Problem 56 - College Algebra with Corequisite Support

Question

For the following exercises, input the left-hand side of the inequality as a \(Y 1\) graph in your graphing utility. Enter \(y 2=\) the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, \(1: a b s(.\) Find the points of intersection, recall \(\left(2^{\text {nd }}\right.\) CALC 5 :intersection, \(1^{\text {st }}\) 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. $$ |4 x+1|-3>2 $$

Step-by-Step Solution

Answer

\((-\infty, -\frac{3}{2}) \cup (1, \infty)\)

Step 1: Graph the expressions

Input the left-hand side of the inequality as a graph into the graphing utility using \(Y_1 = |4x + 1| - 3\). Then enter the right-hand side as \(Y_2 = 2\). These functions will be plotted to help find the points of intersection.

Step 2: Find intersection points

Use a graphing calculator's functionality (e.g., 2nd CALC, option 5: intersection) to find the points where \(Y_1\) and \(Y_2\) intersect. This involves selecting the 1st curve, hitting enter, selecting the 2nd curve, hitting enter, and finally guessing a value to find each point of intersection.

Step 3: Solve the inequality analytically

Set up the equation \(|4x + 1| - 3 = 2\) to solve for intersection points without graphing. Rearrange it to \(|4x + 1| = 5\). This results in two cases: 1. \(4x + 1 = 5\) 2. \(4x + 1 = -5\)

Step 4: Solve each case

Solve both cases separately:1. \(4x + 1 = 5\) - Subtract 1: \(4x = 4\) - Divide by 4: \(x = 1\)2. \(4x + 1 = -5\) - Subtract 1: \(4x = -6\) - Divide by 4: \(x = -\frac{3}{2}\)

Step 5: Determine the solution set

The inequality \(|4x + 1| - 3 > 2\) implies that the values of \(x\) where \(Y_1\) is greater than \(Y_2\) are the solution. Check intervals formed by critical points \(x = -\frac{3}{2}\) and \(x = 1\) on the number line to determine which satisfy the inequality, resulting in two intervals.

Step 6: Express solution in interval notation

Shade the \(x\)-axis between intervals where \(|4x+1|-3 > 2\). Solution is:- For \(x < -\frac{3}{2}\)- For \(x > 1\)The interval notation is: \((-\infty, -\frac{3}{2}) \cup (1, \infty)\).

Key Concepts

Graphing Utilities

Graphing utilities are excellent tools for visualizing mathematical expressions and solving equations. They let us see how an equation behaves and helps find solutions to problems like inequalities.

To use a graphing utility, you'll need to input the expressions as different functions. For absolute value inequalities, like \(Y_1 = |4x + 1| - 3\) and \(Y_2 = 2\), you enter each as a separate equation. Typically, the graphing utility will plot these as different lines or curves on the graph, so you can see where they intersect or where one is greater than the other.

• Go to the function input menu in your graphing calculator.
• Enter each side of the inequality as a separate equation.
• Use the graph feature to visually analyze the functions.

It's essential to become comfortable with the graphing device since it helps verify algebraic solutions and provides a visual confirmation of the concepts behind inequalities.

Interval Notation

Once you've found the solution to an inequality, such as \(|4x + 1| - 3 > 2\), it is often expressed using interval notation. This notation provides a concise way to describe a range of values that satisfies the inequality.

Interval notation uses parentheses \(\text{( } \text{ )}\) for non-inclusive bounds and brackets \[\text{[ } \text{ ]}\] for inclusive bounds. In this problem, our solution set includes two ranges, which were determined by testing intervals between the intersection points \(x = -\frac{3}{2}\) and \(x = 1\).

• Since \(x\) must be less than \(-\frac{3}{2}\) but not equal to \(-\frac{3}{2}\), we use parentheses: \((-\infty, -\frac{3}{2})\).
• Similarly, because \(x\) is greater than \(1\) but not equal to \(1\), we also use parentheses: \((1, \infty)\).

The combined interval notation for this inequality would be \((-\infty, -\frac{3}{2}) \cup (1, \infty)\), indicating both intervals are part of the solution.

Intersection Points

Identifying intersection points is a crucial part of solving systems of equations and inequalities graphically.

In absolute value inequalities, finding where two functions meet on the graph involves finding the points of intersection. For the given problem, we want to know where \(Y_1 = |4x + 1| - 3\) and \(Y_2 = 2\) cross each other on the graph.

• Use your graphing calculator's "intersect" feature to do this efficiently.
• Follow the prompts to select both curves and make a guess near where they intersect to zero in on the precise points.
• For this particular example, the intersection gives us critical \(x\)-values: \(x = -\frac{3}{2}\) and \(x = 1\).

These intersection points help divide the \(x\)-axis into regions where one curve is above or below the other. Based on this understanding, we evaluate which sections satisfy the original inequality.