/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Find the domain of the function.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 26

Find the domain of the function. $$ f(x)=\frac{1}{\sqrt{|x|-x}} $$

Short Answer

Expert verified
The domain of the function \(f(x)=\frac{1}{\sqrt{|x|-x}}\) is \(x \geq 0\), or in interval notation: \([0, \infty)\).

Step by step solution

01

Identify the restrictions for the square root function.

The first restriction comes from the square root in the function: \(\sqrt{|x| - x}\). The value inside the square root must be non-negative, meaning: $$|x| - x \geq 0$$
02

Break down the absolute value function.

Since \(|x|\) is the absolute value of \(x\), we have two cases to consider: Case 1: \(x \geq 0\) In this case, \(|x| = x\), so: $$x - x \geq 0 \Rightarrow 0 \geq 0$$ The inequality is true for all \(x \geq 0\). Case 2: \(x < 0\) In this case, \(|x| = -x\), so: $$-x - x \geq 0 \Rightarrow 2x \leq 0 \Rightarrow x \leq 0$$ The inequality is true for all values of \(x \leq 0\).
03

Combine the cases and find the common solution.

From Step 2, we have two possible solutions: 1. \(x \geq 0\) 2. \(x \leq 0\) These inequalities define the whole real line, meaning the function is defined for all real numbers concerning the square root part.
04

Identify the restrictions for the denominator.

The second restriction comes from the denominator: \(\frac{1}{\sqrt{|x|-x}}\). It will be undefined when \(|x|-x = 0\). Solve for x: $$|x| - x = 0$$ We know that \(|x|-x\) is non-negative from Step 1 and Step 2, so: $$|x| = x$$ This equation is only true when \(x \geq 0\).
05

Determine the domain.

We now have two conditions for the domain of the function: 1. For the square root part, the function is defined for all real numbers: \(\mathbb{R}\). 2. For the denominator, the function is only defined when \(x \geq 0\). Taking both conditions into account, the domain of the function is the intersection of these sets: $$Domain(f(x)) = \{x \in \mathbb{R} : x \geq 0\}$$

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 91影视!

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

Find the zero(s) of the function f to five decimal places. $$ f(x)=x^{3}-9 x+4 $$

Find the exact value of the given expression. $$ \sin ^{-1} \frac{1}{2} $$

Find the inverse of \(f .\) Then sketch the graphs of \(f\) and \(f^{-1}\) on the same set of axes. $$ f(x)=x^{3}+1 $$

Write the expression in algebraic form. $$ \sec \left(\sin ^{-1} x\right) $$

Find \(f^{-1}(a)\) for the function \(f\) and the real number \(a\). $$ f(x)=\frac{3}{\pi} x+\sin x ; \quad-\frac{\pi}{2}
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation
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.