Problem 67 - A Graphical Approach to Precalculus with Limits

Question

Solve each rational inequality by hand. $$\frac{1}{x-3} \leq \frac{5}{x-3}$$

Step-by-Step Solution

Answer

The solution is \((3, \infty)\).

Step 1: Set the Inequality to Zero

Begin by rewriting the inequality such that one side is zero. Subtract \(\frac{5}{x-3}\) from both sides: \ \(\frac{1}{x-3} - \frac{5}{x-3} \leq 0\).

Step 2: Combine the Fractions

Since both terms have the same denominator, combine them into a single fraction: \ \(\frac{1-5}{x-3} = \frac{-4}{x-3}\). So the inequality becomes: \ \(\frac{-4}{x-3} \leq 0\).

Step 3: Determine Critical Points

The critical point of the rational function occurs where the denominator is zero, i.e., at \(x - 3 = 0\) which gives \(x = 3\). The critical point of the expression is \(x = 3\) where the function is undefined.

Step 4: Test Intervals Based on Critical Point

The inequality \(\frac{-4}{x-3}\leq 0\) implies we need to find where this fraction is less than or equal to zero. Use test points from the intervals divided by the critical point \(x = 3\): \- For \(x < 3\), choose \(x = 2\): \(\frac{-4}{2-3} = 4\) which is not \(\leq 0\).\- For \(x > 3\), choose \(x = 4\): \(\frac{-4}{4-3} = -4\) which is \(\leq 0\).

Step 5: Write the Solution Set

Since the inequality holds only for \(x > 3\), and not at \(x = 3\) due to it being undefined, the solution is \((3, \infty)\).

Key Concepts

Critical Points

In rational inequalities, critical points play a crucial role. These are the values of the variable that make the denominator zero or the entire function undefined. Identifying critical points helps in separating the domain of the inequality into different intervals, which can then be tested for solutions. In the given exercise, the critical point arises from solving the equation in the denominator, \( x - 3 = 0 \). This results in a critical point at \( x = 3 \). You must always include these points in your solution analysis because they indicate where the function might change its behavior, such as shifting from positive to negative or vice versa.

Test Intervals

Once we've established the critical points, the next step is to determine the intervals for testing the inequality. By dividing the number line based on these critical points, we create intervals to check for the truth of the inequality. Each interval represents a section where the inequality might hold true or false.
For example, with our critical point \( x = 3 \), the intervals we test are \( x < 3 \) and \( x > 3 \). Using test values within these intervals, like \( x = 2 \) for \( x < 3 \) and \( x = 4 \) for \( x > 3 \), allows us to determine where the inequality is satisfied. Evaluating the original inequality with these test points confirms which intervals are part of the solution set. This systematic approach eliminates any guessing and ensures accuracy.

Combine Fractions

A helpful strategy when working with rational inequalities is to simplify them by combining fractions. When two fractions share the same denominator, they can be merged into one fraction by subtracting or adding the numerators while maintaining the common denominator. This simplification step makes the inequality easier to solve and analyze.
In the exercise, we combine \( \frac{1}{x-3} - \frac{5}{x-3} \) into \( \frac{-4}{x-3} \). This process consolidates the fractions, simplifying the inequality to \( \frac{-4}{x-3} \leq 0 \). Combining fractions in this manner means you're working with a single expression, allowing clearer identification of critical points and making interval testing more straightforward.

Solution Sets

Understanding solution sets is essential in solving inequalities. A solution set indicates the range of values for which the inequality holds true. After testing intervals, the solution set can be expressed using interval notation. It shows where your variable satisfies the given inequality.
In this problem, after checking the intervals, we found that \( x < 3 \) did not satisfy the inequality, but \( x > 3 \) did. Therefore, the solution set is \((3, \infty)\), meaning all numbers greater than 3 make the inequality true. The solution excludes the critical point \( x = 3 \), as the rational expression is undefined there. Remember, in rational inequalities, it is crucial to consider where the function is undefined to describe the solution set accurately.