Problem 65 - Algebra and Trigonometry

Question

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

Solution in interval notation: \((-2, -1) \cup (0, 1]\).

Step 1: Find a common denominator

To solve \(1 + \frac{2}{x+1} \leq \frac{2}{x}\), we start by finding a common denominator for the fractions. The common denominator is \((x+1)x\). Thus, we rewrite the inequality: \(\frac{x(x+1)}{x(x+1)} + \frac{2x}{x(x+1)} \leq \frac{2(x+1)}{x(x+1)}\).

Step 2: Combine the fractions

Combine the fractions into a single inequality over the common denominator: \(\frac{x^2 + x + 2x}{x(x+1)} \leq \frac{2(x+1)}{x(x+1)}\). This simplifies to \(\frac{x^2 + 3x}{x(x+1)} \leq \frac{2x + 2}{x(x+1)}\).

Step 3: Simplify the inequality

Remove the common denominator by subtracting \(\frac{2x + 2}{x(x+1)}\) from both sides: \(\frac{x^2 + 3x - 2x - 2}{x(x+1)} \leq 0\). Simplifying the numerator results in \(\frac{x^2 + x - 2}{x(x+1)} \leq 0\).

Step 4: Factor the numerator

Factor the quadratic expression \(x^2 + x - 2\) in the numerator: \(\frac{(x-1)(x+2)}{x(x+1)} \leq 0\).

Step 5: Determine critical points

Find critical points by setting each factor in the inequality to zero: \(x-1=0\), \(x+2=0\), \(x=0\), and \(x+1=0\). This gives the critical points \(x = 1, -2, 0, -1\).

Step 6: Test intervals around critical points

Test the sign of \(\frac{(x-1)(x+2)}{x(x+1)}\) in each interval determined by the critical points: \((-fty, -2)\), \((-2, -1)\), \((-1, 0)\), \((0, 1)\), and \((1, fty)\). Choose a test point in each interval to determine the sign of the inequality.

Step 7: Write the solution in interval notation

The solution to \(\frac{(x-1)(x+2)}{x(x+1)} \leq 0\) based on testing is \(x \in (-2, -1) \cup (0, 1]\). Points \(-2\) and \(1\) are included due to the non-strict inequality, while \(-1\) and \(0\) are not included as the original denominators would be zero, causing undefined expressions.

Key Concepts

Fraction Simplification

When dealing with inequalities that contain fractions, like the given exercise, simplifying these fractions is often the first step. Simplification involves finding a common denominator between the fractions so that they can be combined or compared easily.

• The target is to express each fraction so it has the same denominator.
• This allows us to directly compare or subtract the fractions effectively.

In the example \(1 + \frac{2}{x+1} \leq \frac{2}{x}\), the common denominator is \((x+1)x\). Converting each term in the inequality to this common denominator allows us to combine them. This step is crucial for simplifying the inequality into a form that is manageable for further solving steps.

Interval Notation

Interval notation is a way of representing the set of solutions for an inequality. It is a concise method to convey which parts of the number line are included in the solution.

• Use parentheses, \((\), to denote that an endpoint is not included.
• Use brackets, \([\), to show that an endpoint is included.

Let's take \((-2, -1)\) as an example of interval notation:
- The numbers between \(-2\) and \(-1\) are included, but the endpoints themselves are not.
The solution set for the given inequality is \((-2, -1) \cup (0, 1]\).
Here, union symbol (\(\cup\)) is used to indicate more than one interval is part of the solution set.

Critical Points

Critical points in the context of inequalities help us determine where the expression changes sign. These points arise when the numerator or the denominator of a simplified fraction equal to zero.

• Numerator turning zero makes the whole fraction zero, a potential part of the solution.
• Denominator zero means the fraction is undefined, hence not part of the solution.

For instance, in \((x-1)(x+2)/(x(x+1)) \leq 0\), the critical points come from solving \((x-1) = 0\), \((x+2) = 0\), \(x = 0\), and \((x+1) = 0\).
These give us \(x = 1, -2, 0, -1\).
Detecting these points is the primary step for testing intervals and determining sign changes in nonlinear inequalities.

Factorization

Factorization breaks down complex expressions into simpler components, allowing us to address each part individually. In inequalities, it helps in identifying critical points and simplifying expressions.

• Look for expressions that can be rewritten as products of simpler expressions.
• A polynomial like \(x^2+x-2\) is factorized to \((x-1)(x+2)\).

Factorization makes the inequality easier to evaluate by isolating terms that can be zero. This process was clearly used when solving our inequality \(\frac{x^2+x-2}{x(x+1)} \leq 0\).
By expressing this as \(\frac{(x-1)(x+2)}{x(x+1)} \leq 0\), we can clearly understand the conditions for the inequality by dealing with each factor separately.

Solution Set Graphing

Graphing the solution set of an inequality provides a visual representation of which values satisfy the inequality. It enhances understanding by depicting intervals on a number line.

• Draw a number line and mark critical points as open or closed dots.
• Shaded regions show where the inequality holds true.

For our inequality, the critical points at \(-2, -1, 0, 1\) define intervals on the number line.
We shade from \(-2\) to \(-1\) and from \(0\) to \(1\), with \(1\) being a closed dot to denote inclusion. Critical points \(-2\) and negative numbers like \(-1, 0\) remain open dots indicating they aren't part of the solution. This graph helps anyone clearly see the results of our interval methods and solution decisions.