is \( \square \). (Type your answer in interval notation.)

Ask by Ramirez Donnelly. in the United States
Jan 27,2025

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Answer

The domain of \( g(x) = \frac{1}{\sqrt{13 + x}} \) is all real numbers greater than \(-13\), which is written as \((-13, \infty)\).

Solution

To determine the domain of the function \( g(x) = \frac{1}{\sqrt{13 + x}} \), we need to ensure that the expression inside the square root is positive and not zero (since it's in the denominator). 1. **Square Root Condition:** \[ 13 + x > 0 \implies x > -13 \] This ensures that the square root is defined and positive. 2. **Denominator Condition:** \[ \sqrt{13 + x} \neq 0 \implies 13 + x \neq 0 \implies x \neq -13 \] This prevents division by zero. Combining both conditions, the domain of \( g(x) \) is all real numbers greater than \(-13\). **Domain:** \( (-13, \infty) \) **Answer:** \((-13, \infty)\)

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To find the domain of the function \( g(x)=\frac{1}{\sqrt{13+x}} \), we must ensure that the expression under the square root is positive since square roots of negative numbers are not defined in the realm of real numbers. Therefore, we have: 1. \( 13+x > 0 \) leads to \( x > -13 \). 2. We also need to ensure that the entire denominator does not equal zero, which means \( \sqrt{13+x} \neq 0 \). This further reinforces the condition \( 13+x > 0 \). So, the solution to the inequality is \( x > -13 \). In interval notation, the domain of \( g(x) \) is \( (-13, \infty) \).

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