Problem 47 - Beginning and Intermediate Algebra

Question

Solve each rational inequality. Graph the solution set and write the solution in interval notation. $$\frac{h-9}{3 h+1} \leq 0$$

Step-by-Step Solution

Answer

The solution to the inequality \(\frac{h-9}{3 h+1} \leq 0\) is given by the interval \([- \frac{1}{3}, 9]\). The graph of the solution set is represented as a shaded interval with a closed circle at both endpoints, indicating that the critical points \(h = -\frac{1}{3}\) and \(h=9\) are included in the solution set.

Step 1: Identify the critical points

To identify the critical points, determine the values of \(h\) for which the expression changes its sign. These occur when the expression is equal to zero or is undefined. Set the expression equal to zero to find points where the expression changes its sign. \(\frac{h-9}{3h+1} = 0\) Multiply both sides by \((3h + 1)\) to eliminate the denominator. \(h-9 = 0\) Add 9 to both sides. \(h = 9\) Now, determine when the expression is undefined by finding when the denominator is equal to zero. \(3h+1 = 0\) Subtract 1 from both sides. \(3h= -1\) Divide by 3. \(h = -\frac{1}{3}\) So, the critical points are \(h = 9\) and \(h = -\frac{1}{3}\).

Step 2: Check the intervals between the critical points

We have 3 intervals to check based on our critical points: \((-\infty, -\frac{1}{3})\), \((-\frac{1}{3}, 9)\), and \((9, \infty)\). Choose a test point from each interval and determine if the inequality is true. Interval \((-\infty, -\frac{1}{3})\): Choose \(h = -1\). \(\frac{-1-9}{3(-1)+1} = \frac{-10}{-2} = 5 > 0\) The inequality is false in this interval. Interval \((-\frac{1}{3}, 9)\): Choose \(h = 0\). \(\frac{0-9}{3(0)+1} = \frac{-9}{1} = -9 < 0\) The inequality is true in this interval. Interval \((9, \infty)\): Choose \(h = 10\). \(\frac{10-9}{3(10)+1} = \frac{1}{31} > 0\) The inequality is false in this interval.

Step 3: Write the interval notation

Based on our analysis in Step 2, the inequality is true in the interval \((-\frac{1}{3}, 9)\). Since the inequality includes the equal sign, we also include the critical point \(h = 9\) in our solution set. The interval notation for the solution is: \(\boxed{[-\frac{1}{3}, 9]}\)

Step 4: Graph the solution set on a number line

To graph the solution set, we will shade the interval \((-\frac{1}{3}, 9)\) and include a closed circle at the endpoints to show that \(h = -\frac{1}{3}\) and \(h = 9\) are included in the solution set. ( [--|-------] -----------|-----|---|-----|----------- -∞ -1/3 9 ∞

Key Concepts

Interval Notation

Interval notation is a way of writing subsets of the real number line. It provides a concise way to describe solution sets for inequalities.

To write an interval, we use brackets and parentheses to denote closed and open ends. Here's a brief guide:

• Closed Interval [a, b]: This interval includes the endpoints a and b. It means all numbers between a and b, including a and b themselves.
• Open Interval (a, b): This interval includes all numbers between a and b but not a and b themselves. Points a and b are excluded.
• Half-Closed/Half-Open Interval [a, b) or (a, b]: Here, only one endpoint is included. For [a, b), a is included, but b is not. Conversely, (a, b] includes b, but not a.

In the rational inequality \( \frac{h-9}{3h+1} \leq 0 \), we use interval notation to express the range of values that satisfy the inequality. When we solved the inequality, we found that it is fulfilled within the interval \( [-\frac{1}{3}, 9] \), indicating that both endpoints are included in this solution set.

Graphing Solution Sets

Graphing solution sets helps us visualize where a solution is valid over a number line.

In plotting the solution, specific symbols are used to communicate which values are included in a set. Here's how you can graph solution sets of rational inequalities.

• Closed Circle: A closed circle on the number line indicates that the endpoint is included in the interval. For instance, at points \( h = -\frac{1}{3} \) and \( h = 9 \), we use closed circles because these values make the inequality \( \frac{h-9}{3h+1} \leq 0 \) true.
• Open Circle: An open circle implies that the endpoint is not part of the solution. In cases where our intervals have these, you use open circles.
• Shading: Once endpoints are addressed, shade the number line portion between them to visualize the included values. For \( [-\frac{1}{3}, 9] \), you shade between \( h=-\frac{1}{3} \) and \( h=9 \).

Graphing pressures a tangible reflection of algebraic solutions and confirms whether these solutions correspond to the inequality.

Critical Points

Critical points in rational inequalities occur where expressions change sign or become undefined. Finding these is a crucial step in solving the inequality.

In our rational inequality \( \frac{h-9}{3h+1} \leq 0 \), we identify critical points through two steps:

• Zeroes of the Numerator: Here, set the numerator \( h-9 \) equal to zero. This gives a critical point: \( h=9 \).
• Undefined Points of Denominator: Set the denominator to zero, \( 3h+1 = 0 \), to find \( h = -\frac{1}{3} \), where the expression becomes undefined.

Once critical points are defined, they inform the intervals over which you test the inequality. Determine if the inequality holds true between these points, which is essential for describing the solution set accurately.