Problem 61 - Algebra A Combined Function

Question

A number minus its reciprocal is less than zero. Find the numbers.

Step-by-Step Solution

Answer

The numbers are in the intervals \(-1 < x < 0\) and \(0 < x < 1\).

Step 1: Understand the problem

We need to find all numbers where the number minus its reciprocal is less than zero. In mathematical form, if the number is \( x \), we need to solve for \( x - \frac{1}{x} < 0 \).

Step 2: Set up the inequality

The inequality given is \( x - \frac{1}{x} < 0 \). To solve this, we can rearrange it to \( x < \frac{1}{x} \).

Step 3: Eliminate the fraction

To eliminate the fraction, multiply the entire inequality by \( x^2 \) (assuming \( x eq 0 \)), resulting in \( x^3 < x \).

Step 4: Rewrite the inequality

The inequality \( x^3 < x \) can be rewritten as \( x^3 - x < 0 \), or equivalently, \( x(x^2 - 1) < 0 \).

Step 5: Factorize the expression

Further factorize \( x(x^2 - 1) < 0 \) to get \( x(x - 1)(x + 1) < 0 \). This shows the critical points are \( x = -1, 0, 1 \).

Step 6: Test intervals

Using the critical points, we test the intervals: \(-\infty < x < -1\), \(-1 < x < 0\), \(0 < x < 1\), and \(x > 1\).- For \(-\infty < x < -1\), choose \(x = -2\): \(-2(-3)(-1) > 0\)- For \(-1 < x < 0\), choose \(x = -0.5\): \((-0.5)(-1.5)(0.5) < 0\)- For \(0 < x < 1\), choose \(x = 0.5\): \((0.5)(-0.5)(1.5) < 0\)- For \(x > 1\), choose \(x = 2\): \((2)(1)(3) > 0\)The inequality holds for \(-1 < x < 0\) and \(0 < x < 1\).

Step 7: Conclude the solution

The set of numbers satisfying \( x - \frac{1}{x} < 0 \) is the union of the intervals \(-1 < x < 0\) and \(0 < x < 1\).

Key Concepts

Reciprocal

In the world of mathematics, particularly in algebra, understanding the term "reciprocal" is key to solving many types of equations and inequalities. The reciprocal of a number is essentially one divided by that number.
- For example, the reciprocal of 2 is \( \frac{1}{2} \), and the reciprocal of \( \frac{3}{4} \) is \( \frac{4}{3} \).- A special characteristic of reciprocals is that when you multiply a number by its reciprocal, the result is always 1. For instance, multiplying 5 by its reciprocal \( \frac{1}{5} \) results in 1. This property is essential in algebra for simplifying expressions and solving equations.
In the exercise at hand, the relationship between a number and its reciprocal forms the basis of the inequality \( x - \frac{1}{x} < 0 \). This statement implies that whatever the value of \( x \) might be, subtracting its reciprocal results in a negative number. Thus, understanding reciprocals helps in grasping the problem's core requirement.

Inequality Solving

Solving inequalities is a fundamental skill in algebra involving finding the values that satisfy the condition set by the inequality. Unlike equations, inequalities use signs like \(<\), \(>\), \(\leq\), and \(\geq\) to describe the relation between expressions instead of equality.
For the inequality \( x - \frac{1}{x} < 0 \), the main task is to determine the values of \( x \) which make the expression satisfy this condition. The key steps include:

• Rearranging the inequality, if necessary, to make it easier to solve.
• Eliminating fractions or other complexities which might hinder simple algebraic manipulation. For instance, multiplying by \( x^2 \) as done in the exercise helps simplify\( x - \frac{1}{x} < 0 \) to \( x^3 < x \).
• Expressing the inequality in a form that can be easily factorized, such as \( x(x-1)(x+1) < 0 \).

By converting complex inequalities into factored forms, you are better positioned to determine the value ranges that satisfy the inequality, which leads us to interval testing.

Interval Testing

Interval testing is a technique used to solve polynomial inequalities by determining which intervals between critical points satisfy the inequality. Once you have factorized the polynomial inequality, you identify the roots or zeros, forming intervals on the number line.
For the inequality \( x(x-1)(x+1) < 0 \), the critical points identified are \( x = -1 \), \( x = 0 \), and \( x = 1 \). Using these, the number line is divided into intervals:

• Hello! Wonderful to see you learning about intervals. The intervals are: \(-\infty < x < -1\), \(-1 < x < 0\), \(0 < x < 1\), and \(x > 1\).
• To determine which intervals satisfy the inequality, select a test point from each interval and substitute it into the inequality.
• If the result is negative, then that interval satisfies the inequality; if positive, it does not.

For instance, choosing \(-0.5\) for the interval \(-1 < x < 0\) gives \((-0.5)(-1.5)(0.5) < 0\), indicating satisfaction of the inequality. This methodical testing across intervals allows you to confidently assert which ranges have solutions to the original inequality, leading to a comprehensive result.