Problem 23 - Algebra A Combined Function

Question

Solve. Write the solution set in interval notation. $$ \frac{x^{2}+6}{5 x} \geq 1 $$

Step-by-Step Solution

Answer

Solution set: (0, 2] ∪ [3, ∞).

Step 1: Rearrange the inequality

First, we want to organize the inequality so that all terms are on one side. Subtract 1 from both sides of the inequality: \( \frac{x^{2}+6}{5x} - 1 \geq 0 \).

Step 2: Combine terms

To combine the terms, find a common denominator, which is \(5x\). Thus, rewrite the inequality: \( \frac{x^{2}+6 - 5x}{5x} \geq 0 \).

Step 3: Set the numerator to zero

Solve \(x^{2} - 5x + 6 = 0\) to find the critical points where the expression equals zero. This is a quadratic equation. Factor it: \((x-2)(x-3) = 0\). Hence, the solutions are \(x = 2\) and \(x = 3\).

Step 4: Identify the undefined point

Set the denominator equal to zero to find any values where the expression is undefined: \(5x = 0\). This implies \(x = 0\) is a point of discontinuity.

Step 5: Test intervals

Divide the number line using the critical points and undefined point: \(x = 0, 2, \text{and } 3\). Test the intervals: (-∞, 0), (0, 2), (2, 3), and (3, ∞). Choose test points within these intervals to check if the inequality holds.1. For (-∞, 0): Test with \(x = -1\) results \(\frac{1 + 6}{-5} < 0\) (False)2. For (0, 2): Test with \(x = 1\) results \(\frac{1+6}{5} - 1 = \frac{2}{5} > 0\) (True)3. For (2, 3): Test with \(x = 2.5\) results \(\frac{(2.5)^2 + 6 - 5(2.5)}{5(2.5)} = \frac{2.25}{12.5} > 0\) (True)4. For (3, ∞): Test with \(x = 4\) results \(\frac{16 + 6 - 20}{20} = \frac{1}{20} > 0\) (True)

Step 6: Conclusion using interval notation

Based on the tested intervals and considering where the expression is defined, the solution set, accounting for \(x = 2, x = 3\) meeting at equality, is (0, 2] ∪ [3, ∞). Note that we exclude 0 because it makes the denominator zero.

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants. These equations represent parabolic graphs on the coordinate plane.
To solve a quadratic equation, one can use factoring, completing the square, or the quadratic formula. In this example, we solved the quadratic equation \( x^2 - 5x + 6 = 0 \) by factoring to get \((x-2)(x-3) = 0\). Factoring transforms the quadratic into a product of linear factors, allowing us to find the roots or solutions at \( x = 2 \) and \( x = 3 \).

• Factoring: It involves looking for two numbers that add to \( b \) and multiply to \( ac \). For \( x^2 - 5x + 6 \), those numbers are -2 and -3.
• Graphical Interpretation: The roots are the x-intercepts of the quadratic's graph. The vertex form and direction (upwards or downwards) of the parabola can also give insight into the nature of the solutions.

Understanding these roots is vital, as they provide critical points in the analysis of quadratic inequalities and help form intervals on a number line.

Interval Notation

Interval notation is a mathematical concept used to describe a set of numbers between two endpoints. It is especially useful in summarizing ranges of solutions for inequalities.
This format uses brackets to indicate whether endpoints are included or not.

• Closed Intervals \([a, b]\): Both endpoints are included.
• Open Intervals \((a, b)\): Both endpoints are excluded.
• Half-open Intervals \((a, b]\) or \([a, b)\): One endpoint is included while the other is not.

For the given inequality, the final solution set is found to be \((0, 2] \cup [3, \infty)\). This means values from just greater than 0 up to and including 2, and from 3 onwards to infinity, satisfy the inequality. The union symbol \( \cup \) denotes a combination of two or more intervals. Let's remember that the interval does not include 0 because the expression becomes undefined at this point.

Rational Expressions

Rational expressions are fractions involving polynomials in their numerator and/or denominator. They require special consideration as they can be undefined at certain points.
In the expression \( \frac{x^2 + 6}{5x} - 1 \), we deal with rational expressions when manipulating the inequality:

• Simplifying: Combine terms by finding a common denominator. Here, the common denominator is \(5x\).
• Undefined Points: Identify when the denominator equals zero. For \( 5x = 0 \), \( x = 0 \) is a point where the expression is undefined.

Handling rational expressions in inequalities requires checking for undefined values, which can affect the interval of definition of the inequality solution. It's crucial to test these values when determining the solution set.

Critical Points

Critical points in the context of inequalities are the values of \( x \) where the expression either equals zero or becomes undefined.
They help to break the number line into intervals for testing where the inequality holds true.

• Finding Critical Points: Solutions to \( x^2 - 5x + 6 = 0 \) (via factoring) and the undefined point \( x = 0 \) from \( 5x = 0 \) are considered critical points.
• Sign Testing: The intervals created by critical points, like \((-\infty, 0)\), \((0, 2)\), \((2, 3)\), and \((3, \infty)\), are tested to determine where the inequality holds true.

Critical points guide us through solving inequalities by signaling changes in the behavior of the expression. They are instrumental in shaping the final solution set and determining which sections of the number line satisfy the original inequality. Remember to test values within these intervals to ensure correctness.