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Chapter 2: Problem 3

Write the domain of the function in interval notation. $$h(x)=\frac{1}{81-x^{2}}$$

Short Answer

Expert verified
The domain is \( (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \).

Step by step solution

01

- Identify restrictions

To determine the domain of the function, first identify any restrictions. The function is undefined where the denominator is zero. Set the denominator equal to zero and solve for the variable: \[ 81 - x^2 = 0 \]
02

- Solve the equation

Solve the equation for \(x\):\[ 81 - x^2 = 0 \]Rearrange it to find: \[ x^2 = 81 \]Take the square root of both sides: \[ x = \pm 9 \]
03

- Determine the exclusions

Given that \( x = 9 \) and \( x = -9 \) make the denominator zero, exclude these values from the domain. These points are the only restrictions.
04

- Write domain in interval notation

Exclude \( x = 9 \) and \( x = -9 \) from the set of all real numbers. The domain in interval notation is: \[ (-\backslash\textbackslash\textbackslash\textbackslashinfty, -9) \backslashtextbackslash\textbackslash\textbackslashcup (-9, 9) \backslashtextbackslash\textbackslash\textbackslashcup (9, \backslash\textbackslash\textbackslash\textbackslashesinfty) \]

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Key Concepts

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

interval_notation
Interval notation is a way to describe the set of all possible values that a function can take. It is written using intervals to specify the range of values. For example, the interval \((-\infty, -9) \cup (-9, 9) \cup (9, \infty)\) means that the function is defined for all values except for \(x = -9\) and \(x = 9\). The symbol \(\infty\) represents infinity, and \(\cup\) is the union symbol, which combines separate intervals into one set.
restrictions_on_domain
The domain of a function consists of all possible input values (x-values) for which the function is defined. Certain values need to be excluded from the domain to avoid undefined expressions. In the given function \(h(x)=\frac{1}{81-x^{2}}\), we need to find where the denominator is zero, as this makes the function undefined. To do this, solve \(81 - x^2 = 0\). The solutions are \(x = \pm 9\), so these values are excluded from the domain. Therefore, the domain excludes \(x = 9\) and \(x = -9\). The domain is written in interval notation as: \((-\infty, -9) \cup (-9, 9) \cup (9, \infty)\).
solving_quadratic_equations
To find the restrictions on the domain of \(h(x)=\frac{1}{81 - x^2}\), we need to solve the quadratic equation \(81 - x^2 = 0\). This can be broken down into the following steps:

  • Rearrange the equation to isolate \(x^2\): \(x^2 = 81\).
  • Take the square root of both sides to solve for \(x\): \(x = \pm 9\).
These steps identify that the values \(x = 9\) and \(x = -9\) make the denominator zero, leading to restrictions on the domain. This specific process of solving quadratic equations helps us determine where the function is undefined, ensuring we provide the correct domain.

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Most popular questions from this chapter

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