/*! 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 68 Find the domain of the function.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 68

Find the domain of the function. $$ h(x)=\sqrt{x-2}-\log _{5}(10-x) $$

Short Answer

Expert verified
The domain of the function is \([2, 10)\).

Step by step solution

01

Analyze the Square Root

For the square root term \( \sqrt{x-2} \), the expression inside the square root cannot be negative. Therefore, \( x-2 \geq 0 \), which simplifies to \( x \geq 2 \). This means that \( x \) must be greater than or equal to 2 to keep the expression defined.
02

Analyze the Logarithm

For the logarithmic term \( \log_{5}(10-x) \), the argument of the logarithm must be greater than zero. Therefore, \( 10-x > 0 \). This simplifies to \( x < 10 \). Hence, \( x \) must be less than 10.
03

Combine the Conditions

To find the domain of the function \( h(x) \), we need to satisfy both conditions derived from the previous steps: \( x \geq 2 \) and \( x < 10 \). These can be combined into a single condition: \( 2 \leq x < 10 \). This interval represents all the values that \( x \) can take to ensure both the square root and logarithm are defined and real.

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.

Understanding the Square Root
When dealing with functions that include a square root, it's important to make sure that the expression inside the square root is not negative. This is because the square root of a negative number is not defined in the set of real numbers.In the function given, we have the square root component as \(\sqrt{x-2}\). To ensure this part of the function is valid, the expression \(x-2\) must be non-negative. This means we solve the inequality \(x-2 \geq 0\). Solving for \(x\), we find \(x \geq 2\), which tells us that \(x\) can be any number from 2 onwards, including 2 itself.This requirement ensures that the square root term stays defined and results in a real number.
Exploring the Logarithm
Logarithms are another key concept that influences the domain of many functions. The argument of a logarithm, which is the part inside the log function, must always be positive. This is because you cannot take the logarithm of zero or a negative number.In our function \(h(x)\), the logarithmic term is \(\log_{5}(10-x)\). For this term to be defined, we must have \(10-x > 0\). Solving this inequality gives us \(x < 10\). Thus, \(x\) must be less than 10 for the logarithm to be valid.This condition ensures that the argument of the log function remains positive, so the value of the logarithm stays defined and gives a real number.
Combining Inequalities for Domain
To find the domain of the function, we need to combine the conditions derived from both the square root and the logarithm parts of the function. The conditions are:
  • For \(\sqrt{x-2}\), we have \(x \geq 2\)
  • For \(\log_{5}(10-x)\), we have \(x < 10\)
By combining these inequalities, we create a domain that meets both requirements simultaneously. We form the interval \(2 \leq x < 10\), which includes all values of \(x\) that ensure both components of the function remain defined. This interval is the permitted range of \(x\) values that keep the entire function real and meaningful. Therefore, the domain of the function \(h(x)\) is \(x\) values between 2 and 10, where 2 is included, and 10 is not.

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

\(25-28=\) These exercises use Newton's Law of Cooling. Cooling Soup A hot bowl of soup is served at a dinner party. It starts to cool according to Newton's Law of Cooling, so its temperature at time \(t\) is given by $$ T(t)=65+145 e^{-0.05 t} $$ where \(t\) is measured in minutes and \(T\) is measured in \(^{\circ} \mathrm{F}\) . (a) What is the initial temperature of the soup? (b) What is the temperature after 10 \(\mathrm{min}\) ? (c) After how long will the temperature be \(100^{\circ} \mathrm{F} ?\)

\(29-43\) . These exercises deal with logarithmic scales. Inverse Square Law for Sound A law of physics states that the intensity of sound is inversely proportional to the square of the distance \(d\) from the source: \(I=k / d^{2} .\) (a) Use this model and the equation $$ B=10 \log \frac{I}{I_{0}} $$ (described in this section) to show that the decibel levels \(B_{1}\) and \(B_{2}\) at distances \(d_{1}\) and \(d_{2}\) from a sound source are related by the equation $$ B_{2}=B_{1}+20 \log \frac{d_{1}}{d_{2}} $$ (b) The intensity level at a rock concert is 120 \(\mathrm{dB}\) at a distance 2 \(\mathrm{m}\) from the speakers. Find the intensity level at a distance of \(10 \mathrm{m} .\)

Bacteria Culture A certain culture of the bacterium Streptococcus \(A\) initially has 10 bacteria and is observed to double every 1.5 hours. (a) Find an exponential model \(n(t)=n_{0} 2^{t / 2}\) for the number of bacteria in the culture after \(t\) hours. (b) Estimate the number of bacteria after 35 hours. (c) When will the bacteria count reach \(10,000 ?\)

Use the definition of the logarithmic function to find \(x\). $$ \begin{array}{ll}{\text { (a) } \log _{4} 2=x} & {\text { (b) } \log _{4} x=2}\end{array} $$

Suppose you鈥檙e driving your car on a cold winter day \(\left(20^{\circ} \mathrm{F} \text { outside) and the engine overheats (at }\right.\) about \(220^{\circ} \mathrm{F}\) ). When you park, the engine begins to cool down. The temperature \(T\) of the engine \(t\) minutes after you park satisfies the equation $$\ln \left(\frac{T-20}{200}\right)=-0.11 t$$ (a) Solve the equation for \(T\) . (b) Use part (a) to find the temperature of the engine after \(20 \min (t=20) .\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

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.