Problem 68 - A Graphical Approach to Precalculus with Limits

Question

Solve each rational inequality by hand. $$\frac{3}{2-x}>\frac{x}{2+x}$$

Step-by-Step Solution

Answer

The solution is \((-\infty, -2) \cup (-2, 2)\).

Step 1: Rewrite the Inequality to a Single Rational Expression

To solve the inequality \( \frac{3}{2-x} > \frac{x}{2+x} \), both sides need to have a common denominator. The common denominator for \( (2-x) \) and \( (2+x) \) is \((2-x)(2+x)\). Rewriting the inequality using this common denominator, we have:\[\frac{3(2+x)}{(2-x)(2+x)} > \frac{x(2-x)}{(2-x)(2+x)}\]Combine the fractions:\[\frac{3(2+x) - x(2-x)}{(2-x)(2+x)} > 0\]

Step 2: Expand and Simplify the Numerator

Next, we expand and simplify the numerator \(3(2+x) - x(2-x)\):\[3(2+x) = 6 + 3x\]\[x(2-x) = 2x - x^2\]Substituting back, the expression becomes:\[6 + 3x - (2x - x^2) \]Simplifying this gives:\[6 + 3x - 2x + x^2 = x^2 + x + 6\]

Step 3: Solve the Rational Inequality

The inequality now is:\[\frac{x^2 + x + 6}{(2-x)(2+x)} > 0\]To find where this expression is greater than zero, identify critical points where the expression can change signs: roots of the numerator are none (as C < 0 in the discriminant formula) and roots of the denominator, which are 2 and -2.

Step 4: Test Intervals Around Critical Points

Using the critical points \(-2\) and \(2\), the number line is divided into intervals: \((-\infty, -2)\), \((-2, 2)\), and \((2, \infty)\). For each interval, choose a test point and check if the inequality holds:- For \((-\infty, -2)\), choose \(-3\): Plugging into the expression gives a positive result.- For \((-2, 2)\), choose \(0\): Plugging into the expression gives a positive result.- For \((2, \infty)\), choose \(3\): Plugging into the expression gives a negative result.

Step 5: Determine the Solution Set

The expression \( \frac{x^2 + x + 6}{(2-x)(2+x)} > 0 \) is true on intervals where the test results were positive. Therefore, the solution set is:\[(-\infty, -2) \cup (-2, 2)\]Carefully exclude the points where the expression is undefined (\(x = 2\) and \(x = -2\)) by using parentheses.

Key Concepts

Common Denominator

To solve a rational inequality, the first important step is to rewrite it into a single rational expression by finding a common denominator. In our case, we are dealing with the inequality \( \frac{3}{2-x} > \frac{x}{2+x} \). Here, each term has a different denominator, which are \((2-x)\) and \((2+x)\) respectively.

To combine these into a single fraction, we find the common denominator by multiplying the individual denominators, resulting in \((2-x)(2+x)\). This allows us to rewrite each term with this common denominator:

- Multiply the numerator \(3\) by \( (2+x)\) to get \(3(2+x)\)
- Multiply the numerator \(x\) by \((2-x)\) to get \(x(2-x)\)

Combining these under the common denominator gives us a new inequality: \[\frac{3(2+x) - x(2-x)}{(2-x)(2+x)} > 0\]This step is crucial as it allows us to compare the expressions by simplifying them into a single format.

Critical Points

Once we have rewritten the inequality, the next task is to identify "critical points". Critical points occur where the expression might change signs - these are typically found where the numerator or denominator equal zero.

In the expression:\[\frac{x^2 + x + 6}{(2-x)(2+x)} > 0\]we first examine the numerator \(x^2 + x + 6\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find that there are no real roots, as the discriminant \(b^2 - 4ac = 1^2 - 4 \times 1 \times 6 = -23\) is negative. Thus, the numerator has no critical points.

Next, examine the denominator \((2-x)(2+x)\). Setting each factor to zero, we find critical points at \(x = 2\) and \(x = -2\). These critical points indicate where the expression is undefined, and can thus influence sign changes in the inequality.

Interval Testing

After identifying critical points, the next step is to use "interval testing". This approach checks how the expression behaves in segments divided by the critical points.

For the inequality\[\frac{x^2 + x + 6}{(2-x)(2+x)} > 0\]the critical points \(x = -2\) and \(x = 2\) divide the number line into three intervals:

• \((-\infty, -2)\)
• \((-2, 2)\)
• \((2, \infty)\)

Choose a test value from each interval to see whether it satisfies the inequality:

• Choose \(-3\) from the interval \((-\infty, -2)\). Substituting this into the inequality gives a positive result.
• Choose \(0\) from the interval \((-2, 2)\). Substituting this gives a positive result.
• Choose \(3\) from the interval \((2, \infty)\). Substituting this gives a negative result.

Interval testing helps determine where the inequality holds true, revealing that it holds in the intervals \((-\infty, -2)\) and \((-2, 2)\).

Numerator Expansion

In solving the inequality, it was essential to focus on expanding and simplifying the numerator. This process contributes to correctly combining and analyzing the expression. The critical expression is:\[3(2+x) - x(2-x)\]

First, expand each term separately:

• \(3(2+x)\) which expands to \(6 + 3x\)
• \(x(2-x)\) which expands to \(2x - x^2\)

Next, substitute these expansions into the original numerator and simplify:\[6 + 3x - (2x - x^2) = 6 + 3x - 2x + x^2\]After simplifying, you obtain:\[x^2 + x + 6\]

This expansion and simplification show the path for analyzing the inequality. Thoroughly understanding how to expand and simplify expressions is vital in working with rational inequalities.