91Ó°ÊÓ

Interval notation is a way of describing a set of numbers on a number line. It's used to show the solution set of an inequality, like the one in our exercise. Here's how it works:

When we determine the intervals where an inequality holds, we use parentheses or brackets to express those ranges:

• Round brackets ( ) are used when the number is not included in the interval, such as for strict inequalities like < or >.
• Square brackets [ ] mean that the number is included, suitable for ≤ or ≥ inequalities.

In our exercise, after determining the solution to the inequality, we found that it holds for the interval \-2 ≤ x ≤ 3\. This interval includes the endpoints, so we write it in interval notation as \[-2, 3\].

This concise notation quickly tells you that x can be any number from -2 to 3, inclusive.

Quadratic factorization is a method used to simplify quadratic expressions by expressing them as the product of two simpler binomials. To factor a quadratic like \(x^2 - x - 6\), we did the following:

1. **Identify a, b, and c**: For \(ax^2 + bx + c\), here \(a = 1\), \(b = -1\), \(c = -6\).
2. **Find two numbers**: These numbers need to multiply to \(ac = -6\) and add to \(b = -1\). The numbers -3 and +2 satisfy both conditions.

Thus, the quadratic \(x^2 - x - 6\) factors into \((x - 3)(x + 2)\).

This factorization allows us to easily identify the roots, which are where the expression equals zero, hence solving \((x - 3)(x + 2) = 0\) gives roots at \(x = 3\) and \(x = -2\). Factorization reveals the key points crucial for solving related inequalities.

A common denominator is critical when working with inequalities involving fractions, like in our exercise. It allows us to eliminate fractions and simplify the problem.

1. **Identify denominators**: Here, we had \(\frac{6}{x-1}\) and \(\frac{6}{x}\).
2. **Determine the common denominator**: For these fractions, the simplest common denominator is \(x(x-1)\).
3. **Eliminate fractions**: Multiply every term by this common denominator. This step clears the fractions, transforming the original inequality into a more manageable form:

\((x(x-1))\left(\frac{6}{x-1} - \frac{6}{x}\right) \geq (x(x-1))\cdot 1\) simplifies to \(6x - 6(x-1) \geq x(x-1)\).

Eliminating the fractions is a key first step that paves the way to solve the inequality using simpler algebraic methods.

Sign analysis is essential in determining intervals where a factored inequality holds true. Here's how it's applied:

1. **Identify key points**: These are the roots obtained from factorization, \(x = -2\) and \(x = 3\). They divide the number line into intervals.
2. **Test intervals**: Choose test points from each interval formed by the roots, such as \(x = -3\), \(x = 0\), and \(x = 4\).
3. **Evaluate sign of each interval**: Substitute these test points into the inequality \((x - 3)(x + 2) \leq 0\).

• For \((x - 3)(x + 2)\):
  • At \(x = -3\), the result is positive.
  • At \(x = 0\), the result is negative.
  • At \(x = 4\), the result is positive.

This tells us that the inequality holds true in the interval \((-2, 3)\), because that’s where the sign is negative or zero.

4. **Check endpoints**: Ensure the endpoints \(-2\) and \(3\) are included if \(\leq\) or \(\geq\), as they also satisfy the inequality. This completes the solution, identifying the valid x-values.