Problem 55 - Algebra and Trigonometry

Question

Using Distances to Solve Absolute Value Inequalities Recall that \(|a-b|\) is the distance between \(a\) and \(b\) on the number line. For any number \(x\) , what do \(|x-1|\) and \(|x-3|\) represent? Use this interpretation to solve the inequality \(|x-1|<|x-3|\) geometrically. In general, if \(a

Step-by-Step Solution

Answer

For \(|x-1|<|x-3|\), the solution is \(x < 2\). For \(a < b\), \(|x-a|<|x-b|\) means \(x < \frac{a+b}{2}\).

Step 1: Interpret Geometrically

The expression \(|x-1|\) represents the distance between a number \(x\) and the number 1 on the number line. Similarly, \(|x-3|\) represents the distance between \(x\) and the number 3.

Step 2: Compare Distances

For the inequality \(|x-1|<|x-3|\), we are looking for the values of \(x\) where the distance from \(x\) to 1 is less than the distance from \(x\) to 3.

Step 3: Examine the Critical Point

On the number line, x is exactly halfway between 1 and 3 at the point 2. At this point, the distances \(|x-1|\) and \(|x-3|\) are equal. Therefore, the critical point is \(x = 2\).

Step 4: Determine the Solution Region

For \(|x-1| < |x-3|\), we need \(x\) location closer to 1 than to 3. This is true for all values of \(x\) that are to the left of 2, i.e., \(x < 2\).

Step 5: Generalize the Solution

When \(a < b\), \(|x-a| < |x-b|\) implies that \(x\) should be closer to \(a\) than \(b\). The solution is therefore \(x < \frac{a+b}{2}\).

Key Concepts

Distance on Number Line

When dealing with absolute value inequalities, understanding the concept of distance is crucial. Absolute value expressions like \(|x-a|\) physically represent the distance between a variable \(x\) and a fixed point \(a\) on the number line.
For instance, the expression \(|x-1|\) signifies how far \(x\) is from the point 1, while \(|x-3|\) signifies the distance from point 3. This literal understanding of distance allows you to visualize inequalities as statements about one distance being greater, less, or equal to another.
Recognizing these distances makes it easier to solve inequalities since you can visualize where one point is closer relative to another. Let's explore how this concept is utilized further in inequality solutions.

Geometric Interpretation

Geometry offers a helpful way to interpret and solve absolute value inequalities. By placing numbers on a number line, we gain a clear visual representation of the distances involved.
Consider \(|x-1|<|x-3|\): here, we want all points \(x\) where the distance to 1 is less than the distance to 3. To simplify, this inequality asks which points are nearer to 1 than they are to 3.

• The critical point that divides the regions of the inequality lies at the midpoint between 1 and 3, which is 2.
• Any \(x\) values that fall to the left of 2 satisfy the condition \(|x-1|<|x-3|\), as they will be physically closer to 1 than to 3 on the number line.

This interpretation aligns with your geometric insight into absolute value inequalities.

Critical Points

Critical points play a pivotal role in solving absolute value inequalities. These points serve as the boundaries that delineate where inequalities switch from true to false and vice versa.
When examining \(|x-1|<|x-3|\), the critical point of interest is where \(x\) would mathematically balance the distances to 1 and 3. Located at 2, the critical point represents an equilibrium where the two distances, \(|x-1|\) and \(|x-3|\), become equal.
By understanding critical points, you can quickly determine regions where inequalities hold, aiding in finding the overall solution.

Inequality Solutions

Solving absolute value inequalities involves determining which regions of the number line satisfy the distance conditions. Using the critical points as a starting reference, you can identify where inequalities hold.
For the inequality \(|x-1|<|x-3|\), we determined that the solution is all \(x\) values less than the critical point 2. Thus, the solution set is \(x<2\). This tells us that \(x\) should be closer to 1 than to 3.
In a broader context, if given \(a < b\), the inequality \(|x-a|<|x-b|\) suggests that \(x\) should be closer to \(a\), and the critical line dividing the two regions falls at the midpoint, \(x < \frac{a+b}{2}\). This general rule succinctly captures inequality solutions, providing a straightforward method for solving these kinds of problems.