Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the solution sets of the given inequalities. $$ \left|2+\frac{5}{x}\right|>1 $$

Step-by-Step Solution

Answer

The solution set is \((-5, -\frac{5}{3})\).

Step 1: Rewrite the Inequality

The inequality is given as \(|2+\frac{5}{x}| > 1\). To solve this, we first consider it as two separate inequalities. The expression inside the absolute value, \(2+\frac{5}{x}\), must be greater than 1 or less than -1. This gives us two separate inequalities: \(2+\frac{5}{x} > 1\) and \(2+\frac{5}{x} < -1\).

Step 2: Solve the First Inequality

Start with \(2+\frac{5}{x} > 1\). Solve for \(x\):\[2 + \frac{5}{x} > 1 \]Subtract 2 from both sides:\[ \frac{5}{x} > -1 \]Multiply both sides by \(x\) (being careful about the sign of \(x\)): \[5 > -x \]This can be rewritten as: \[x > -5 \].

Step 3: Solve the Second Inequality

Now, consider \(2+\frac{5}{x} < -1\). Solve for \(x\):\[2 + \frac{5}{x} < -1 \]Subtract 2 from both sides of the inequality: \[ \frac{5}{x} < -3 \]Multiply both sides by \(x\) (consider possible sign flip): \[5 < -3x \]Rewrite it as \[3x < -5\], then divide both sides by 3: \[x < -\frac{5}{3} \].

Step 4: Combine Solutions

From the two inequalities, we have the solutions \(x > -5\) and \(x < -\frac{5}{3}\). Combine these into a single solution set by taking their intersection. Since \(x\) must satisfy both simultaneously, the overlapping interval is \((-5, -\frac{5}{3})\).

Step 5: Analyze and Test Points

Check endpoints and test with some values within and outside the determined intervals to confirm the solution. Since the original inequality \(\left|2+\frac{5}{x}\right| > 1\) includes no equality, ensure the denominator does not reach zero (so \(x eq 0\)). Values like \(x = -4\) should work while \(x = 0\) must not be included in the solution.

Key Concepts

Absolute Value Inequalities

Absolute value inequalities might seem complex at first glance. But they are simply a way to describe the distance of a number from zero on a number line. Any expression within the absolute value symbols can either be positive or negative. For instance, the inequality \(\left|2+\frac{5}{x}\right| > 1\) suggests that the expression \(2+\frac{5}{x}\) could either be more than one unit away from zero in the positive direction or more than one unit away in the negative direction.

The absolute value concept tells us that we break the inequality into two separate inequalities: \(2+\frac{5}{x} > 1\) or \(2+\frac{5}{x} < -1\).

By splitting the absolute value inequality into these two conditions, we can comfortably solve each inequality separately. This approach effectively handles the dual nature of distances described by the absolute value.

Solution Sets

Solution sets represent the range or ranges of values that satisfy an inequality. For someone solving an inequality like \(\left|2+\frac{5}{x}\right| > 1\), finding the solution set involves determining which values of x make the inequality true.

In this context, solving means identifying x's that ensure \(2+\frac{5}{x}\) is either beyond 1 or less than -1. Once you derive intervals like \(-5, -\frac{5}{3}\), these intervals form the solution set.

It's important to examine endpoints closely since they are derived from inequalities without equality. They typically specify open intervals, meaning the endpoint values themselves are not included in the solution set.

• Ensure that no operation, like division by zero, invalidates part of the solution.
• Critical points, like zero, need exclusion if they create an undefined state, maintaining the solution's validity.

Calculus Step-by-Step Solutions

Step-by-step calculus solutions involve a methodical approach to solving equations and inequalities, as seen in this example. Here's a basic outline:

• **Rewrite the Original Inequality:** Analyze and separate the inequality to handle the absolute value. Create two separate inequalities to work with.
• **Solve Each Inequality Individually:** Carefully solve each inequality, paying attention to algebraic principles, especially when multiplying or dividing by negative values, which flips inequality signs.
• **Combine and Analyze the Result:** After solving, combine the results into a single coherent solution set, capturing the values that satisfy the original inequality.
• **Test for Validity:** Substitute values from each interval into the original inequality to ensure they truly satisfy all conditions. This prevents erroneous conclusions due to overlooked factors like division by zero.

By thoroughly following each step, students can confidently tackle complex calculus problems, reinforcing their understanding as they verify all potential solutions.