Problem 49 - Algebra and Trigonometry

Question

\(33-66\) . Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ x^{3}-4 x>0 $$

Step-by-Step Solution

Answer

The solution is \((-\infty, -2) \cup (0, 2) \cup (2, \infty)\).

Step 1: Factor the Inequality

First, factor the given inequality \(x^3 - 4x > 0\). We can factor out the common term \(x\) from the expression to get \(x(x^2 - 4) > 0\). Next, recognize that \(x^2 - 4\) can be factored further as \((x - 2)(x + 2)\). Thus, the inequality becomes \(x(x - 2)(x + 2) > 0\).

Step 2: Identify Critical Points

Set each factor of the inequality \(x(x - 2)(x + 2)\) equal to zero to find the critical points. This gives the equations \(x = 0\), \(x - 2 = 0\), and \(x + 2 = 0\). Solving these gives the critical points \(x = 0\), \(x = 2\), and \(x = -2\).

Step 3: Create Test Intervals

Use the critical points from Step 2 to divide the number line into intervals: \((-\infty, -2)\), \((-2, 0)\), \((0, 2)\), and \((2, \infty)\).

Step 4: Test Each Interval

Choose a test point from each interval and substitute it into the inequality \((x(x - 2)(x + 2) > 0)\) to determine if the inequality holds.- For \((-\infty, -2)\), use \(x = -3\): \((-3)((-3) - 2)((-3) + 2) = -3 \text{ positive} > 0.\)- For \((-2, 0)\), use \(x = -1\): \((-1)((-1) - 2)((-1) + 2) = 2 > 0.\)- For \((0, 2)\), use \(x = 1\): \(1(1-2)(1+2) = -3 < 0.\)- For \((2, \infty)\), use \(x = 3\): \(3(3-2)(3+2) = 15 > 0.\)

Step 5: Determine the Solution Set

From Step 4, the inequality holds true for intervals \((-\infty, -2)\), \((-2, 0)\), and \((2, \infty)\). Therefore, the solution for \(x^3 - 4x > 0\) is \((-\infty, -2) \cup (0, 2) \cup (2, \infty)\).

Step 6: Graph the Solution Set

To graph the solution, plot the number line and shade the regions \((-\infty, -2)\), \((0, 2)\), and \((2, \infty)\). Do not include the points \(x = -2, 0, 2\) as they are where the inequality equals zero.

Key Concepts

Factoring Polynomials

Factoring polynomials is a valuable technique when dealing with nonlinear inequalities. Our goal is to break down complex expressions into simpler factors that can be easily managed. For example, consider the inequality given in the problem: \[ x^3 - 4x > 0 \]To solve it, we first look for common terms. Here, we can factor out an \(x\). This gives us:\[ x(x^2 - 4) > 0 \]Next, notice that \(x^2 - 4\) is a difference of squares, which can further be factored into:\[ (x - 2)(x + 2) \]This fully factors the original expression into:\[ x(x - 2)(x + 2) > 0 \]Factoring simplifies the expression and allows us to move on to finding critical points and testing intervals easily.

Critical Points

Critical points are essential in analyzing expressions like \( x(x - 2)(x + 2) > 0 \). We find these by setting each factor equal to zero, which indicates where the expression either equals zero or changes sign.

• For \(x = 0\), we have the point where the expression becomes zero.
• For \(x - 2 = 0\), solve to get \(x = 2\).
• For \(x + 2 = 0\), solve to get \(x = -2\).

Together, these roots \(-2, 0,\) and \(2\) are our critical points.Understanding these points allows us to test the intervals formed between them, ultimately helping us solve the inequality.

Test Intervals

Once we establish critical points, we divide the real number line into several intervals to determine where the inequality holds true. These intervals are formed between and beyond our critical points:

• \((-\infty, -2)\)
• \((-2, 0)\)
• \((0, 2)\)
• \((2, \infty)\)

Choosing a test point from each interval, substitute it back into the factored inequality to check if the entire expression results in a positive value, indicating that the inequality is satisfied in that interval. For example:- Using \(x = -3\) for \((-\infty, -2)\), we calculate \[-3(-3 - 2)(-3 + 2) > 0\], which holds.- Test the others similarly to find which intervals satisfy the inequality.Through this method, we verify where the solution set lies.

Interval Notation

The final step in solving a nonlinear inequality is representing the solution in interval notation. Once we identify the intervals where the inequality holds, we compile them into a concise expression.For the inequality \(x^3 - 4x > 0\), our solution intervals were:

• \((-\infty, -2)\)
• \((0, 2)\)
• \((2, \infty)\)

Thus, the solution set described using interval notation is:\((-\infty, -2) \cup (0, 2) \cup (2, \infty)\)Interval notation is especially useful because it provides a simple, visual way to communicate sets of numbers on a number line effectively.