Problem

Determine if Continuous f(x) = square root of 4x^2-32

The problem presented asks for an analysis of whether the function \( f(x) = \sqrt{4x^2 - 32} \) is continuous. A continuous function is one that does not have any gaps, jumps, or points of discontinuity in its domain. The question essentially requires you to examine the mathematical properties of the function, particularly within its domain of definition, to determine whether it meets the criteria for continuity. You will have to consider where the function is defined, which depends on the values of \( x \) for which the expression underneath the square root, \( 4x^2 - 32 \), is non-negative, since the square root of a negative number is not defined in the set of real numbers.

$f \left(\right. x \left.\right) = \sqrt{4 x^{2} - 32}$

Answer

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Solution:

Step 1: Identify the domain where the function is defined.

  • Step 1.1: To find the domain, ensure the inside of the square root, $4x^2 - 32$, is non-negative.

    • $4x^2 - 32 \geq 0$
  • Step 1.2: Isolate $x$ in the inequality.

    • Step 1.2.1: Add $32$ to both sides.

      • $4x^2 \geq 32$
    • Step 1.2.2: Divide the inequality by $4$.

      • Step 1.2.2.1: Divide both sides by $4$.

        • $\frac{4x^2}{4} \geq \frac{32}{4}$
      • Step 1.2.2.2: Simplify the equation.

        • Step 1.2.2.2.1: Reduce the fraction.

          • $x^2 \geq \frac{32}{4}$
        • Step 1.2.2.3: Simplify the right side.

          • $x^2 \geq 8$
    • Step 1.2.3: Take the square root of both sides.

      • $\sqrt{x^2} \geq \sqrt{8}$
    • Step 1.2.4: Simplify further.

      • Step 1.2.4.1: Address the absolute value.

        • $|x| \geq \sqrt{8}$
      • Step 1.2.4.2: Simplify the square root of $8$.

        • Step 1.2.4.2.1: Express $8$ as $2^2 \cdot 2$.

          • $|x| \geq \sqrt{2^2 \cdot 2}$
        • Step 1.2.4.2.2: Extract square root of perfect squares.

          • $|x| \geq 2\sqrt{2}$
    • Step 1.2.5: Convert the absolute value inequality into a piecewise statement.

      • Step 1.2.5.1: Consider when $x$ is non-negative.

        • $x \geq 2\sqrt{2}$
      • Step 1.2.5.2: Consider when $x$ is negative.

        • $-x \geq 2\sqrt{2}$
      • Step 1.2.6: Combine the solutions.

        • $x \geq 2\sqrt{2}$ or $x \leq -2\sqrt{2}$
  • Step 1.3: Write the domain in interval and set-builder notations.

    • Interval Notation: $(-\infty, -2\sqrt{2}] \cup [2\sqrt{2}, \infty)$
    • Set-Builder Notation: $\{x | x \leq -2\sqrt{2} \text{ or } x \geq 2\sqrt{2}\}$

Step 2: Determine the continuity of the function.

  • Since the domain is not all real numbers, the function $\sqrt{4x^2 - 32}$ is not continuous over the entire real number line.

Knowledge Notes:

  • Square Roots and Domain: The domain of a function involving a square root includes all values of $x$ for which the expression under the square root (the radicand) is non-negative. This is because the square root of a negative number is not defined in the set of real numbers.

  • Inequalities: When solving inequalities, the same operations that are applied to one side must be applied to the other side to maintain the balance of the inequality. However, when multiplying or dividing by a negative number, the direction of the inequality must be reversed.

  • Absolute Value: The absolute value of a number represents its distance from zero on the number line, regardless of direction. When solving inequalities involving absolute values, one must consider both the positive and negative scenarios.

  • Interval Notation: Interval notation is a way to describe the set of numbers that fall within a certain range on the number line. It uses parentheses to denote open intervals and brackets to denote closed intervals.

  • Set-Builder Notation: Set-builder notation is a concise way to express a set by specifying a property that its members must satisfy. It often uses a vertical bar or colon to separate the variable from the condition.

  • Continuity: A function is continuous at a point if the function is defined at that point, the limit of the function exists at that point, and the limit equals the function's value. A function is continuous over an interval if it is continuous at every point in that interval.

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