Problem 20 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ \frac{2}{1-x} \leq 1 $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -1] \cup (1, \infty)\).

Step 1: Move the constant to one side

Subtract 1 from both sides of the inequality:\[ \frac{2}{1-x} -1 \leq 0 \]

Step 2: Combine into a single fraction

Find a common denominator and combine the terms:\[ \frac{2}{1-x} - \frac{1-x}{1-x} \leq 0 \]which simplifies to:\[ \frac{2 - (1-x)}{1-x} \leq 0 \]and ultimately to:\[ \frac{1+x}{1-x} \leq 0 \]

Step 3: Determine critical points

Identify points where the expression equals zero or is undefined. Set the numerator equal to zero:\[ 1 + x = 0 \]Solving this, we get:\[ x = -1 \]The denominator is zero when:\[ 1 - x = 0 \]Solving this, we get:\[ x = 1 \]These are our critical points: \( x = -1 \) and \( x = 1 \).

Step 4: Test intervals around critical points

Evaluate the inequality in the intervals determined by the critical points: \((-\infty, -1)\), \((-1, 1)\), and \((1, \infty)\).For \(x \in (-\infty, -1)\): Choose \(x = -2\), \[ \frac{1 + (-2)}{1 - (-2)} = \frac{-1}{3} < 0 \]For \(x \in (-1, 1)\): Choose \(x = 0\), \[ \frac{1 + 0}{1 - 0} = 1 > 0 \]For \(x \in (1, \infty)\): Choose \(x = 2\), \[ \frac{1 + 2}{1 - 2} = \frac{3}{-1} = -3 < 0 \]

Step 5: Include critical points where inequality holds

Since the inequality is \( \frac{1+x}{1-x} \leq 0 \), include the points where the inequality equals zero, except where it's undefined. \( x = -1 \) should be included and \( x = 1 \) should not be included due to the denominator becoming zero. So, combining the intervals and points we have:\[ (-\infty, -1] \cup (1, \infty) \].

Step 6: Illustrate the solution on the real number line

On the number line, draw a solid dot at \( x = -1 \) and shade everything to the left to \(-\infty\). Leave \( x = 1 \) unshaded and shade everything to the right to \( \infty \).

Key Concepts

inequalities in math

Inequalities are mathematical expressions that show the relationship of one quantity being greater than, less than, greater than or equal to, or less than or equal to another quantity. For example, the inequality \(\frac{2}{1-x} \leq 1\) means that the expression \(\frac{2}{1-x}\) is always less than or equal to 1. They are essential for representing realistic scenarios where values need to fall within a certain range.
Inequalities use the symbols:

• \(<\) for less than
• \(>\) for greater than
• \(\leq\) for less than or equal to
• \(\geq\) for greater than or equal to

When solving inequalities, it's important to remember:

• Reversing the inequality sign when multiplying or dividing by a negative number.
• Testing intervals between critical points to determine where the inequality holds true.
• Including or excluding boundary points based on the inequality type (strict or non-strict).

rational expressions

Rational expressions are fractions where the numerator and denominator are polynomials. In our given problem, \(\frac{2}{1-x} -1 \leq 0\), the expression \(\frac{2}{1-x}\) is a rational expression.
Here are some key aspects of handling rational expressions:

• Simplify the expression by combining like terms or factoring when possible.
• Identify points where the expression is undefined, i.e., where the denominator equals zero.
• Find common denominators when adding or subtracting rational expressions.

In our solution, we converted \(\frac{2}{1-x} -1 \leq 0\) to a single fraction \(\frac{1+x}{1-x} \leq 0\) by finding a common denominator. This process helps in visualizing and solving the inequality more easily.

interval notation

Interval notation is a way of representing a set of numbers on the number line. This is very useful in solutions to inequalities because it provides a compact way to describe ranges of values.
Here's how interval notation works:

• Use '(' or ')' to indicate that an endpoint is not included (open interval). For example, \( (a, b) \) represents all numbers between a and b, excluding a and b.
• Use '[' or ']' to indicate that an endpoint is included (closed interval). For example, \( [a, b] \) represents all numbers between a and b, including a and b.
• Use 'u' (union) to combine multiple intervals. For instance, \( (-\infty, -1] \cup (1, \infty) \) represents all numbers less than or equal to -1, and greater than 1.

In our example, the final solution set was written as \( (-\infty, -1] \cup (1, \infty) \), which means all values less than or equal to -1 and all values greater than 1, capturing both parts where the inequality holds true but excluding the undefined point (x=1).