91Ó°ÊÓ

Understanding inequalities is crucial when finding the domain of a function. When dealing with functions, inequalities can help us determine the set of possible inputs (x-values) that will keep the function valid. For example, in expressions like \( 6x - 2 \geq 0 \), the inequality shows us the conditions under which the expression inside a square root remains non-negative. In this case, ensuring that \( x \geq \frac{1}{3} \) means only working with values of \( x \) that make the square root result in a real number.To solve an inequality, we approach it as a step-by-step process similar to solving regular equations, but with careful attention to the direction of the inequality sign. This often involves operations such as one-side additions, subtractions, or even multiplications and divisions, while ensuring the inequality holds through each step.

Square roots are only defined for non-negative numbers. This means that for any expression like \( \sqrt{6x - 2} \), the condition \( 6x - 2 \geq 0 \) must be met. If this condition is not satisfied, the square root will not yield a real number and hence be invalid in standard mathematical contexts used in schooling.To ensure the entire expression under the square root is non-negative, you can calculate the critical points using inequality. Solving \( 6x - 2 \geq 0 \) gives us values of \( x \) where the square root remains valid, which is essential when defining the domain of a function involving square roots. Thus, the rule of thumb is that the expression inside a square root must be zero or positive.

Fractional expressions introduce another layer of conditions when finding a function's domain. A critical requirement is ensuring the denominator never equals zero, as division by zero is undefined in mathematics. For the function in question, \( \frac{6}{4 - \sqrt{6x - 2}} \), we need to ensure that \( 4 - \sqrt{6x - 2} eq 0 \).If the denominator ends up equaling zero, this would make the expression undefined. For instance, we calculate \( \sqrt{6x - 2} = 4 \) to find the offending \( x \) that makes the denominator zero. Solving gives us \( x = 3 \), which tells us to exclude \( x = 3 \) from the domain. Such restrictions are typical in maintaining legitimacy and definition within the realm of real numbers.

For a function to be real-valued, it must produce real numbers as outputs for every input in its domain. This is a fundamental requirement, especially in educational settings where complex numbers are yet to be introduced.In the context of our function, both the square root and fraction components impose conditions on \( x \) to ensure all results are real numbers. By solving the inequality \( 6x - 2 \geq 0 \) and checking \( 4 - \sqrt{6x-2} eq 0 \), we derive a domain where the function reliably calculates real values: \( x \in \left[ \frac{1}{3}, 3 \right) \cup (3, \infty) \). Separating the domain like this ensures that every permissible \( x \) yields a legitimate, sensible answer from the function.