Problem 4 - Algebra A Combined Function

Question

Write the solution set in interval notation. $$ (x+4)(x-1)>0 $$

Step-by-Step Solution

Answer

\((-\infty, -4) \cup (1, \infty)\)

Step 1: Find Critical Points

The inequality is \((x+4)(x-1)>0\).To determine the critical points, we set each factor equal to zero: 1. \(x + 4 = 0\) leads to \(x = -4\).2. \(x - 1 = 0\) leads to \(x = 1\). Thus, the critical points are \(x = -4\) and \(x = 1\). These points will divide the number line into intervals.

Step 2: Test Intervals Between Critical Points

The critical points \(-4\) and \(1\) divide the number line into three intervals: 1. \((-\infty, -4)\)2. \((-4, 1)\)3. \((1, \infty)\) We need to test the sign of the expression in these intervals. Choose a test point from each interval and substitute it into the inequality.

Step 3: Evaluate Interval 1: \((-\infty, -4)\)

Choose a test point, like \(x = -5\). Substitute into \((x+4)(x-1)\): \((-5+4)(-5-1) = (-1)(-6) = 6 > 0\).This interval satisfies the inequality because the product is positive.

Step 4: Evaluate Interval 2: \((-4, 1)\)

Choose a test point, like \(x = 0\). Substitute into \((x+4)(x-1)\): \((0+4)(0-1) = (4)(-1) = -4 < 0\).This interval does not satisfy the inequality because the product is negative.

Step 5: Evaluate Interval 3: \((1, \infty)\)

Choose a test point, like \(x = 2\). Substitute into \((x+4)(x-1)\):\((2+4)(2-1) = (6)(1) = 6 > 0\).This interval satisfies the inequality because the product is positive.

Step 6: Combine the Satisfying Intervals

The intervals that satisfy the inequality are \((-\infty, -4)\) and \((1, \infty)\). Since the inequality is strictly greater than zero \((>0)\), the critical points \(x=-4\) and \(x=1\) are not included in the solution set. Thus, the solution in interval notation is \((-\infty, -4) \cup (1, \infty)\).

Key Concepts

Inequalities

Inequalities are expressions that compare two values, typically involving variables. They express constraints or conditions that show how two values relate in size or magnitude. For example,

• the inequality \(x > 5\) implies that \(x\) is greater than 5.
• In the inequality \((x+4)(x-1)>0\), the inequality suggests the product of \(x+4\) and \(x-1\) is greater than zero.

When working with inequalities, we usually aim to find the range of values for \(x\) that satisfy the condition. To solve these, we can identify critical points where the expression is equal to zero, and test intervals to determine where the inequality holds true.

Critical Points

Critical points play a crucial role in solving inequalities. They are the values of \(x\) that make each factor in the inequality equal to zero. In our example, the inequality was \((x+4)(x-1)>0\). To find critical points:

• We set \(x+4 = 0\), leading to the critical point \(x = -4\).
• We set \(x-1 = 0\), leading to the critical point \(x = 1\).

Once identified, these critical points divide the number line into distinct intervals. Since these points make the product zero, the inequality does not hold exactly at these points when the inequality requires positivity (greater than zero). Understanding critical points is essential because they help us establish the intervals for testing where the inequality is satisfied.

Number Line

The number line is a crucial visualization tool for solving inequalities. It provides an intuitive way to partition the entire set of real numbers into intervals based on critical points discovered from the inequality.
Given the critical points \(-4\) and \(1\), the number line is split into three intervals:

• \((-\infty, -4)\)
• \((-4, 1)\)
• \((1, \infty)\)

Each interval is tested to check whether it satisfies the inequality \((x+4)(x-1)>0\). The number line helps to visually separate these intervals, making it easier to identify solution sets and understand why certain ranges fit the inequality's conditions while others do not.

Interval Testing

Interval testing is the method used to determine where an inequality is true along a number line. After identifying critical points and dividing the number line into intervals, we choose a test point from each interval and substitute it back into the inequality.

• For the interval \((-\infty, -4)\), choosing \(x = -5\) results in a positive value, meaning the inequality holds in this interval.
• For \((-4, 1)\), choosing \(x = 0\) results in a negative value, which does not satisfy the inequality.
• For \((1, \infty)\), choosing \(x = 2\) results in a positive value, meaning the inequality holds here too.

By testing these intervals, we confirm which sections of the number line satisfy the inequality. The final solution will be the union of intervals where the inequality is true. This approach ensures a thorough check of possible solutions across all relevant values of \(x\).