Problem 53 - Algebra and Trigonometry

Question

Find the domain of the function. $$ g(x)=\sqrt[4]{x^{2}-6 x} $$

Step-by-Step Solution

Answer

The domain of \( g(x) \) is \( x \in (-\infty, 0] \, \cup \, [6, \infty) \).

Step 1: Understanding the Function

The function given is \( g(x) = \sqrt[4]{x^2 - 6x} \). This is a fourth root function, which requires the expression inside the root to be greater than or equal to zero to produce real numbers.

Step 2: Set the Expression Non-negative

Set the expression inside the fourth root to be greater than or equal to zero: \[ x^2 - 6x \geq 0 \] This inequality ensures that the fourth root is defined and real.

Step 3: Solve the Inequality

Solve the inequality \( x^2 - 6x \geq 0 \). Factor the expression: \[ x(x - 6) \geq 0 \] Determine where the expression is zero: \( x = 0 \) and \( x = 6 \). Create a number line to test intervals around these points.

Step 4: Test Intervals and Determine the Sign of Each Interval

Choose test points from intervals: 1. For \( x < 0 \): Let's test \( x = -1 \), substitute into \( x(x-6) \) to get \( (-1)(-1-6) = 7 \), which is positive. 2. For \( 0 < x < 6 \): Let's test \( x = 3 \), resulting in \( 3(3-6) = -9 \), which is negative. 3. For \( x > 6 \): Let's test \( x = 7 \), resulting in \( 7(7-6) = 7 \), which is positive.

Step 5: Construct the Domain

From the test, the function is non-negative in the intervals \(( -\infty, 0 ] \cup [ 6, \infty )\). Therefore, the domain of \( g(x) \) is \( x \in (-\infty, 0] \, \cup \, [6, \infty) \).

Key Concepts

Fourth Root Function

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.

Inequality Solving

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

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.