Problem 25 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ \frac{x^{2}-x}{x^{2}+2 x} \leq 0 $$

Step-by-Step Solution

Answer

The solution is \([-2,0) \cup [0,1] \).

Step 1: Identify the Expression

We start with the inequality \(\frac{x^2-x}{x^2+2x} \leq 0\). The goal is to find the values of \(x\) that make this inequality true.

Step 2: Factor the Numerator and Denominator

Factorize both the numerator and the denominator. The numerator \(x^2 - x\) can be factored as \(x(x-1)\), and the denominator \(x^2 + 2x\) as \(x(x+2)\). The inequality becomes \(\frac{x(x-1)}{x(x+2)} \leq 0\).

Step 3: Define Critical Points

The critical points come from setting the numerator and denominator equal to zero. For \(x(x-1) = 0\), the critical points are \(x = 0\) and \(x = 1\). For \(x(x+2) = 0\), the critical points are \(x = 0\) and \(x = -2\) (since the denominator can't be zero in a rational expression, these are restricted from solutions).

Step 4: Determine Test Intervals

Use the critical points \(-2, 0,\) and \(1\) to create intervals: \((-\infty, -2), (-2, 0), (0, 1), (1, \infty)\). Choose test points in each interval to see if the inequality \(\frac{x(x-1)}{x(x+2)} \leq 0\) holds.

Step 5: Test Each Interval

Test points:- Interval \((-\infty, -2)\): Choose \(x = -3\). The expression evaluates to positive.- Interval \((-2, 0)\): Choose \(x = -1\). The expression evaluates to negative.- Interval \((0, 1)\): Choose \(x = 0.5\). The expression evaluates to negative.- Interval \((1, \infty)\): Choose \(x = 2\). The expression evaluates to positive.

Step 6: Incorporate Critical Points

Point \(x = -2\) is undefined due to the denominator being zero. At \(x = 0\) and \(x = 1\), the expression evaluates to zero, which satisfies the \(\leq 0\) condition.

Step 7: Write Solution in Interval Notation

The solution is the union of intervals where the inequality holds: \([-2,0) \cup [0,1] \).

Key Concepts

Critical Points

In the context of inequalities, critical points are specific values of the variable at which the expression undergoes a change in direction or behavior. These points are derived from the numerator and denominator of a rational expression. By solving the equations set by making them equal to zero, these points are identified.

• Numerator critical points occur when the numerator is zero.
• Denominator critical points appear when the denominator equals zero, but are generally excluded from valid solutions because division by zero is undefined.

To find the critical points in our inequality, we factored the numerator and denominator to yield:

• The numerator: \(x(x-1) = 0\) gives us the critical points \(x = 0\) and \(x = 1\).
• The denominator: \(x(x+2) = 0\) provides the points \(x = 0\) and \(x = -2\), but since denominators must not be zero, these values are restricted.

Once we have our critical points, we can better understand how the expression behaves across different intervals.

Test Intervals

Test intervals are ranges that help determine where the inequality is true or false. Once you have the critical points, they divide the number line into intervals. Each interval is then tested with a random point called a 'test point' to see if it satisfies the original inequality.

• The intervals created from our critical points \(-2, 0, 1\) are: \((-fty, -2)\), \((-2, 0)\), \((0, 1)\), and \((1, fty)\).
• Select a test point from each interval and replace it in the inequality.

If the inequality holds true with that test point, then it holds for the whole interval. In this exercise, for example:

• For \((-fty, -2)\) using \(x = -3\), the test result was positive.
• For \((-2, 0)\) using \(x = -1\), the test result was negative.
• For \((0, 1)\) using \(x = 0.5\), the test result was negative.
• For \((1, fty)\) with \(x = 2\), the test result was positive.

From this, the intervals \((-2, 0)\) and \((0, 1)\) showed that the inequality \(\leq 0\) holds true.

Factoring

Factoring involves breaking down a complex expression into simpler components. This simplification makes solving equations, especially inequalities, much easier. The goal is to rewrite an expression as a product of its factors.

• For our inequality, the numerator \(x^2-x\) is expressed as \(x(x-1)\) by factoring out \(x\).
• The denominator \(x^2+2x\) is factored as \(x(x+2)\).

Factoring is a crucial step in identifying critical points since it directly leads to functions where expressions can potentially reach zero or become undefined. Once factored, the inequality can be rewritten in a simpler, more manageable form that is easier to analyze for critical points and test intervals.

Interval Notation

Interval notation is a symbolic representation of a set of numbers along a number line. It is a concise and clear way to denote intervals over which an inequality holds true. In this exercise, after identifying test intervals and checking where the inequality is met, we conclude with interval notation:

• The brackets \([]\) indicate that an endpoint is included in the interval, here representing closed intervals.
• The parentheses are applied to show that an endpoint is not included, showcasing open intervals.

Our solution reads as \([-2,0) \cup [0,1]\), which means all values from \(-2\) to \(0\) are included except \(0\), and all values from \(0\) to \(1\) including \(0\) and \(1\). This denotes all the \(x\) values making the original inequality true.