91Ó°ÊÓ

A fourth root function involves taking the fourth root of a given expression. Generally, the fourth root is written as \( \sqrt[4]{x} \). For a fourth root function like \( g(x) = \sqrt[4]{x^2 - 6x} \), the expression inside the root must be non-negative, meaning it is greater than or equal to zero. This is because the fourth root of a negative number is not defined within the set of real numbers. Therefore, when you are tasked with finding the domain of a fourth root function, you should first set the expression inside the root to be \( \geq 0 \). Consider it as solving an inequality to ensure the result of the expression under the root is valid, producing a real number for any value of \( x \) within the domain.

To determine where the expression inside the fourth root is non-negative, you solve the inequality \( x^2 - 6x \geq 0 \). Here is how you can achieve that:

• First, factor the quadratic expression inside: \( x(x - 6) \geq 0 \).
• This means the expression is zero at points \( x = 0 \) and \( x = 6 \).

These are the boundary points. The goal is now to determine the sign of the expression \( x(x - 6) \) in the intervals divided by these boundary points. Recognizing how the factors change sign will guide which intervals satisfy the inequality. Consider any \( x \) value less than, between, and greater than these mentioned boundary points to test the expression outcomes.

Interval testing involves checking intervals divided by critical points (roots of the expression), to see where your inequality holds true. Once we've identified the critical points, \(x = 0\) and \(x = 6\), the number line can be divided into three intervals:

• \(( -\infty, 0 )\)
• \(( 0, 6 )\)
• \(( 6, \infty )\)

For each interval, select a test point, substitute back into the factorized expression \( x(x-6) \), and check the sign:

• In \(( -\infty, 0 )\), selecting \( x = -1 \), produces \(( -1)(-1-6) = 7\), which is positive.
• In \(( 0, 6 )\), using \( x = 3 \) gives \(3(3-6) = -9\), resulting in a negative value.
• In \(( 6, \infty )\), taking \( x = 7 \) results in \( 7(7-6) = 7 \), which is positive.

This means \( g(x) = \sqrt[4]{x^2 - 6x} \) is real and non-negative in the intervals \(( -\infty, 0 ] \cup [ 6, \infty )\). Hence, these intervals represent the function's domain.