Problem 83 - A Graphical Approach to Precalculus with Limits

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

For part (a), solutions are \(x = -3\) and \(x = \frac{5}{3}\). For (b) and (c), analyze graph intervals.

Step 1: Understand the Equation

The equation \(|f(x)| = |g(x)|\) implies the absolute values of two expressions are equal. We interpret this as two separate cases: both sides being equal or their negatives being equal.

Step 2: Define the Cases

For \(|A| = |B|\), the cases are where either \(A = B\) or \(A = -B\). Here, \(A = x - \frac{1}{2}\) and \(B = \frac{1}{2}x - 2\).

Step 3: Solve Case 1: \(x - \frac{1}{2} = \frac{1}{2}x - 2\)

Subtract \(\frac{1}{2}x\) from both sides: \(x - \frac{1}{2}x = -2 + \frac{1}{2}\). Simplify to get \(\frac{1}{2}x = -\frac{3}{2}\). Solve for \(x\) by multiplying both sides by 2: \(x = -3\).

Step 4: Solve Case 2: \(x - \frac{1}{2} = -\left(\frac{1}{2}x - 2\)\)

Distribute the negative sign: \(x - \frac{1}{2} = -\frac{1}{2}x + 2\). Add \(\frac{1}{2}x\) to both sides and simplify: \(x + \frac{1}{2}x = 2 + \frac{1}{2}\), resulting in \(\frac{3}{2}x = \frac{5}{2}\). Solve for \(x\): multiply both sides by \(\frac{2}{3}\) to get \(x = \frac{5}{3}\).

Step 5: Validate the Solutions Graphically

Plot the graphs of \(|x - \frac{1}{2}|\) and \(|\frac{1}{2}x - 2|\) on the same coordinate plane. See where the graphs intersect to confirm the solutions \(x = -3\) and \(x = \frac{5}{3}\).

Step 6: Solve \(|f(x)| > |g(x)|\)

Consider the inequality \(|x - \frac{1}{2}| > |\frac{1}{2}x - 2|\). After examining the critical points, the solutions to this inequality will fall within specific intervals based on the behaviors of the graphs. For \(|f(x)| > |g(x)|\), find intervals where the graph of \(|x - \frac{1}{2}|\) is above \(|\frac{1}{2}x - 2|\).

Step 7: Solve \(|f(x)| < |g(x)|\)

Consider the inequality \(|x - \frac{1}{2}| < |\frac{1}{2}x - 2|\). Examine the graphs and identify intervals where the graph of \(|\frac{1}{2}x - 2|\) is above \(|x - \frac{1}{2}|\). These intervals provide the solution to this inequality.

Key Concepts

Analytical Solution

Solving absolute value equations analytically involves breaking down the equation into manageable pieces. For an equation like \(|f(x)| = |g(x)|\), we must consider the scenarios where the expressions inside the absolute values can be equal or opposites. In our case, \(f(x) = x - \frac{1}{2}\) and \(g(x) = \frac{1}{2}x - 2\).

• Consider \(x - \frac{1}{2} = \frac{1}{2}x - 2\), which solves to \(x = -3\).
• Then consider \(x - \frac{1}{2} = -\left(\frac{1}{2}x - 2\right)\), leading to \(x = \frac{5}{3}\).

In summary, solving analytically means narrowing down potential solutions through equalities and negations.

Graphical Representation

Visualizing absolute value functions can be enlightening, especially for confirming analytical solutions. By graphing \(|x - \frac{1}{2}|\) and \(|\frac{1}{2}x - 2|\), the points of intersection highlight where the equations equal each other. This graphical method acts as a visual check of our solutions.
Plot both functions on a coordinate plane:

• \(|x - \frac{1}{2}|\) is a V-shaped graph centered at \(x = \frac{1}{2}\).
• \(|\frac{1}{2}x - 2|\) creates another V-shape that alters its angle based on \(x\) coefficients.

The intersections at \(x = -3\) and \(x = \frac{5}{3}\) confirm our analytical findings.

Inequalities in Precalculus

Dealing with inequalities like \(|f(x)| > |g(x)|\) or \(|f(x)| < |g(x)|\) involves understanding where one function overtakes the other on a graph. The critical points are where the graphs of \(|x - \frac{1}{2}|\) and \(|\frac{1}{2}x - 2|\) intersect, as these help segment the number line into intervals.
To tackle \(|x - \frac{1}{2}| > |\frac{1}{2}x - 2|\):

• Identify regions on the graph where the first curve is above the second one.
• This occurs between critical points and can be confirmed by testing points in these intervals.

Similarly, \(|x - \frac{1}{2}| < |\frac{1}{2}x - 2|\) indicates intervals where the second graph is higher, achievable through a similar examination of the graph. These graphical insights make solving absolute inequalities manageable.