Problem 56 - Algebra and Trigonometry

Question

Find the domain of the function. $$ G(x)=\sqrt{x^{2}-9} $$

Step-by-Step Solution

Answer

The domain of \( G(x) \) is \( (-\infty, -3] \cup [3, \infty) \).

Step 1: Understanding the Function

The function given is \( G(x) = \sqrt{x^2 - 9} \). To find the domain, we need to determine all values of \( x \) for which the expression inside the square root is non-negative. This is because the square root of a negative number is not defined in the set of real numbers.

Step 2: Setting the Condition for the Square Root

Since the expression under the square root must be greater than or equal to zero, we set up the inequality: \( x^2 - 9 \geq 0 \). Solving this inequality will give us the values of \( x \) that make the function \( G(x) \) defined.

Step 3: Solving the Inequality

The inequality \( x^2 - 9 \geq 0 \) can be rewritten as \((x-3)(x+3) \geq 0\). This means we need to find the intervals where the product of \((x-3)\) and \((x+3)\) is non-negative.

Step 4: Finding the Critical Points

The critical points are \( x = 3 \) and \( x = -3 \). These points divide the real number line into three intervals: \((-\infty, -3)\), \([-3, 3]\), and \((3, \infty)\). We will test points from each interval to determine where the product \((x-3)(x+3)\) is non-negative.

Step 5: Testing Intervals

1. For \( x < -3 \), pick \( x = -4 \): \((-4-3)(-4+3) = (-7)(-1) = 7 > 0\).2. For \( -3 \leq x \leq 3 \), pick \( x = 0 \): \((0-3)(0+3) = (-3)(3) = -9 < 0\).3. For \( x > 3 \), pick \( x = 4 \): \((4-3)(4+3) = (1)(7) = 7 > 0\).

Step 6: Constructing the Domain

From the testing, \((x-3)(x+3)\) is non-negative when \( x \in (-\infty, -3] \cup [3, \infty) \). Therefore, the domain of \( G(x) \) is all real numbers \( x \) such that \( x \leq -3 \) or \( x \geq 3 \).

Key Concepts

Exploring Real Numbers

The domain of a function consists of all possible input values, typically represented by real numbers. In mathematics, the set of real numbers is vast and includes:

• Natural numbers: Counting numbers like 1, 2, 3, etc.
• Integers: Whole numbers including negative numbers, zero, and positive numbers.
• Rational numbers: Numbers that can be expressed as a fraction of two integers, such as 1/2 or -4/3.
• Irrational numbers: Numbers that cannot be expressed as fractions, like \(\pi\) or \(\sqrt{2}\).

Real numbers are key when determining the domain of functions because they define which values are allowed as inputs. In the context of our function, it's important to note that the expression inside the square root must be a real number. For \(G(x) = \sqrt{x^2 - 9}\), this means finding values of \(x\) that keep the expression under the square root non-negative. Real numbers help us set the range for \(x\) so that \(G(x)\) remains real and defined.

Understanding Inequalities

An inequality expresses that one quantity is larger or smaller than another. In mathematics, inequalities use symbols like:

• \(>\) for greater than.
• \(<\) for less than.
• \(\geq\) for greater than or equal to.
• \(\leq\) for less than or equal to.

To determine the domain for the function \(G(x) = \sqrt{x^2 - 9}\), we work with the inequality \(x^2 - 9 \geq 0\). This ensures the value inside the square root is non-negative. Solving this inequality helps identify which \(x\) values keep the expression valid. By breaking this expression into \((x-3)(x+3) \geq 0\), we find critical points that segment the real number line. Testing these segments provides insight into which ranges satisfy the inequality, thus defining the domain of the function: \((-\infty, -3] \cup [3, \infty)\).

The Square Root Function

The square root function involves finding a number that, when multiplied by itself, gives the original number back. Square roots are crucial for defining functions like \(G(x) = \sqrt{x^2 - 9}\). Here are vital points about square roots:

• The square root of a positive number is always real and positive, except the square root of zero which is zero.
• Square roots don't apply to negative numbers in the set of real numbers, as they require imaginary numbers.
• The symbol \(\sqrt{}\) denotes the principal square root, which is the non-negative root.

For \(G(x)\), we need \(x^2 - 9\) to be zero or positive so that the square root yields a real number. The role of the square root function here is prominent in finding the input values where \(G(x)\) is defined, ensuring that the function does not involve the square root of negative numbers. This process shows how square roots restrict the domain to values of \(x\) making the expression non-negative, confirming the domain \((-\infty, -3] \cup [3, \infty)\).