Problem 38 - Calculus with Concepts in Calculus

Question

Solve the inequality. $$ \frac{2 x^{2}-1}{\left(1-x^{2}\right)^{1 / 2}}<0 $$

Step-by-Step Solution

Answer

The solution is \((-\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}})\).

Step 1: Identify the inequality

We have the inequality \( \frac{2x^2 - 1}{\sqrt{1 - x^2}} < 0 \). This inequality suggests that the numerator must be negative and the denominator must be positive for the entire expression to be less than zero.

Step 2: Analyze the numerator

The numerator is \( 2x^2 - 1 \). Set it less than zero to find when it is negative: \( 2x^2 - 1 < 0 \). Solving this inequality gives \( x^2 < \frac{1}{2} \), so \( -\sqrt{\frac{1}{2}} < x < \sqrt{\frac{1}{2}} \).

Step 3: Analyze the denominator

The denominator is \( \sqrt{1 - x^2} \). It is undefined for \( 1 - x^2 < 0 \) and zero when \( 1 - x^2 = 0 \). It is positive when \( 1 - x^2 > 0 \), i.e., \( x^2 < 1 \). This gives \( -1 < x < 1 \).

Step 4: Find the overlapping interval

Combine the intervals from both the numerator and the denominator: the numerator interval is \( -\sqrt{\frac{1}{2}} < x < \sqrt{\frac{1}{2}} \); the denominator interval is \( -1 < x < 1 \). Both conditions are satisfied in the interval \( -\sqrt{\frac{1}{2}} < x < \sqrt{\frac{1}{2}} \).

Step 5: Solution interval

The solution to the inequality is the interval \( \left(-\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}}\right) \), where the inequality is satisfied.

Key Concepts

Numerical Analysis

Numerical analysis can be seen as the scientific study of methods to solve mathematical problems numerically. In the context of solving inequalities such as \( \frac{2x^2 - 1}{\sqrt{1 - x^2}} < 0 \), numerical methods help us understand the behavior of functions, particularly when you can't easily solve the inequality analytically or on paper. Here’s why it's essential:

• Verification: Numerical methods can verify solutions obtained analytically and ensure we haven't missed edge cases or critical points.
• Approximations: It helps in approximating roots or bounds of the inequality, especially when dealing with irrational numbers like \( \sqrt{0.5} \).
• Complex Functions: For more complex irrational functions, numerical methods can be crucial for obtaining a solution.

In this problem, you can use numerical tools to approximate \( \sqrt{0.5} \), enhancing understanding and ensuring accuracy. Remember, while analytical methods provide exact solutions, numerical analysis is indispensable for practical, approximated insights.

Interval Notation

Interval notation is a succinct way of representing subsets of real numbers. When solving inequalities, as in our problem \( \frac{2x^2 - 1}{\sqrt{1 - x^2}} < 0 \), understanding interval notation is vital.

• Basics: An interval is denoted by using parentheses \(( )\) for open ends, and brackets \([ ]\) for closed ends. For instance, \( (-1, 1) \) means all numbers between \(-1\) and \(1\), not including \(-1\) and \(1\) themselves.
• Open vs Closed: Intervals may exclude boundary points (open) or include them (closed). The open interval \( \left(-\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}}\right) \) excludes the endpoints.
• Union and Intersection: Sometimes, inequalities may result in multiple intervals that need to be expressed using unions \((\cup)\) or intersections \((\cap)\).

In this context, interval notation beautifully captures the solution of the inequality and highlights the range where the expression holds true, making it ideal for conveying both simple and complex solution sets.

Calculus Concepts

Understanding calculus concepts is crucial in tackling inequalities involving continuous functions and expressions. In our problem \( \frac{2x^2 - 1}{\sqrt{1 - x^2}} < 0 \), calculus can assist in several ways:

• Continuity and Domain: For a smooth understanding of the problem, we need to consider the domain of the function, especially where the denominator \( \sqrt{1 - x^2} \) exists. Calculus ensures we note that the function is only continuous where the denominator is positive and defined.
• Critical Points: These are points where the first derivative might be zero or undefined. Although not directly applied here, knowing about critical points is helpful for locating minimums, maximums, or points of inflection that can affect inequality solutions.
• Signs and Change: Calculus helps us study intervals where the function is increasing or decreasing, and where it crosses the x-axis. It allows us to mark changes in function behavior, which is crucial for identifying where our inequality holds true.

Understanding these concepts enables students to explore the inequality beyond surface-level algebra, providing a more profound insight into the mathematical structures governing the solution.