Problem 49 - Precalculus Mathematics for Calculus

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$(x+2)(x-1)(x-3) \leq 0$$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -2]\) and \([1, 3]\).

Step 1: Identify Key Points

The inequality \((x+2)(x-1)(x-3) \leq 0\) is zero at the points where each of the factors equals zero. Thus, solve \(x+2 = 0\), \(x-1 = 0\), and \(x-3 = 0\) to find key points: \(x = -2\), \(x = 1\), and \(x = 3\).

Step 2: Determine Intervals

The key points divide the number line into intervals: \((-\infty, -2)\), \((-2, 1)\), \((1, 3)\), and \((3, \infty)\). We will test the sign of the expression in each interval.

Step 3: Test Each Interval

Select test points from each interval (e.g., \(x = -3\) from \((-\infty, -2)\), \(x = 0\) from \((-2, 1)\), \(x = 2\) from \((1, 3)\), and \(x = 4\) from \((3, \infty)\)). Substitute these into the expression:- For \(x = -3\), \((-1)(-4)(-6) = -24\), negative.- For \(x = 0\), \((2)(-1)(-3) = 6\), positive.- For \(x = 2\), \((4)(1)(-1) = -4\), negative.- For \(x = 4\), \((6)(3)(1) = 18\), positive.

Step 4: Analyze Equality and Solution Interval

The inequality includes equal to zero: \((x+2)(x-1)(x-3) \leq 0\). Thus, points where the expression equals zero (\(x = -2, 1, 3\)) are included in the solution set. From Step 3, identify intervals with non-positive values: \((-\infty, -2]\), \([1, 3]\).

Step 5: Express Solution in Interval Notation

Combine the results from Step 4 to write the solution set. The inequality is satisfied for \((-\infty, -2]\) and \([1, 3]\) in interval notation.

Key Concepts

Interval Notation

Interval notation is a way to express a range of values or solutions for a variable. It's very useful in inequalities, as it provides a clear and concise way to present the solution set. For example, if a solution includes a range from negative infinity to -2, and also includes -2, we use \((-\infty, -2]\). The parentheses \(()\) indicate that the endpoint isn't included, while square brackets \([]\) mean the endpoint is included.

For the inequality \( (x+2)(x-1)(x-3) \leq 0 \), our solution uses interval notation to show where the inequality holds true. The solution is \( (-\infty, -2] \) and \([1, 3] \). This means that all values less than or equal to -2, and between 1 and 3, satisfy the inequality.

When writing in interval notation:

• Use \(()\) for open intervals where endpoints aren't included.
• Use \([]\) for closed intervals where endpoints are included.
• Combine intervals using a union symbol (usually implicit).

Inequality Solutions

Solving inequalities is a core skill in algebra that allows us to understand ranges of values for which a condition is true. The process involves finding sets of values which, when substituted into the inequality, make it true.

In the case of \( (x+2)(x-1)(x-3) \leq 0 \), solving it requires several steps:

• Identify where the product equals zero. This helps find boundary points.
• Divide the number line into intervals based on these points.
• Test each interval to find where the inequality holds.
• Pay attention to whether the inequality includes equals, which in this case, it does.

By testing values in each interval, we can determine which parts of the number line make the inequality non-positive. This allows us to express these ranges as solutions.

Nonlinear Inequalities

A nonlinear inequality involves a polynomial of degree higher than one, such as \( (x+2)(x-1)(x-3) \leq 0 \). They're different from linear inequalities because of the variable's power.

The key steps to solving nonlinear inequalities include:

• Factor the expression, making it easier to find zeros and intervals.
• Use these zeros to divide the number line into several sections.
• Check each interval with a test point.

Nonlinear inequalities often have multiple intervals where solutions might be found. The sign of the expression changes across each boundary point, which is why test points are crucial.

These inequalities can represent real-world situations where thresholds or limits apply, such as in physics for defining ranges of safe operation.