Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Express the solution set of the given inequality in interval notation and sketch its graph. $$ (2 x-3)(x-1)^{2}(x-3)>0 $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, 1) \cup (3, \infty)\).

Step 1: Identify Critical Points

The inequality \((2x-3)(x-1)^2(x-3)>0\) has factors \(2x-3\), \((x-1)^2\), and \(x-3\). Setting each factor equal to zero gives critical points:1. \(2x-3=0\) yields \(x=\frac{3}{2}\).2. \((x-1)^2=0\) yields \(x=1\).3. \(x-3=0\) yields \(x=3\).These are the points where the expression could change sign. Hence, critical points are \(x=1\), \(x=\frac{3}{2}\), and \(x=3\).

Step 2: Determine Sign in Intervals

We divide the number line into intervals using the critical points: \((-\infty, 1)\), \((1, \frac{3}{2})\), \((\frac{3}{2}, 3)\), and \((3, \infty)\). Now, we test a point in each interval:1. \(x=0\) for \((-\infty, 1)\): \((2(0)-3)(0-1)^2(0-3) = (-3)(1)(-3) = 9\), positive.2. \(x=1.25\) for \((1, \frac{3}{2})\): \((2(1.25)-3)(1.25-1)^2(1.25-3) = (-0.5)(0.25)^2(-1.75)\), negative.3. \(x=2\) for \((\frac{3}{2}, 3)\): \((2(2)-3)(2-1)^2(2-3) = (1)(1)(-1) = -1\), negative.4. \(x=4\) for \((3, \infty)\): \((2(4)-3)(4-1)^2(4-3) = (5)(3)^2(1) = 45\), positive.

Step 3: Construct Solution Set

Based on sign analysis:- \((-\infty, 1)\) gives a positive product.- \((1, \frac{3}{2})\) gives a negative product.- \((\frac{3}{2}, 3)\) gives a negative product.- \((3, \infty)\) gives a positive product.The inequality is \(>0\), so solutions are from intervals that produce a positive result.Thus, the solution set is \((-\infty, 1) \cup (3, \infty)\).

Step 4: Sketch the Graph

Plot the number line and mark the critical points \(x=1\), \(x=\frac{3}{2}\), and \(x=3\). Shade the regions \((-\infty, 1)\) and \((3, \infty)\) to indicate the solution set where the inequality holds.This represents where the product is positive.

Key Concepts

Critical Points

In solving inequalities such as \((2x-3)(x-1)^2(x-3)>0\), critical points are where each factor of the expression equals zero. These are the pivot points where the sign of the expression could change. Critical points are found by setting each factor to zero and solving for \(x\). For our example:

• \(2x-3=0\) leads to \(x=\frac{3}{2}\).
• \((x-1)^2=0\) gives \(x=1\).
• \(x-3=0\) results in \(x=3\).

These points are crucial as they form the boundaries of the intervals where the sign of the inequality might shift. Recognizing and pinpointing these critical points is the first step in analyzing any inequality.

Interval Notation

Interval notation is a compact way to represent solutions to inequalities on a number line. After identifying the critical points, these points divide the number line into intervals. For our inequality,

• The critical points \(x=1\), \(x=\frac{3}{2}\), and \(x=3\) create the intervals \((-\infty, 1)\), \((1, \frac{3}{2})\), \((\frac{3}{2}, 3)\), and \((3, \infty)\).

Interval notation uses parentheses \(()\) to indicate that the endpoints are not included and square brackets \([]\) to include endpoints. Since our inequality is \(>0\), we only consider intervals where the product is positive. Thus, the solution in interval notation becomes \((-\infty, 1) \cup (3, \infty)\). Be mindful of using "\(\cup\)" to denote separate intervals where the solutions do not overlap.

Sign Analysis

Sign analysis involves testing points from each interval created by the critical points, to determine where the expression is positive or negative. It's like checking which side of the critical point leads to a positive or negative product.

• Choose test points like \(x=0\) for \((-\infty, 1)\), \(x=1.25\) for \((1, \frac{3}{2})\), \(x=2\) for \((\frac{3}{2}, 3)\), and \(x=4\) for \((3, \infty)\).

Calculate the signs of the expression for each interval based on these test values. Points in \((-\infty, 1)\) and \((3, \infty)\) both result in a positive product when substituted back into the inequality. This step solidifies which intervals contribute solutions, aligning with the inequality's requirement of being greater than zero.

Graph Sketching

Graph sketching for inequalities helps visualize where the inequality holds true on a number line. Here's how you can represent it:

• Draw the number line and place marks at each of the critical points \(x=1\), \(x=\frac{3}{2}\), and \(x=3\).
• Shade the intervals \((-\infty, 1)\) and \((3, \infty)\) given they resulted in a positive sign, highlighting these as the solution areas.

Graph sketching complements sign analysis by providing a visual confirmation of the solution set. It's a straightforward way to illustrate where on the number line the inequality holds true, thus reinforcing your understanding through a clear and precise image.