Problem 48 - Algebra and Trigonometry

Question

\(33-66\) . Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ (x+2)(x-1)(x-3) \leq 0 $$

Step-by-Step Solution

Answer

The solution in interval notation is \([-2, 3]\).

Step 1: Identify Critical Points

To solve the given inequality \((x+2)(x-1)(x-3) \leq 0\), we first find the critical points by setting each factor equal to zero. This gives us the equations \(x+2=0\), \(x-1=0\), and \(x-3=0\). Solving for \(x\), the critical points are \(x=-2\), \(x=1\), and \(x=3\).

Step 2: Determine Intervals

Based on the critical points \(-2\), \(1\), and \(3\), we divide the number line into intervals: \((-\infty, -2)\), \((-2, 1)\), \((1, 3)\), and \((3, \infty)\).

Step 3: Test Sign in Each Interval

For each interval determined in the previous step, choose a test point to determine the sign of \((x+2)(x-1)(x-3)\):- Test point \(x=-3\) for \((-\infty, -2)\): \((-)(-)(-) = -\) so negative.- Test point \(x=0\) for \((-2, 1)\): \((+)(-)(-) = +\) so positive.- Test point \(x=2\) for \((1, 3)\): \((+)(+)(-) = -\) so negative.- Test point \(x=4\) for \((3, \infty)\): \((+)(+)(+) = +\) so positive.

Step 4: Analyze Inequality and Critical Points

We are looking for intervals where the expression is less than or equal to zero (\(\leq 0\)). From the sign testing:- The expression is negative from \((-\infty, -2)\) and \((1, 3)\).- At \(x=-2\), \(x=1\), and \(x=3\), the expression is equal to zero.- Therefore, include \(-2\), \(1\), and \(3\) as part of the solution. Combine intervals having negative or zero: \([-2, 1]\) and \([1, 3]\).

Step 5: Write the Solution in Interval Notation

Combine the intervals where the inequality holds: \([-2, 3]\).

Step 6: Graph the Solution Set

Draw a number line and shade the region from \(-2\) to \(3\), inclusive. Both endpoints \(-2\) and \(3\) are included as they make the inequality equal to zero.

Key Concepts

Critical Points

Critical points in the context of solving nonlinear inequalities are the values of the variable that satisfy the equation when the expression is set to zero. For an inequality such as \((x+2)(x-1)(x-3) \leq 0\), these are the points where the expression changes sign. To find these points, set each factor equal to zero and solve for the variable:

• \(x+2=0\) gives \(x=-2\)
• \(x-1=0\) gives \(x=1\)
• \(x-3=0\) gives \(x=3\)

With these critical points, you can further analyze and divide the number line into distinct intervals, which is a crucial step in solving nonlinear inequalities.

Interval Notation

Interval notation is a way of writing subsets of the real number line. It provides a concise way to express ranges of values. For the solution of a nonlinear inequality, once you have determined which intervals satisfy the inequality, you use interval notation to express the solution set. In the inequality \((x+2)(x-1)(x-3) \leq 0\), the solution set is all \(x\) values between and including the critical points where the expression is negative or zero. This can be written in interval notation as \([-2, 3]\), where the square brackets indicate that the endpoints \(-2\) and \(3\) are included. These intervals are based on critical points and testing the sign of the expression across those intervals.

Sign Test

The sign test is essential for determining the nature of the product or quotient expression across different intervals on the number line. After identifying critical points, you choose test points from each interval to substitute into the inequality. For the inequality \((x+2)(x-1)(x-3) \leq 0\), the intervals are \((-\infty,-2)\), \((-2,1)\), \((1,3)\), and \((3,\infty)\). Pick a test point from each, such as:

• \(x=-3\) from \((-\infty, -2)\)
• \(x=0\) from \((-2, 1)\)
• \(x=2\) from \((1, 3)\)
• \(x=4\) from \((3, \infty)\)

Evaluate the sign of the expression at each test point to determine whether the product is positive or negative. This helps in identifying which intervals satisfy the inequality.

Solution Set

The solution set of a nonlinear inequality consists of all the values that satisfy the inequality. To compile this set after performing the sign test and evaluating the critical points, note where the original inequality holds true. In our example, for \((x+2)(x-1)(x-3) \leq 0\), the expression is less than or equal to zero in intervals where it's negative or nullified at the critical points. Thus, the solution set is \([-2, 3]\), including the boundary points because the inequality is non-strict. This set represents all possible \(x\) values that make the inequality valid.

Number Line

A number line graphically represents the ranges of solutions for inequalities. By marking the critical points on the number line and examining the intervals between them through a sign test, one can visually interpret the regions where the inequality holds.Here's how you would draw it for \((x+2)(x-1)(x-3) \leq 0\):

• Mark the critical points \(-2\), \(1\), and \(3\) on the line.
• Shade the region from \(-2\) to \(3\)
• Use a closed circle or bracket on \(-2\) and \(3\) to indicate inclusion.

This shaded portion shows all parts of the number line that satisfy the inequality, illustrating the solution set in a clear and visual manner.