Problem 59 - Algebra and Trigonometry

Question

\(33-66\) . Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ 1+\frac{2}{x+1} \leq \frac{2}{x} $$

Step-by-Step Solution

Answer

The solution is \([-2, 1]\).

Step 1: Find a Common Denominator

To solve the inequality \(1+\frac{2}{x+1} \leq \frac{2}{x}\), first we need to find a common denominator to combine the fractions. The common denominator for \(x+1\) and \(x\) is \((x+1)x\).

Step 2: Rewrite the Inequality

Rewrite each fraction using the common denominator:\[1 + \frac{2x}{(x+1)x} \leq \frac{2(x+1)}{(x+1)x}\]This simplifies to:\[1 + \frac{2x}{x^2+x} \leq \frac{2x+2}{x^2+x}\]

Step 3: Combine Terms

Bring all terms to one side:\[1 + \frac{2x}{x^2+x} - \frac{2x+2}{x^2+x} \leq 0\]Simplify:\[1 - \frac{2}{x^2+x} \leq 0\]

Step 4: Solve the Resulting Inequality

Subtract 1 from both sides:\[- \frac{2}{x(x+1)} \leq -1\]Multiply throughout by -1 (this reverses the inequality sign):\[\frac{2}{x(x+1)} \geq 1\]

Step 5: Set Up a Quartic Inequality

Multiply both sides by \(x(x+1)\), considering sign changes if needed:\[2 \geq x(x+1)\]Rewriting gives:\[0 \geq x^2 + x - 2\]

Step 6: Solve Quadratic Inequality

Factor the quadratic:\[x^2 + x - 2 = (x-1)(x+2)\]The inequality becomes:\[(x-1)(x+2) \leq 0\]

Step 7: Determine the Solution Intervals

The critical points occur where \(x-1=0\) or \(x+2=0\), which are \(x=1\) and \(x=-2\). Test intervals between these points: - For \(x < -2\), both factors are negative, making the product positive.- For \(-2 < x < 1\), one factor is negative, making the product negative.- For \(x > 1\), both factors are positive, making the product positive.

Step 8: Express the Solution in Interval Notation

Since we want \((x-1)(x+2) \leq 0\), the solution interval is where the product is less than or equal to zero, which is \([-2, 1]\).

Step 9: Graph the Solution Set

The graph of the solution set includes the interval from \(-2\) to \(1\), with closed circles at both ends to indicate that these endpoints are included:----[=======]----Here the '=' signs show the solution set \([-2, 1]\).

Key Concepts

Solving Inequalities

Solving inequalities involves finding all possible values of a variable that satisfy a comparison like "less than," "greater than," or "equal to."
When dealing with nonlinear inequalities like fractions, we must be careful with certain steps:

• Find a Common Denominator: This allows us to eliminate fractions and simplify the expressions.
• Relocate Terms: Bring all terms to one side to compare against zero.
• Flip the Inequality Sign: If you multiply or divide by a negative number, remember to reverse the inequality sign.

Once simplified, you will often encounter a polynomial inequality, which requires factors to be analyzed over different intervals.

Interval Notation

Interval notation is a concise way of expressing a range of values, often used for solutions to inequalities. Here's how it works:
For example, if the solution of an inequality is that a variable falls between two numbers, such as \([-2, 1]\), in interval notation this means:

• \

Graphing Solutions

Graphing solutions involves plotting the intervals where the inequality holds true. This visual model helps easily identify where the conditions of the inequality are met.
To graph the solution of \([-2, 1]\):

• Number Line Approach: Draw a simple horizontal line representing all possible values of the variable.
• Closed Interval: Use closed circles at the ends, -2 and 1, indicating these numbers are included.
• Shade the Interval: Shade the section between -2 and 1, showing all the numbers that satisfy the inequality.

By following these steps, you create a clear and visual representation of the range that your variable can take to satisfy the given inequality.