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Chapter 3: Problem 78

What is the domain of the function \(\left(1+x^{2}\right)^{1 / 8} ?\)

Short Answer

Expert verified
The domain of the function \(\left(1+x^{2}\right)^{1 / 8}\) is all real numbers, expressed as the interval notation \((-\infty, \infty)\) or in set notation \(\{x: x \in \mathbb{R}\}\), since \(1+x^2\) is always non-negative for all x.

Step by step solution

01

Identify the expression inside the root

The expression inside the 8th root is \(1+x^2\). Our goal is to find values of x such that this expression is non-negative.
02

Determine when the expression inside the root is non-negative

The expression is \(1+x^2\). We want to find when this is greater than or equal to 0. Since \(x^2\) is always non-negative (meaning \(x^2 \ge 0\) for all x), and adding 1 to a non-negative number will still yield a non-negative number, then \(1+x^2\) will also always be non-negative.
03

Write the domain of the function

Since \(1+x^2\) is always non-negative for all x, the domain of the function \(\left(1+x^{2}\right)^{1 / 8}\) is all real numbers, which can be written as the interval notation \((-\infty, \infty)\) or in set notation \(\{x: x \in \mathbb{R}\}\).

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Most popular questions from this chapter

Suppose \(x\) is a positive number and \(m, n,\) and \(p\) are positive integers. Using only the definitions of roots and integer powers, explain why $$ \left(x^{1 / m}\right)^{n}=\left(x^{1 /(m p)}\right)^{n p}. $$

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