Problem 51 - Precalculus Mathematics for Calculus

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$(x-4)(x+2)^{2}<0$$

Step-by-Step Solution

Answer

The solution is \((-2, 4)\), excluding \(-2\) and \(4\).

Step 1: Determine Critical Points

To find the critical points of the inequality \((x-4)(x+2)^2 < 0\), set each factor equal to zero. This gives us the equations \(x-4 = 0\) and \((x+2)^2 = 0\). Solving these equations, we find the critical points: \(x = 4\) and \(x = -2\).

Step 2: Test Intervals Between Critical Points

The critical points \(-2\) and \(4\) divide the number line into three intervals: \((-\infty, -2)\), \((-2, 4)\), and \((4, \infty)\). We need to test a point from each interval to determine where the inequality holds. For example:- Choose \(x = -3\) in the interval \((-\infty, -2)\), which gives \((x-4)(x+2)^2 = (7)(1) = 7 > 0\).- Choose \(x = 0\) in the interval \((-2, 4)\), which gives \((x-4)(x+2)^2 = (-4)(4) = -16 < 0\).- Choose \(x = 5\) in the interval \((4, \infty)\), which gives \((x-4)(x+2)^2 = (1)(49) = 49 > 0\).

Step 3: Evaluate the Critical Points Themselves

We need to check the inequality at the critical points \(x = -2\) and \(x = 4\) to determine if they are included in the solution.- At \(x = -2\), \((x-4)(x+2)^2 = (-6)(0) = 0\).- At \(x = 4\), \((x-4)(x+2)^2 = (0)(36) = 0\).Since we have \(<0\) not \(\leq 0\), neither \(-2\) nor \(4\) are included in the solution.

Step 4: Write the Solution in Interval Notation

Based on the test intervals and the evaluation of critical points, the inequality is satisfied only in the interval \((-2, 4)\). Therefore, the solution in interval notation is \((-2, 4)\).

Step 5: Graph the Solution Set

The graph of the solution set \((-2, 4)\) is a number line with an open interval between \(-2\) and \(4\). Use open circles (or parentheses) at both ends to indicate that these points are not included in the solution set.

Key Concepts

Interval Notation

Interval notation is a way of representing a range of numbers along the number line. It uses brackets or parentheses to define the start and end of an interval. The type of bracket or parenthesis indicates whether the endpoints are included in the interval. This can help visualize the solutions of inequalities.

In interval notation, we use:

• Parentheses "\(( )\)" to exclude the endpoints. This means the endpoint values are not part of the solution.
• Brackets "\([ ]\)" to include the endpoints, meaning the values are part of the solution.

For our specific exercise, the solution was expressed as \((-2, 4)\). In this case, both endpoints \(-2\) and \(4\) are not included in the solution set because we have an open interval. We determined this by testing points and evaluating the inequality, given that the solution must satisfy \((x - 4)(x + 2)^2 < 0\). The graph of this interval would show an open segment from \(-2\) to \(4\) with open circles at these points, indicating their exclusion.

Critical Points

Critical points play a crucial role in solving inequalities. They are the values of the variable where the expression equals zero. Identifying critical points helps us to divide the number line into different regions where we can determine the sign of the function.

For the inequality \((x - 4)(x + 2)^2 < 0\), we identified the critical points by setting each factor equal to zero. This gave the equations \(x - 4 = 0\) and \((x + 2)^2 = 0\), leading to critical points at \(x = 4\) and \(x = -2\).

These critical points are important because they define the boundaries on the number line where the sign of the product may change. By testing intervals between these critical points, we can determine where the inequality expression is negative, positive, or zero.

Solution Sets

The solution set of an inequality is the set of all values that satisfy the inequality. For nonlinear inequalities, the solution often involves ranges of values rather than individual numbers.

To find the solution set for \((x - 4)(x + 2)^2 < 0\), we divided the number line using our critical points \(-2\) and \(4\). This division created intervals that we tested to verify the sign of the expression in each.

After testing, we found that only the interval \((-2, 4)\) resulted in the inequality being satisfied (i.e., the product was negative).

Hence, the solution set in this case includes all real numbers \(x\) that lie between \(-2\) and \(4\), exclusive of the endpoints. Thus, the solution set is expressed in interval notation as \((-2, 4)\) and visually represented as a line segment between \(-2\) and \(4\) with open circles to indicate that these points are excluded from the set.