Problem 58 - College Algebra

Question

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter \(\mathrm{Y} 2=\) the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, liabs ( Find the points of intersection, recall \((2^{\text { nd }}\) CALC 5:intersection, lst curve, enter, } \(2^{\text { nd }}\) curve, enter, guess, enter). Copy a sketch of the graph and shade the \(x\) -axis for your solution set to the inequality. Write final answers in interval notation. $$ |x+2| \geq 5 $$

Step-by-Step Solution

Answer

The solution is \((-\infty, -7] \cup [3, \infty)\).

Step 1: Identify the Inequality

We have the inequality \(|x + 2| \geq 5\). This means we need to find the set of \(x\) values where the absolute value of \(x + 2\) is greater than or equal to 5.

Step 2: Split the Absolute Value into Two Inequalities

The inequality \(|x + 2| \geq 5\) can be split into two separate inequalities:1. \(x + 2 \geq 5\)2. \(x + 2 \leq -5\)

Step 3: Solve the First Inequality

Solve \(x + 2 \geq 5\) by subtracting 2 from both sides:\[x \geq 3\]

Step 4: Solve the Second Inequality

Solve \(x + 2 \leq -5\) by subtracting 2 from both sides:\[x \leq -7\]

Step 5: Combine Solutions and Use Interval Notation

Combine the solutions from Steps 3 and 4. Since the original inequality is \(|x+2| \geq 5\), the solution is disjoint and includes both parts:\[x \in (-\infty, -7] \cup [3, \infty)\]

Step 6: Graph the Functions and Find Intersection Points

In your graphing utility, input the left side of the inequality as \(Y1 = |x + 2|\) and the right side as \(Y2 = 5\). Use the graphing utility to find the intersection points of these two graphs, which will confirm the solution set found algebraically.

Step 7: Shade the Solution Set on the x-axis

Draw a number line to represent the \(x\)-axis and shade the regions \((-\infty, -7]\) and \([3, \infty)\) to represent the solution set. These regions indicate where the graph of \(Y1\) is above or touching \(Y2\).

Key Concepts

Absolute Value

Understanding absolute value is crucial when solving inequalities involving expressions like \( |x + 2| \geq 5 \). The absolute value of a number indicates its distance from zero on the number line, regardless of direction.
So, \( |x + 2| \geq 5 \) means we are looking for all values of \( x \) such that the expression inside the absolute value is at least 5 units away from zero. This can happen in two scenarios:

• The expression inside, \( x+2 \), is greater than or equal to 5, moving to the right on the number line.
• Or, the expression is less than or equal to -5, moving to the left.

By considering these two scenarios, we split our inequality into two separate inequalities: \( x + 2 \geq 5 \) and \( x + 2 \leq -5 \). These situations help us understand where the solution sets will be when solving for \( x \).

Graphing Utilities

Graphing utilities play a critical role in visualizing inequalities and confirming algebraic solutions. When working with absolute value inequalities, one powerful approach is to graph the inequalities to better understand the relationship between the expressions.
In this scenario, input the left side of the inequality as \( Y1 = |x + 2| \) and the right side as \( Y2 = 5 \)in your graphing utility. Graphing these functions allows you to visually inspect where the graph \( |x+2| \) is at least as high as the line \( y = 5 \).
This visual inspection provides a straightforward way to verify your solution, reinforcing your algebraic findings.'

Interval Notation

Interval notation is a concise way to represent a set of values, which is extremely useful when expressing solutions to inequalities. After solving inequalities involving absolute values, we often need to present the solution in interval notation.
For example, after solving \( |x+2| \geq 5 \), we find two separate intervals for \( x \):

• \( x \leq -7 \)
• or \( x \geq 3 \)

This means all \( x \)-values lower than or equal to -7, or equal to or greater than 3, satisfy the inequality. So, we write this in interval notation as \((-\infty, -7] \cup [3, \infty)\).
The use of round brackets \(()\) and square brackets \([]\) helps indicate whether the endpoints are included or not in the solution sets.

Intersection Points

Intersection points on a graph reveal crucial information about the solutions to an inequality. When using a graphing utility, these points help you confirm where two function graphs meet.
In our case, for \( |x+2| = 5 \), the graphs \( Y1 = |x+2| \) and \( Y2 = 5 \)represent the expression and the boundary of the inequality, respectively.
You will find intersection points where these graphs meet. In this example, the graphs intersect at \( x = -7 \) and \( x = 3 \).
These points are essential because they mark the boundaries where the inequality holds true, helping you shade the correct regions on the \( x \)-axis in your graph.