Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Express the solution set of the given inequality in interval notation and sketch its graph. $$ \frac{3}{x+5}>2 $$

Step-by-Step Solution

Answer

The solution is \((-5, -3.5)\).

Step 1: Set Up the Inequality

Start by writing the inequality: \( \frac{3}{x+5} > 2 \). We need to manipulate this to find the values of \( x \) that satisfy the inequality.

Step 2: Move Terms to One Side

Subtract 2 from both sides of the inequality: \( \frac{3}{x+5} - 2 > 0 \). Rewrite this as a single fraction: \( \frac{3}{x+5} - \frac{2(x+5)}{x+5} > 0 \).

Step 3: Simplify the Expression

Combine the fractions: \( \frac{3 - 2(x+5)}{x+5} > 0 \). This simplifies to \( \frac{3 - 2x - 10}{x+5} > 0 \). Further, simplify to \( \frac{-2x - 7}{x+5} > 0 \).

Step 4: Find Critical Points

Find where the expression equals zero or is undefined. Set the numerator equal to zero: \(-2x - 7 = 0 \Rightarrow x = -\frac{7}{2} \).Set the denominator equal to zero: \(x + 5 = 0 \Rightarrow x = -5 \).The critical points are \( x = -\frac{7}{2} \) and \( x = -5 \).

Step 5: Test Intervals

Test intervals around the critical points (-∞, -5), (-5, -3.5), and (-3.5, ∞). Substitute a number from each interval into the expression \( \frac{-2x - 7}{x+5} \): 1. Test \( x = -6 \) in (-∞, -5): \( \frac{-2(-6) - 7}{-6+5} = \frac{5}{-1} = -5 \), which is not positive; thus, this interval does not satisfy.2. Test \( x = -4 \) in (-5, -3.5): \( \frac{-2(-4) - 7}{-4+5} = \frac{1}{1} = 1 \), which is positive; thus, this interval satisfies.3. Test \( x = 0 \) in (-3.5, ∞): \( \frac{-2(0) - 7}{0+5} = \frac{-7}{5} = -1.4 \), which is not positive; thus, this interval does not satisfy.

Step 6: Write Solution in Interval Notation

The only interval where the inequality holds is (-5, -3.5). It is important to note that \( x = -5 \) is undefined, so it cannot be included in the solution set.

Step 7: Graph the Solution

On a number line, draw an open circle at \( x = -5 \) (since it's not included in the solution), then shade the region between \( x = -5 \) and \( x = -3.5 \). Finally, put a parenthesis around both \( -5 \) and \( -3.5 \), representing the interval \((-5, -3.5)\).

Key Concepts

Interval Notation

When we solve inequalities, we often express the solution set using interval notation. Interval notation is a way to describe a set of numbers that satisfy an inequality, showing the beginning and end of the interval.
For instance, the open interval \((-5, -3.5)\) denotes all numbers greater than \(-5\) and less than \(-3.5\).
This means:

• The number \(-5\) is not included in the solution set, which is indicated by the use of a parenthesis.
• Similarly, the number \(-3.5\) is also not included, denoted by another parenthesis.

Interval notation is particularly useful because it conveys both the range of values and whether the endpoints are included or not. Remember, parentheses \((-\infty, -3)\) indicate that endpoints are not part of the solution, whereas brackets \([-3, 5]\) show that endpoints are included in the solution set.

Critical Points

Critical points are where the inequality changes from being true to false, or vice versa. To identify them, we focus on where the related expression is equal to zero or where it becomes undefined.
In our example, this happens when:

• The numerator is zero: solve \(-2x - 7 = 0\) to find \(x = -\frac{7}{2}\).
• The denominator is zero: solve \(x + 5 = 0\) to find \(x = -5\).

These values, \(x = -\frac{7}{2}\) and \(x = -5\), are termed critical points because they determine potential boundaries or changes in the solution set. Critical points help break down the number line into testable segments, guiding us in determining where the inequality holds true.

Test Intervals

Test intervals are sections of the number line created using critical points. These intervals help us determine where the inequality is satisfied.
Once critical points are identified, we segment the number line into intervals:

• One interval is \((-\infty, -5)\).
• Another is \((-5, -3.5)\).
• The third is \((-3.5, \infty)\).

Within each interval, choose any value to test the inequality. This helps check which intervals satisfy the inequality:

• If the result is positive, the inequality holds true for that interval.
• If not, it does not satisfy the requirement.

For example, when \(x = -4\) in the interval \((-5, -3.5)\), the inequality holds, as the result is positive. Thus, \((-5, -3.5)\) is included in the solution set, vividly showing where the inequality holds true on a graph.