Question

Solve each inequality. Express your solution in interval notation. (a) \(1.99<\frac{1}{x}<2.01\) (b) \(2.99<\frac{1}{x+2}<3.01\)

Short answer

(a): \( \left( \frac{1}{2.01}, \frac{1}{1.99} \right) \); (b): \( \left( \frac{1}{3.01} - 2, \frac{1}{2.99} - 2 \right) \).

Step-by-step solution

Understand the Compound Inequality

In part (a), the inequality given is \( 1.99 < \frac{1}{x} < 2.01 \). It is a compound inequality which means we need to find the values of \( x \) that make both parts of the inequality true at the same time.

Solve the Left Part of the Inequality

To solve \( 1.99 < \frac{1}{x} \), we consider two cases based on the sign of \( x \):1. If \( x > 0 \), multiply both sides by \( x \) (being positive, it doesn't reverse the inequality) to get \( x < \frac{1}{1.99} \).2. If \( x < 0 \), multiply both sides by \( x \) (being negative, it reverses the inequality) to get \( x > \frac{1}{1.99} \). Hence, for this inequality \( x \) cannot be positive.

Solve the Right Part of the Inequality

Similarly, solve \( \frac{1}{x} < 2.01 \):1. If \( x > 0 \), multiply by \( x \) to get \( x > \frac{1}{2.01} \).2. If \( x < 0 \), multiply by \( x \) (inequality reverses) to get \( x < \frac{1}{2.01} \), again \( x \) cannot be positive.

Combine Solutions to Find Intersection

Combine the results:For \( x < 0 \): \( x > \frac{1}{1.99} \) and \( x < \frac{1}{2.01} \) do not overlap since \( \frac{1}{1.99} > \frac{1}{2.01} \), hence there is no solution for \( x < 0 \).For \( x > 0 \): \( x < \frac{1}{1.99} \) and \( x > \frac{1}{2.01} \), therefore the interval is \( \left( \frac{1}{2.01}, \frac{1}{1.99} \right) \).

Repeat Analysis for Part (b)

Consider the inequality \( 2.99 < \frac{1}{x+2} < 3.01 \). This requires a similar breakdown:Substitute \( y = x+2 \), solve for \( y \), and then substitute back to find \( x \). Similar to before, dealing separately with \( y > 0 \) and \( y < 0 \).

Solve for the 'y' Inequalities in Part (b)

For \( 2.99 < \frac{1}{y} \):1. If \( y > 0 \), \( y < \frac{1}{2.99} \).2. If \( y < 0 \), \( y > \frac{1}{2.99} \), implying no solution for \( y < 0 \).For \( \frac{1}{y} < 3.01 \):1. If \( y > 0 \), \( y > \frac{1}{3.01} \).2. If \( y < 0 \), \( y < \frac{1}{3.01} \), again no solution for \( y < 0 \).

Combine 'y' Solutions for Part (b) and Express as 'x'

Combining \( y > 0 \) results in the solution: \( \frac{1}{3.01} < y < \frac{1}{2.99} \).Substitute back \( y = x + 2 \) to get \( \frac{1}{3.01} < x + 2 < \frac{1}{2.99} \).Solve to express in terms of \( x \):\( \frac{1}{3.01} - 2 < x < \frac{1}{2.99} - 2 \).Which results in the interval \( \left( \frac{1}{3.01} - 2, \frac{1}{2.99} - 2 \right) \).

Key concepts

Compound Inequalities

Compound inequalities are equations that contain two inequalities joined together by an 'and' or 'or' statement. For example, the expression \( 1.99 < \frac{1}{x} < 2.01 \) is a compound inequality. It means that the value of the variable \( x \) must satisfy both conditions simultaneously. The "and" condition requires both inequalities to be true at the same time. This often results in a range of values that \( x \) can take.

In compound inequalities with "and", we look for the intersection of the solutions, meaning we find the set of values that satisfy both inequalities. If there is no overlap, then there is no solution for that condition. When facing compound inequalities, it’s crucial to solve each part separately and then find where they intersect.

Interval Notation

Interval notation is a mathematical construct used to represent the set of solutions for an inequality in a concise way. It provides an easy-to-understand format for expressing ranges of values that a variable can take. For example, if we determine that a solution exists between two numbers, say \( a \) and \( b \), we can write \((a, b)\) if neither endpoint is included, or \([a, b]\) if both are included.

In the range \((a, b)\), the parentheses indicate that \(a\) and \(b\) are not included in the solution set. If the inequality sign allows endpoints to be included, brackets \([\ ]\) are used instead. This form of notation is particularly useful in compound inequalities where the solution results in a continuous span of numbers, as seen in the intervals given in our example \( \left( \frac{1}{2.01}, \frac{1}{1.99} \right) \).

Inequality Reversal

When solving inequalities, the direction of the inequality sign can reverse depending on the operations performed on it. This is particularly sensitive when multiplying or dividing by a negative number. For instance, if we have the inequality \( -x < a \) and we divide or multiply through by \( -1 \), the inequality sign will flip, yielding \( x > -a \).

Reversal is crucial in solving parts of compound inequalities separately. In the provided problem, when considering possible values of \( x \) or \( y \) being negative, multiplying by the variable itself automatically triggers a reversal of the inequality sign. Hence, paying attention to the sign of the variable helps correctly navigate between different solution intervals.

Solving Inequalities Step-by-Step

Breaking down inequalities into manageable steps is key to finding the right solution. First, interpret the compound inequality and understand the individual components that need to be addressed. You tackle each segment separately, like the \( 1.99 < \frac{1}{x} \), and determine the conditions they each impose on \( x \).

Handle each side of the inequality with the same set of rules. Remember to account for possible scenarios (e.g., \( x > 0 \) and \( x < 0 \)) and solve under these conditions separately, verifying each solution.

• For \( x > 0 \), handle inequalities without flipping signs.
• For \( x < 0 \), remember that multiplying or dividing affects the inequality direction.

Finally, combine valid solutions to find the complete set adhering to the compound inequality's "and" or "or" logic, and then express this result using interval notation, as seen in deriving \( \left( \frac{1}{3.01} - 2, \frac{1}{2.99} - 2 \right) \).