Problem 39 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ x^{3}>x $$

Step-by-Step Solution

Answer

The solution is \((-1, 0) \cup (1, \infty)\).

Step 1: Move All Terms to One Side

Start by moving all terms to the left-hand side of the inequality to get it in standard form: x^3 - x > 0.

Step 2: Factor the Expression

Factor the expression on the left-hand side of the inequality. x^3 - x = x(x^2 - 1).

Step 3: Further Factorization

Notice that the quadratic expression can be further factored using the difference of squares: x(x^2 - 1) = x(x - 1)(x + 1).

Step 4: Identify Critical Points

Find the critical points by setting each factor equal to zero:\(x = 0\), \(x - 1 = 0\) leading to \(x = 1\), and \(x + 1 = 0\) leading to \(x = -1\). These points split the number line into intervals.

Step 5: Test Intervals

Test the intervals created by the critical points to determine the sign of the expression in each interval:- Interval \((-\infty, -1)\): Choose \(x = -2\): \((-2)(-2 - 1)(-2 + 1) = (-2)(-3)(-1) = -6\) (negative)- Interval \((-1, 0)\): Choose \(x = -0.5\): \((-0.5)(-0.5 - 1)(-0.5 + 1) = (-0.5)(-1.5)(0.5) = 0.375\) (positive)- Interval \((0, 1)\): Choose \(x = 0.5\): \((0.5)(0.5 - 1)(0.5 + 1) = (0.5)(-0.5)(1.5) = -0.375\) (negative)- Interval \((1, \infty)\): Choose \(x = 2\): \((2)(2 - 1)(2 + 1) = (2)(1)(3) = 6\) (positive).

Step 6: Determine the Solution

The inequality \(x(x-1)(x+1) > 0\) is positive in the intervals \((-1, 0)\) and \((1, \infty)\).Therefore, the solution, expressed in interval notation, is \((-1, 0) \cup (1, \infty)\).

Key Concepts

Factoring

Factoring is a powerful algebraic tool that simplifies expressions, making equations and inequalities easier to solve. In this exercise, we begin with the expression on the left-hand side of the inequality \(x^3 - x\) and factor it into a more manageable form. This involves looking for common factors among terms.

First, notice that each term shares a factor of \(x\). Factoring \(x\) out gives \(x(x^2 - 1)\). But our work isn’t done yet! The expression \(x^2 - 1\) is a "difference of squares", a special pattern that can be further factored into \((x - 1)(x + 1)\). Thus, the full factorization of \(x^3 - x\) is \(x(x-1)(x+1)\).

This factoring simplifies the inequality, making it easier to find where it changes sign, which is crucial for solving it. Factoring breaks down complex expressions into simpler, multiplicative components. This is essential for solving inequalities by identifying important values like critical points.

Interval Notation

Once an inequality is solved, expressing the solution in interval notation provides a clear and concise representation of all the values that satisfy the inequality.

In our problem, we found the intervals where \(x(x-1)(x+1)>0\). These intervals were determined after testing the signs of the expression in various segments determined by the critical points. We discovered the expression is positive over the intervals \((-1, 0)\) and \((1, \infty)\).

Using interval notation, we represent solutions as a union of intervals, indicating any value within these sections satisfies the inequality. Thus, the solution is \((-1, 0) \cup (1, \infty)\). Note:

• Round brackets \(( )\) indicate that endpoint values are not included.
• Use \(\cup\) to join separate intervals.

This format quickly communicates a range of values, essential for clearly conveying solutions without long, complex descriptions.

Critical Points

Critical points are the values of \(x\) where the expression changes sign, which are identified by setting each factor from the factored inequality to zero. Solving each of these factor equations provides the boundaries of intervals on the number line.

In our exercise, by setting \(x=0\), \(x-1=0\), and \(x+1=0\), we determined the critical points \(x = 0\), \(x = 1\), and \(x = -1\). These points divide the number line into distinct regions.

The role of critical points is pivotal because they indicate where the inequality could change from true to false, or vice versa. By testing points within these regions, you can determine whether the expression is positive or negative, leading to a comprehensive solution.