Problem 50 - Algebra and Trigonometry

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ (x-5)(x-2)(x+1)>0 $$

Step-by-Step Solution

Answer

The solution is \((-1, 2) \cup (5, \infty)\).

Step 1: Identify Critical Points

First, we find the critical points by setting each factor of the inequality to zero. The critical points are the values of \(x\) where the expression equals zero.- \(x-5=0\) gives \(x=5\)- \(x-2=0\) gives \(x=2\)- \(x+1=0\) gives \(x=-1\)Thus, the critical points are \(x=-1, x=2, x=5\).

Step 2: Test Intervals

The critical points divide the number line into four intervals: \((-fty, -1), (-1, 2), (2, 5), (5, fty)\). We need to test a point from each interval in the inequality to determine where the product \((x-5)(x-2)(x+1) > 0\).- For interval \((-fty, -1)\), test point \(x = -2\): \((-2-5)(-2-2)(-2+1) = (-7)(-4)(-1) = -28 < 0\).- For interval \((-1, 2)\), test point \(x = 0\): \((0-5)(0-2)(0+1) = (-5)(-2)(1) = 10 > 0\).- For interval \((2, 5)\), test point \(x = 3\): \((3-5)(3-2)(3+1) = (-2)(1)(4) = -8 < 0\).- For interval \((5, fty)\), test point \(x = 6\): \((6-5)(6-2)(6+1) = (1)(4)(7) = 28 > 0\).

Step 3: Construct Solution Set

Based on the test points, the inequality \((x-5)(x-2)(x+1) > 0\) holds for the intervals \((-1, 2)\) and \((5, fty)\). Therefore, the solution set in interval notation is \((-1, 2) \cup (5, fty)\).

Step 4: Graph the Solution Set

To graph the solution set, plot an open circle at \(x = -1\), \(x = 2\), and \(x = 5\) since the inequality is 'greater than' and not 'greater than or equal to'.Shade the region on the number line from \(-1\) to \(2\) and from \(5\) to infinity, excluding the points \(-1, 2,\) and \(5\).

Key Concepts

Critical Points

In solving nonlinear inequalities, finding the critical points is crucial. Critical points occur where each factor of the inequality equals zero, marking the boundaries between intervals. For the inequality \[(x-5)(x-2)(x+1) > 0\],setting each factor to zero gives:

• \(x-5=0 \) yields \(x=5\)
• \(x-2=0 \) yields \(x=2\)
• \(x+1=0 \) yields \(x=-1\)

Thus, the critical points are \(x = -1, 2,\) and \(5\). These points divide the number line into distinct regions, helping us test the inequality across different sections.

Interval Notation

Interval notation is a concise way to describe the sets of solutions. It efficiently shows where an inequality holds true on the number line. Here, the critical points create four intervals: \((-\infty, -1), (-1, 2), (2, 5), (5, \infty)\).
After testing each interval, we know that the inequality holds true in \((-1, 2) \ \text{and}\ (5, \infty)\).
Interval notation uses parentheses \(( )\) for open boundaries, indicating that the endpoints are not included in the solution. Thus, the solution in interval notation is written as \((-1, 2) \cup (5, \infty)\).
The union symbol \( \cup \) joins disjoint solution sets.

Inequality Solution

Solving an inequality, particularly a nonlinear one, involves finding where the product of expressions changes sign. First, identify critical points, then test intervals between them.
For each interval, choose a test point. Plug this point into the inequality to determine whether the result is positive or negative.

• For \((-\infty, -1)\), a test at \(x = -2\) results in a negative outcome.
• For \((-1, 2)\), a test at \(x = 0\) gives a positive outcome.
• For \((2, 5)\), a test at \(x = 3\) results in a negative outcome.
• For \((5, \infty)\), a test at \(x = 6\) results in a positive outcome.

Thus, the solution set where the inequality holds is\((-1, 2) \cup (5, \infty)\).

Graphing Inequalities

Graphing inequalities visually represents where solutions exist on the number line. With our inequality\((x-5)(x-2)(x+1) > 0\), we place open circles at critical points \(-1, 2, \ \text{and}\ 5\) indicating that these points are not included.
Then, shade the line segments that satisfy the inequality:

• From \(-1\) to \(2\) is shaded.
• From \(5\) to infinity is shaded.

Each shaded part represents an interval where the inequality holds true. Remember that the open circles emphasize that endpoints like \(-1, 2, \ \text{and}\ 5\) are not part of the solution set.