Solution: We analyze the expression by considering the critical points where expressions inside absolute values change sign: $ x = \frac{5}{2} $ and $ x = -3 $. We consider three intervals:
["Title: Analyzing Absolute Value Expressions: Understanding Critical Points and Interval Behavior", "When solving equations or analyzing functions involving absolute values, one of the most powerful strategies is identifying the critical points where the expression inside the absolute value changes sign. These key locations often divide the real number line into intervals where the expression behaves predictably—either positive, negative, or zero. For expressions like $ |f(x)| $, recognizing these critical points allows for a clear, case-by-case analysis. In this article, we explore the essential method of analyzing absolute value expressions by examining the critical points $ x = \frac{5}{2} $ and $ x = -3 $, which occur when the inner function changes sign. We’ll break down three key intervals defined by these points and demonstrate how to solve or evaluate absolute value expressions reliably.", "---", "### Understanding Absolute Value Expressions", "An absolute value expression $ |g(x)| $ represents the distance of $ g(x) $ from zero on the number line—always producing a non-negative result. When $ g(x) $ changes sign, the expression’s behavior changes: it behaves like $ g(x) $ when $ g(x) \geq 0 $, and $ -g(x) $ when $ g(x) < 0 $. For these reasons, absolute value functions are not smooth at the points where $ g(x) = 0 $, creating potential “corners” in their graphs.", "To analyze such expressions thoroughly, the critical points—where $ g(x) = 0 $—are indispensable. These divide the domain into intervals where $ g(x) $ maintains consistent sign. Evaluating the expression in each interval removes ambiguity by specifying whether $ g(x) $ is positive, negative, or zero.", "---", "### Step 1: Identify Critical Points", "For the expression $ |x - \frac{5}{2}| $ and $ |x + 3| $, we find the critical points by solving where the inner functions equal zero:", "- $ x = \frac{5}{2} = 2.5 $\n- $ x = -3 $", "These points divide the number line into three distinct intervals:", "1. $ (-\infty, -3) $\n2. $ -3 < x < \frac{5}{2} $\n3. $ x > \frac{5}{2} $", "Each interval offers a consistent sign for the expressions inside absolute values, enabling precise evaluation.", "---", "### Step 2: Analyze Each Interval", "Let’s evaluate the behavior in each interval, using sign analysis and evaluating the expression without absolute values.", "---", "#### Interval 1: $ x < -3 $", "Pick a test point, e.g., $ x = -4 $", "- $ x - \frac{5}{2} = -4 - 2.5 = -6.5 < 0 $ → $ |x - \frac{5}{2}| = -(x - \frac{5}{2}) = -x + \frac{5}{2} $\n- $ x + 3 = -4 + 3 = -1 < 0 $ → $ |x + 3| = -(x + 3) = -x - 3 $", "Expression:\n$ |x - \frac{5}{2}| + |x + 3| = (-x + \frac{5}{2}) + (-x - 3) = -2x - \frac{1}{2} $", "---", "#### Interval 2: $ -3 < x < \frac{5}{2} $", "Test $ x = 0 $", "- $ x - \frac{5}{2} = -2.5 < 0 $ → $ |x - \frac{5}{2}| = -x + \frac{5}{2} $\n- $ x + 3 = 3 > 0 $ → $ |x + 3| = x + 3 $", "Expression:\n$ |x - \frac{5}{2}| + |x + 3| = (-x + \frac{5}{2}) + (x + 3) = \frac{5}{2} + 3 = \frac{11}{2} $", "Here, the absolute values “cancel” the sign change, creating a constant value across this interval.", "---", "#### Interval 3: $ x > \frac{5}{2} $", "Test $ x = 3 $", "- $ x - \frac{5}{2} = 0.5 > 0 $ → $ |x - \frac{5}{2}| = x - \frac{5}{2} $\n- $ x + 3 = 6 > 0 $ → $ |x + 3| = x + 3 $", "Expression:\n$ |x - \frac{5}{2}| + |x + 3| = (x - \frac{5}{2}) + (x + 3) ="]
