Problem 71 - 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 \(|f(x)|<|g(x)|\) $$\left|x-\frac{1}{2}\right|=\left|\frac{1}{2} x-2\right|$$

Step-by-Step Solution

Answer

Solutions: (a) x = -3, x = 5/3; (b) x in (-∞, -3) ∪ (5/3, ∞); (c) x in (-3, 5/3).

Step 1: Set Up the Equation Cases

For the equation \( \left| x - \frac{1}{2} \right| = \left| \frac{1}{2}x - 2 \right| \), we need to analyze the cases based on the absolute values:1. \( x - \frac{1}{2} = \frac{1}{2}x - 2 \)2. \( x - \frac{1}{2} = -\left( \frac{1}{2}x - 2 \right) \)3. \(-\left( x - \frac{1}{2} \right) = \frac{1}{2}x - 2 \)4. \(-\left( x - \frac{1}{2} \right) = -\left( \frac{1}{2}x - 2 \right) \)

Step 2: Solve Case 1

Solve \( x - \frac{1}{2} = \frac{1}{2}x - 2 \):Rearrange the equation:\[x - \frac{1}{2}x = -2 + \frac{1}{2} \\frac{1}{2}x = -\frac{3}{2}\]Divide by \( \frac{1}{2} \):\[ x = -3 \]

Step 3: Solve Case 2

Solve \( x - \frac{1}{2} = -\left( \frac{1}{2}x - 2 \right) \):Expand the right side:\( x - \frac{1}{2} = -\frac{1}{2}x + 2 \)Rearrange the equation:\[x + \frac{1}{2}x = 2 + \frac{1}{2} \\frac{3}{2}x = \frac{5}{2}\]Divide by \( \frac{3}{2} \):\[ x = \frac{5}{3} \]

Step 4: Solve Case 3

Solve \(-\left( x - \frac{1}{2} \right) = \frac{1}{2}x - 2 \):Expand the left side:\( -x + \frac{1}{2} = \frac{1}{2}x - 2 \)Rearrange the equation:\[-x - \frac{1}{2}x = -2 - \frac{1}{2} \-\frac{3}{2}x = -\frac{5}{2}\]Divide by \( -\frac{3}{2} \):\[ x = \frac{5}{3} \]

Step 5: Solve Case 4

Solve \(-\left( x - \frac{1}{2} \right) = -\left( \frac{1}{2}x - 2 \right) \):Expand both sides:\(-x + \frac{1}{2} = -\left( \frac{1}{2}x - 2 \right) \)\( -x + \frac{1}{2} = -\frac{1}{2}x + 2 \)Rearrange the equation:\[-x + \frac{1}{2}x = 2 - \frac{1}{2} \-\frac{1}{2}x = \frac{3}{2}\]Divide by \( -\frac{1}{2} \):\[ x = -3 \]

Step 6: Verify the Solutions

The solutions obtained from the previous steps are \( x = -3 \) and \( x = \frac{5}{3} \). Verify these by substituting into the original equation:For \( x = -3 \):\( | -3 - \frac{1}{2} | = | \frac{1}{2} \times -3 - 2 | \) results in both sides being 3.5.For \( x = \frac{5}{3} \):\( | \frac{5}{3} - \frac{1}{2} | = | \frac{1}{2} \times \frac{5}{3} - 2 | \) results in both sides being \( \frac{5}{6} \).Both solutions satisfy the original equation.

Step 7: Solve |f(x)| > |g(x)|

To solve \( \left| x - \frac{1}{2} \right| > \left| \frac{1}{2}x - 2 \right| \), consider the critical points \( x = -3 \) and \( x = \frac{5}{3} \). Test intervals:1. For \( x < -3 \), choose \( x = -4 \): \( 4.5 > 4 \) is true.2. For \( -3 < x < \frac{5}{3} \), choose \( x = 0 \): \( 0.5 < 2 \) is false.3. For \( x > \frac{5}{3} \), choose \( x = 2 \): \( 1.5 > 1 \) is true.Hence, \( x \in (-\infty, -3) \cup \left( \frac{5}{3}, \infty \right) \).

