Problem 51 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

Solution: \((-\infty, -1) \cup [3, \infty)\).

Step 1: Identify Critical Points

To solve the inequality \( \frac{x-3}{x+1} \geq 0 \), we first find the critical points by setting the numerator and denominator equal to zero. This gives us the equations \( x-3=0 \) and \( x+1=0 \), which solve to \( x=3 \) and \( x=-1 \), respectively.

Step 2: Determine Intervals Based on Critical Points

The critical points \( x = 3 \) and \( x = -1 \) divide the number line into three intervals: \((-\infty, -1)\), \((-1, 3)\), and \((3, \infty)\). We will test each interval to determine where the inequality \( \frac{x-3}{x+1} \geq 0 \) holds true.

Step 3: Test Intervals

For the interval \((-\infty, -1)\), choose a test point like \(x=-2\): \(\frac{-2-3}{-2+1} = \frac{-5}{-1} = 5\), which is positive. For \((-1, 3)\), choose \(x=0\): \(\frac{0-3}{0+1} = -3\), which is negative.For \((3, \infty)\), choose \(x=4\): \(\frac{4-3}{4+1} = \frac{1}{5}\), which is positive.

Step 4: Include Critical Points in Solution

Since the inequality is \( \geq 0 \), include points where the fraction is equal to zero. At \(x=3\), \(\frac{3-3}{3+1}=0\). Thus, include \( x=3 \) in the solution. The point \(x=-1\) is undefined for the function (division by zero), so it is not included.

Step 5: Write Solution in Interval Notation

Combine the results from the tests: The inequality is satisfied on the intervals \( (-\infty, -1) \) and \( (3, \infty) \), plus \{3\} from the zero value. Therefore, the solution in interval notation is \( (-\infty, -1) \cup [3, \infty) \).

Step 6: Graph the Solution

To graph the solution set: draw an open circle at \( x = -1 \) (since it’s undefined), a closed dot at \( x = 3 \) (since it’s included), and shade the regions \((-\infty, -1)\) and \([3, \infty)\) on the number line.

Key Concepts

Understanding Interval Notation

Interval notation is a way of expressing a set of numbers along a number line. It is particularly helpful for delineating within which intervals a mathematical expression holds true, especially in inequalities. Let's break it down:

• Open Interval: An open interval is represented using parentheses, like \((a, b)\). This means that all numbers from \(a\) to \(b\) are included, except for \(a\) and \(b\) themselves.
• Closed Interval: Conversely, a closed interval uses square brackets, for example, \[a, b\]. This interval includes \(a\) and \(b\), as well as all the numbers in between.
• Mixed Intervals: Sometimes, you'll encounter intervals that are partially closed, like \[a, b)\]. Here, it includes \(a\) but not \(b\).
• Infinity: Since infinity isn't a number you can reach, intervals involving infinity are always open, like \((a, \,\infty)\).

When expressing solutions in interval notation, think about which values make the inequality true. This concise method gives a clear picture of the number set. Remember, you can also combine intervals using a union, denoted by the \(\cup\) symbol. Such combinations represent a solution set comprising multiple intervals, as we saw with the union of \((-\infty, -1) \,\cup\, [3, \infty)\).

Determining Critical Points

Critical points are vital in solving nonlinear inequalities because they often signal changes in the truth of an expression. Here's why they matter:

• Finding Critical Points: For rational inequalities, locate critical points by setting both the numerator and denominator to zero. These points are where the inequality expression can potentially equal zero or reach undefined states.
• Role in Problem Solving: The critical points divide the number line into segments wherein the expression changes from positive to negative or vice versa. By identifying these segments, you know where to focus when testing intervals.
• Example: In \(\frac{x-3}{x+1} \geq 0\), we found \(x=3\) (numerator zero, makes the expression zero) and \(x=-1\) (denominator zero, makes the expression undefined). This set the foundation for testing intervals and forming a solution.

Using critical points not only structures the solving process but also narrows down the specific areas of the number line that will be part of the solution.

Mastering Inequality Testing

Testing intervals is a crucial step to determine where a nonlinear inequality holds true. By choosing test points from the intervals defined by critical points, you can quickly assess the positivity or negativity of the expression.

Steps in Inequality Testing

1. Break the number line into regions using the obtained critical points.
2. Pick any number (test point) within each region.3. Substitute this number into the original inequality.
4. Determine if the result strengthens or weakens your inequality (i.e., whether it makes the expression positive or negative).

Let’s Illustrate

• Example: For the region \((-\infty, -1)\), choose \(x = -2\). Substituting, we find \(\frac{-5}{-1} = 5\), a positive result, indicating this region satisfies our inequality.
• Choice of Test Point: Test points are at your discretion, as long as they lie within the interval. \(x=0\) for \((-1, 3)\) and \(x=4\) for \((3, \infty)\) serve well in illustrating outcomes for these intervals' expressions.
• Completion: Repeating this for all intervals ensures a comprehensive understanding of where your solution holds true.

By using test points, you can effectively map out where and why each segment of your intervals fulfills the inequality, honing in on a solid solution.