The domain of the function \(\mathrm{f}(\mathrm{x})=[1 / \sqrt{(|\mathrm{x}|-\mathrm{x})]}\) is (a) \((-\infty, \infty)\) (b) \((0, \infty)\) (c) \((-\infty, 0)\) (d) \((-\infty, \infty)-\\{0\\}\)

Short Answer

Expert verified
The domain of the function \(f(x) = \cfrac{1}{\sqrt{|x| - x}}\) is (c) \((- \infty, 0)\), as the function is defined for all negative values of x.

Step by step solution

01

Identify restrictions

In the given function, \(f(x) = \cfrac{1}{\sqrt{|x| - x}}\), the restrictions we need to consider are: 1. The denominator of a fraction must not be equal to zero. 2. The number inside the square root should be non-negative. Now we will analyze the expressions that could cause these restrictions. The denominator of the function is \(\sqrt{|x|-x}\) and we need to make sure it's not equal to 0.
02

Identify possible values of the argument of the square root

We have the argument of the square root: \(|x|-x\). Let's analyze it for different values of x. 1. If \(x\) is positive, \(|x|=x\) and the argument will be: \[x-x=0\] 2. If \(x\) is negative, \(|x|=-x\) and the argument will be: \[-x-x=-2x\] 3. If \(x\) is zero, \(|0|=0\) and the argument will be: \[0-0=0\]
03

Check the restrictions for the argument of the square root

Now let's check the restrictions: 1. For the positive values of \(x\), the argument will be 0. Therefore, the function is not defined for positive values of \(x\). 2. For the negative values of \(x\), the argument will be \(-2x\), which is always non-negative and not equal to 0. So, the function is defined for negative values of \(x\). 3. For \(x=0\), the argument will be 0. Therefore, the function is not defined for \(x=0\). Based on our analysis, the domain of the function \(f(x) = \cfrac{1}{\sqrt{|x| - x}}\) consists of all negative values of x. The correct answer is (c) \((- \infty, 0)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\mathrm{f}(\mathrm{x})=\sec \mathrm{x}+\tan \mathrm{x}, \mathrm{g}(\mathrm{x})=[(\tan \mathrm{x}) /\\{1-\sec (\mathrm{x})\\}]\) Statement \(-1 \mathrm{~g}\) is an odd function Statement \(-2 \mathrm{f}\) is neither an odd function nor an even function (a) Statement lis true (b) Statement 2 is true (c) 1 and 2 both are true (d) 1 and 2 both are false

If \(\mathrm{f}(\mathrm{x})\) is a polynomial satisfying \(\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(1 / \mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{f}(1 / \mathrm{x})\) and \(\mathrm{f}(3)=28\) then \(\mathrm{f}(4)=\ldots \ldots\) (a) 65 (b) 17 (c) 63 (d) 15

In a collage of 400 students every students read 5 newspapers and every newspaper is read by 80 students. The number of newspaper is (a) 25 (b) at the most 20 (c) at the most 25 (d) at least 25

The set \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap\left(\mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right)^{\prime} \cap \mathrm{C}^{\prime}\) equals (a) \(\mathrm{B} \cap \mathrm{C}^{\prime}\) (b) \(\mathrm{B} \cup \mathrm{C}^{\prime}\) (c) \(\mathrm{A} \cap \mathrm{C}\) (d) \(\mathrm{A} \cup \mathrm{C}\)

In a certain town \(30 \%\) families own a scooter and \(40 \%\) on a car \(50 \%\) own neither a scooter nor a car 2000 families own both a scooter and car consider the following statements in this regard (1) \(20 \%\) families own both scooter and car (2) \(35 \%\) families own either a car or a scooter (3) 10000 families live in town. Which of the above statements are correct? (a) 2 and 3 (b) 1,2 and 3 (c) 1 and 2 (d) 1 and 3

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free