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The domain of the function f(x)=1/(sqrt(...

The domain of the function `f(x)=1/(sqrt(|x|-x))` is: (1) `(-oo,oo)` (2) `(0,oo` (3) `(-oo,""0)` (4) `(-oo,oo)"-"{0}`

A

`(-oo,oo) ~{0}`

B

`(-oo,oo)`

C

`(0,oo)`

D

(-oo,0)`

Text Solution

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The correct Answer is:
To find the domain of the function \( f(x) = \frac{1}{\sqrt{|x| - x}} \), we need to ensure that the expression inside the square root is non-negative and that the denominator is not equal to zero. Let's break this down step by step. ### Step 1: Analyze the expression inside the square root The expression inside the square root is \( |x| - x \). We need to determine when this expression is non-negative: 1. **Case 1:** When \( x \geq 0 \) - Here, \( |x| = x \). - Thus, \( |x| - x = x - x = 0 \). - The square root \( \sqrt{0} \) is defined, but the denominator becomes zero, which makes the function undefined. 2. **Case 2:** When \( x < 0 \) - Here, \( |x| = -x \). - Thus, \( |x| - x = -x - x = -2x \). - We need \( -2x \geq 0 \) for the square root to be defined. - This simplifies to \( x \leq 0 \). ### Step 2: Determine the conditions for the function to be defined From the analysis: - For \( x \geq 0 \), the function is undefined because the denominator becomes zero. - For \( x < 0 \), the expression \( -2x \) is non-negative, which means the square root is defined. ### Step 3: Conclusion about the domain The function \( f(x) \) is defined for all negative values of \( x \). Therefore, the domain of the function is: \[ (-\infty, 0) \] ### Final Answer The correct option is: (3) \( (-\infty, 0) \) ---

To find the domain of the function \( f(x) = \frac{1}{\sqrt{|x| - x}} \), we need to ensure that the expression inside the square root is non-negative and that the denominator is not equal to zero. Let's break this down step by step. ### Step 1: Analyze the expression inside the square root The expression inside the square root is \( |x| - x \). We need to determine when this expression is non-negative: 1. **Case 1:** When \( x \geq 0 \) - Here, \( |x| = x \). ...
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