Problem 74 - A Graphical Approach to College Algebra

Question

An equation of the form \(|f(x)|=|g(x)|\) is given. (a) Solve the equation analytically and support the solution graphically. (b) Solve \(|f(x)|>|g(x)|\) (c) Solve \(|f(x)|<|g(x)|\) $$|6 x+9|=|6 x-3|$$

Step-by-Step Solution

Answer

\(x = -1\) for equality; \(x \in (-\frac{3}{2}, -1)\) for greater; \(x \in (-1, \frac{1}{2})\) for less.

Step 1: Understand the Expression

The given equation is \[|6x+9| = |6x-3|\]To solve this, we analyze where the expressions within the absolute values change signs, which happens when each of them equals zero.

Step 2: Find Critical Points

Set each expression inside the absolute value to zero:1. \(6x + 9 = 0\) gives us \(x = -\frac{3}{2}\).2. \(6x - 3 = 0\) gives us \(x = \frac{1}{2}\).The critical points are \(x = -\frac{3}{2}\) and \(x = \frac{1}{2}\), which we will use to determine the sign changes.

Step 3: Analyze Intervals Between Critical Points

The critical points create three intervals to examine:1. \((-finity, -\frac{3}{2})\)2. \((-\frac{3}{2}, \frac{1}{2})\)3. \((\frac{1}{2}, finity)\)On each interval, determine the sign of \(6x+9\) and \(6x-3\).

Step 4: Solve the Equation on Each Interval

1. **Interval \((-finity, -\frac{3}{2})\):** Both \(6x+9\) and \(6x-3\) are negative. Therefore, \[|6x+9| = - (6x+9), \quad |6x-3| = -(6x-3) \] Solving \[-(6x+9) = -(6x-3) \] simplifies to \(-9 = -3\), which is false; no solution.2. **Interval \((-\frac{3}{2}, \frac{1}{2})\):** Here, \(6x+9\) is positive and \(6x-3\) is negative. Therefore, \[|6x+9| = 6x+9, \quad |6x-3| = -(6x-3) \] Solving \[6x+9 = -(6x-3) \] results in \(12x = -12\), so \(x = -1\).3. **Interval \((\frac{1}{2}, finity)\):** Both are positive, therefore, \[|6x+9| = 6x+9, \quad |6x-3| = 6x-3 \] Solving \[6x + 9 = 6x-3 \] results in \(9 = -3\), which is false; no solution.

Step 5: Confirm Solution and Support Graphically

The only solution to \[|6x+9| = |6x-3|\]was found on the interval \((-\frac{3}{2}, \frac{1}{2})\), which is \(x = -1\). A graph of \( |6x+9|\) and \( |6x-3|\) will intersect only at this point.

Step 6: Solve \(|6x+9| > |6x-3|\)

We need to analyze the same intervals as before:1. **Interval \((-finity, -\frac{3}{2})\):** There's no equality. 2. **Interval \((-\frac{3}{2}, \frac{1/2})\):** Comparing, we find \(x \in (-\frac{3}{2}, -1)\) satisfies this inequality.3. **Interval \((\frac{1}{2}, finity)\):** There's no equality, hence no solution.

Step 7: Solve \(|6x+9| < |6x-3|\)

Inequality must hold on parts of the evaluated intervals:1. **Interval \((-finity, -\frac{3}{2})\):** There's no inequality that holds.2. **Interval \((-\frac{3}{2}, \frac{1}{2})\):** Here \(x \in (-1, \frac{1}{2})\) satisfies this inequality.3. **Interval \((\frac{1/2}, finity)\):** Actually, this inequality can never be satisfied.

Step 8: Conclusion of Analytical Solutions

For \(|6x+9| = |6x-3|\), we find \(x = -1\).For \(|6x+9| > |6x-3|\), we find \(x \in (-\frac{3}{2}, -1)\).For \(|6x+9| < |6x-3|\), we find \(x \in (-1, \frac{1}{2})\).

Key Concepts

Inequalities

When dealing with inequalities involving absolute values, such as \(|f(x)| > |g(x)|\) or \(|f(x)| < |g(x)|\), it is essential to break down the problem into intervals based on the critical points where the expression changes signs. By setting each term inside the absolute value to zero, you can find the critical points which help dictate where each function changes from positive to negative or vice versa.
Thus, the solution strategy centers on assessing each interval and testing whether the inequality holds.

• Always identify the intervals created by your critical points. This setup will guide which forms the absolute values take.
• Analyze each interval separately. Determine which side of the inequality holds true by substituting points from each interval into the inequality.
• Only intervals where the inequality holds true are part of the solution.

This systematic analysis ensures each potential solution is valid for the given inequality. Remember that a graph of each function can also confirm the regions where one function surpasses the other.

Critical Points

The concept of critical points is crucial when solving absolute value equations or inequalities. They are the values of \(x\) that make the expression within the absolute value zero.
Understanding how to find these points is vital because they indicate where the nature of the function changes.
For the equation \(|6x+9| = |6x-3|\), find the critical points by solving:

• \(6x+9 = 0\) results in \(x = -\frac{3}{2}\).
• \(6x-3 = 0\) leads to \(x = \frac{1}{2}\).

Once located, these points divide the number line into intervals which are then analyzed for function inequality or identity.
Critical points mark changes in behavior: crossing a critical point often shifts the expression's sign inside the absolute value. Hence, understanding them unlocks easier solutions to both equations and inequalities.

Graphical Analysis

Graphical analysis serves as a powerful tool in understanding absolute value equations and inequalities. It provides a visual representation, allowing us to "see" solutions and verify our analytical findings. For example, plotting the functions \(|6x+9|\) and \(|6x-3|\) helps us quickly identify where and if they intersect.
Key steps in graphical analysis include:

• Plotting each function individually. Absolute value graphs are typically V-shaped.
• Looking at the intersection points or the regions where one graph is above the other, which correspond to solutions for inequalities.
• Comparing visual answers to analytical calculations to ensure consistency.

By employing this dual approach, students can deepen their understanding of where and why solutions exist. Graphical analysis not only verifies solutions but also enhances intuition for handling absolute value problems.