Step 8: Solve |f(x)| < |g(x)|

To solve \( \left| x - \frac{1}{2} \right| < \left| \frac{1}{2}x - 2 \right| \), using the same critical points as above, test the intervals:1. For \( x < -3 \), choose \( x = -4 \): \( 4.5 < 4 \) is false.2. For \( -3 < x < \frac{5}{3} \), choose \( x = 0 \): \( 0.5 < 2 \) is true.3. For \( x > \frac{5}{3} \), choose \( x = 2 \): \( 1.5 < 1 \) is false.Hence, \( x \in (-3, \frac{5}{3}) \).

Key Concepts

Graphical Solutions

Graphical solutions offer a visual insight into mathematical problems, especially with absolute value equations. Let's consider the exercise where we solve an equation of the form \( \left| x - \frac{1}{2} \right| = \left| \frac{1}{2} x - 2 \right| \). To understand this equation graphically, we plot the functions \( y_1 = |x - \frac{1}{2}| \) and \( y_2 = |\frac{1}{2}x - 2| \) on a coordinate plane.

• Each absolute value function creates a distinct V-shaped graph.
• The vertices of these graphs are significant points where the internal expression of the absolute values are zero.
• For \( y_1 \), the vertex appears at \( x = \frac{1}{2} \), and for \( y_2 \), it occurs at \( x = 4 \).

By observing these V-shapes, solutions to the equation are found where the graphs intersect, giving points like \( x = -3 \) and \( x = \frac{5}{3} \) as shown in the analytical solution, aligning with where each absolute value expression results in equivalent values.
Graphing offers students an intuitive method to infer solutions and comprehend the relationship between different parts of the equation.

Inequalities

Inequalities involving absolute values can be intimidating but become manageable with the right approach. When dealing with inequalities such as \( \left| x - \frac{1}{2} \right| > \left| \frac{1}{2} x - 2 \right| \) or \( \left| x - \frac{1}{2} \right| < \left| \frac{1}{2} x - 2 \right| \), the solutions relate to where one expression surpasses another in magnitude.
To solve \( \left| x - \frac{1}{2} \right| > \left| \frac{1}{2} x - 2 \right| \):

• Consider critical points from where the solutions equalize, \( x = -3 \) and \( x = \frac{5}{3} \).
• These points divide the x-axis into segments namely: \( (-\infty, -3) \), \( (-3, \frac{5}{3}) \), and \( (\frac{5}{3}, \infty) \).
• Analyze each segment by testing values (e.g., \( x = -4, 0, 2 \)) to see if the inequality holds.

The viable solution for \( |f(x)| > |g(x)| \) is \(-\infty < x < -3 \) and \( x > \frac{5}{3} \).
Solving \( \left| x - \frac{1}{2} \right| < \left| \frac{1}{2} x - 2 \right| \) would involve a similar approach, giving the solution \( -3 < x < \frac{5}{3} \). Inequalities enable us to uncover ranges where constraints hold true and can be verified graphically.

Case Analysis

Case analysis forms the backbone of solving absolute value equations analytically. With our example \( \left| x - \frac{1}{2} \right| = \left| \frac{1}{2} x - 2 \right| \), the strategy involves breaking the equation into separate cases, accounting for the varied behaviors of the absolute value expressions.
Each absolute value splits into two linear expressions because they consider both positive and negative values of the inside expression:

• Case 1: \( x - \frac{1}{2} = \frac{1}{2}x - 2 \)
• Case 2: \( x - \frac{1}{2} = -\left( \frac{1}{2}x - 2 \right) \)
• Case 3: \( -\left( x - \frac{1}{2} \right) = \frac{1}{2}x - 2 \)
• Case 4: \( -\left( x - \frac{1}{2} \right) = -\left( \frac{1}{2}x - 2 \right) \)

Solving each case separately allows us to identify all potential solutions, as shown with results \( x = -3 \) and \( x = \frac{5}{3} \).
This method respects the nature of absolute values and provides a structured way to decipher complex equations, ensuring students understand the necessity to consider all conditions where the equation might hold true.