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Chapter 1: Problem 104

Domain Determine the values of the variable for which the expression is defined as a real number. $$\sqrt[4]{\frac{1-x}{2+x}}$$

Short Answer

Expert verified
The domain is \((-2, 1]\).

Step by step solution

01

Understand the conditions for a real root

The expression given is a fourth root, \(\sqrt[4]{\frac{1-x}{2+x}}\). For a fourth root to be defined as a real number, the radicand, \(\frac{1-x}{2+x}\), must be non-negative. This means \(\frac{1-x}{2+x} \geq 0\).
02

Analyze the inequality \(\frac{1-x}{2+x} \geq 0\)

To solve \(\frac{1-x}{2+x} \geq 0\), we identify where the expression equals zero or changes sign. The fraction is zero when the numerator is zero, i.e., \(1-x = 0\) or \(x = 1\). The fraction is undefined when the denominator is zero, i.e., \(2+x = 0\) or \(x = -2\). Evaluate these critical points and intervals around them to determine where the fraction is non-negative.
03

Evaluate intervals based on critical points

Consider the intervals based on critical points, \(x = -2\) and \(x = 1\):- For \(x < -2\), choose a test point, say \(x = -3\): \(\frac{1 - (-3)}{2 + (-3)} = \frac{4}{-1} = -4\), which is negative.- For \(-2 < x < 1\), choose \(x = 0\): \(\frac{1 - 0}{2 + 0} = \frac{1}{2}\), which is positive.- For \(x > 1\), choose \(x = 2\): \(\frac{1 - 2}{2 + 2} = \frac{-1}{4}\), which is negative.Thus, the expression is non-negative only for \(-2 < x \leq 1\).
04

Conclude the domain definition

From the analysis, conclude that the domain of the expression is all values of \(x\) in the interval \((-2, 1]\). These are the values for which \(\sqrt[4]{\frac{1-x}{2+x}}\) remains defined as a real number.

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Key Concepts

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

Radicand non-negativity
When dealing with expressions that involve roots, it's essential to ensure the expression inside the root, known as the radicand, meets specific conditions to remain defined. For the given expression, \[ \sqrt[4]{\frac{1-x}{2+x}} \]to be a real number, the radicand, \( \frac{1-x}{2+x} \), must be non-negative.In mathematical terms, this means that the radicand must satisfy the inequality:\[ \frac{1-x}{2+x} \geq 0 \]This condition ensures that only real values, not imaginary ones, are considered. Imaginary numbers arise when taking even roots of negative numbers, which we want to avoid in our scenario.
Inequalities
Solving inequalities is akin to solving equations, but requires careful handling of the inequality sign. For our radicand \( \frac{1-x}{2+x} \geq 0 \), we need to find where this fraction is non-negative. A systematic way to tackle this problem involves two steps:
  • Identify when the numerator is zero, which gives us \( x = 1 \).
  • Determine when the denominator is zero, making the fraction undefined; here, \( x = -2 \).
With these critical points, explore the different intervals they create on the number line: \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \). By testing points from each interval in the inequality, we verify if \( \frac{1-x}{2+x} \) is non-negative in those intervals.
Interval notation
Once the intervals where the inequality holds true are identified, express the solution using interval notation. Interval notation is a compact way of describing a range of values, useful for communicating domains of functions.In our scenario, testing the points around the critical values \( x = -2 \) and \( x = 1 \) showed that the expression is non-negative only between them. Specifically, this is true from \( x = -2 \) (not inclusive) to \( x = 1 \) (inclusive), written as the interval:\[ (-2, 1] \]Remember:
  • "(" or ")" means the endpoint is not included ("open interval").
  • "[" or "]" means the endpoint is included ("closed interval").
In this way, interval notation concisely communicates the domain of the function defined as real numbers.

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Most popular questions from this chapter

Complete the following tables. What happens to the \(n\) th root of 2 as \(n\) gets large? What about the \(n\) th root of \(\frac{1}{2} ?\) $$\begin{array}{|r|r|}\hline n & 2^{1 / n} \\\\\hline 1 & \\\2 & \\\5 & \\\10 & \\\100 & \\\\\hline\end{array}$$ $$\begin{array}{|r|r|} \hline n & \left(\frac{1}{2}\right)^{1 / n} \\\\\hline 1 & \\\2 & \\\5 & \\\10 & \\\100 & \\\\\hline\end{array}$$ Construct a similar table for \(n^{1 / n}\). What happens to the \(n\) th root of \(n\) as \(n\) gets large?

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Simplify the expression. (a) \(\left(\frac{x^{8} y^{-4}}{16 y^{4 / 3}}\right)^{-1 / 4}\) (b) \(\left(\frac{4 s^{3} t^{4}}{s^{2} t^{9 / 2}}\right)^{-1 / 2}\)

Scientific Notation Write each number in scientific notation. (a) \(129,540,000\) (b) \(7,259,000,000\) (c) 0.0000000014 (d) 0.0007029

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