/*! 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 126 Write the domain in interval not... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 126

Write the domain in interval notation. $$ d(x)=\log \left(\frac{1}{\sqrt{x+8}}\right) $$

Short Answer

Expert verified
The domain is \( (-8, \infty) \).

Step by step solution

01

Identify domain restrictions for logarithmic functions

Logarithmic functions are only defined for positive arguments. Therefore, the argument of the logarithm must be greater than 0. Set up the inequality: \( \frac{1}{\root 2 \left( x+8 \right) } > 0 \).
02

Solve the inequality for the argument

Since the reciprocal and the square root are always positive for positive arguments, the expression inside the square root (\( x+8 \)) must be positive. Set up the inequality: \( x + 8 > 0 \).
03

Solve for \(x\)

Isolate \(x\) by subtracting 8 from both sides: \( x > -8 \). Therefore, the domain of the function is all values of \(x\) greater than \(-8\).
04

Write the domain in interval notation

Translate the inequality \( x > -8 \) into interval notation. The domain in interval notation is \( (-8, \infty) \).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interval Notation
Interval notation is a way of representing a set of numbers along a number line. It is particularly useful for expressing domains and solutions to inequalities.
In interval notation, intervals are written using brackets or parentheses:
  • Parentheses \( ( \text ) \) mean the endpoint is not included (open interval).
  • Brackets \( [ \text ] \) mean the endpoint is included (closed interval).
For the exercise, we need to express the domain of the logarithmic function using interval notation.
The domain was found to be all values of \ x \ greater than \ -8 \ (\( x > -8 \)). This is represented as \ (-8, \infty) \ in interval notation, signifying that the domain includes all numbers greater than \ -8 \ but not \ -8 \ itself.
Solving Inequalities
Solving inequalities involves finding all possible values of a variable that satisfy an inequality.
Let鈥檚 break it down:
  • **Identify the inequality:** From the exercise, we know \ x + 8 > 0 \ needs to be satisfied for the logarithm to be defined.
  • **Isolate the variable:** Subtract 8 from both sides to solve for \ x \:
    \( x > -8 \)
By solving this inequality, we determine the range of values for \ x \ that makes the expression valid.
This ensures our logarithmic function's argument is always positive, as it must be for the function to be defined.
Domain Restrictions
Domain restrictions specify the values of \ x \ for which a function is defined.
For logarithmic functions, the argument must be positive because the logarithm of a non-positive number is undefined.
  • In the given function, \ log \( \frac{1}{\sqrt{x+8}} \) , the argument \( \frac{1}{\sqrt{x+8}} \) must be greater than 0.
  • Since the fraction and square root are inherently positive, it simplifies to \ x + 8 > 0 \ .
This domain restriction ensures the argument of the logarithmic function is positive.
Understanding and identifying these restrictions is crucial as they maintain the validity of the function across its domain.

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

Determine if the statement is true or false. For each false statement, provide a counterexample. For example, \(\log (x+y) \neq \log x+\log y\) because \(\log (2+8) \neq \log 2+\log 8\) (the left side is 1 and the right side is approximately 1.204 ). $$ \log _{5}\left(\frac{1}{x}\right)=\frac{1}{\log _{5} x} $$

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(e^{2 x}-6 e^{x}-16=0\)

The population of Canada \(P(t)\) (in millions) since January \(1,1900,\) can be approximated by $$P(t)=\frac{55.1}{1+9.6 e^{-0.02515 t}}$$ where \(t\) is the number of years since January 1,1900 . a. Evaluate \(P(0)\) and interpret its meaning in the context of this problem. b. Use the function to predict the Canadian population on January \(1,2015 .\) Round to the nearest million. c. Use the function to predict the Canadian population on January 1,2040 . d. Determine the year during which the Canadian population will reach 45 million. e. What value will the term \(\frac{9.6}{e^{0.02515 t}}\) approach as \(t \rightarrow \infty\) ? f. Determine the limiting value of \(P(t)\).

Solve for the indicated variable. \(N=N_{0} e^{-0.025 t}\) for \(t\) (used in chemistry)

After a new product is launched the cumulative sales \(S(t)\) (in \(\$ 1000) t\) weeks after launch is given by: $$S(t)=\frac{72}{1+9 e^{-0.36 t}}$$ a. Determine the cumulative amount in sales 3 weeks after launch. Round to the nearest thousand. b. Determine the amount of time required for the cumulative sales to reach \(\$ 70,000\). c. What is the limiting value in sales?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

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.