Problem 48 - Algebra A Combined Function

Question

Solve each inequality. Write the solution set in interval notation. $$ \frac{4 x}{x-3} \geq 5 $$

Step-by-Step Solution

Answer

The solution set in interval notation is \((3, 15]\).

Step 1: Move the terms

We begin by subtracting 5 from both sides of the inequality to bring it to the form: \[ \frac{4x}{x - 3} - 5 \geq 0. \]

Step 2: Combine terms over a common denominator

Rewrite \(-5\) as a fraction with the same denominator to combine: \[ \frac{4x}{x - 3} - \frac{5(x - 3)}{x - 3} \geq 0. \] This results in: \[ \frac{4x - 5x + 15}{x - 3} \geq 0, \] or \[ \frac{-x + 15}{x - 3} \geq 0. \]

Step 3: Identify critical points

Critical points occur where the numerator and denominator are zero. Set \(-x + 15 = 0\) (numerator zero) gives \(x = 15\), and \(x - 3 = 0\) (denominator zero) gives \(x = 3\).

Step 4: Test intervals

Test the sign of the expression \(\frac{-x + 15}{x - 3}\) in each interval created by the critical points: \((-\infty, 3), (3, 15), (15, \infty)\). Choose test values such as \(x=0\), \(x=4\), and \(x=16\) for each interval.

Step 5: Determine signs in each interval

For \((-\infty, 3)\), using \(x = 0\): \(\frac{-0 + 15}{0 - 3} = -5\) (negative). For \((3, 15)\), using \(x = 4\): \(\frac{-4 + 15}{4 - 3} = 11\) (positive). For \((15, \infty)\), using \(x = 16\): \(\frac{-16 + 15}{16 - 3} = -\frac{1}{13}\) (negative).

Step 6: Derive solution intervals

The inequality is \(\geq 0\), so we include intervals where the expression is positive. From step 5, this is \((3, 15)\). Additionally, the critical points where \(\frac{-x + 15}{x - 3} = 0\) are considered: \(x = 15\), making the solution \((3, 15]\).

Step 7: Consider domain restrictions

The expression is undefined at \(x = 3\) due to division by zero, so \(x = 3\) is not included in the solution set.

Step 8: Write the solution in interval notation

Combine the information from previous steps to write the solution in interval notation: \((3, 15]\).

Key Concepts

Interval Notation

Interval notation is a way to express the solution set of inequalities, making it compact and clear. To do this, we define intervals that describe all possible values that satisfy the inequality. There are a few symbols and conventions to keep in mind when using interval notation:

• Parentheses \((\) and \()\) are used to exclude endpoints, indicating that the value is not part of the solution set.
• Brackets \([\) and \(]\) are used to include endpoints, showing that the value is part of the solution set.
• The symbols \((-\infty, \infty)\) represent all real numbers, noting that infinity is not a number but a concept and cannot be included in the set.

For the given problem, the final solution in interval notation is \((3, 15]\), indicating that the solution includes all numbers between 3 and 15, with 15 included because it makes the inequality true.

Critical Points

Critical points are key values that determine the boundaries of the test intervals in inequality problems. They occur where the expression is equal to zero or is undefined. Finding these points is a crucial step in solving rational inequalities.
In the provided exercise, critical points came from the numerator and the denominator:

• The numerator \(-x + 15\) equals zero at \(x = 15\).
• The denominator \(x - 3\) equals zero at \(x = 3\).

By identifying these critical points, we can establish the intervals \((-fty, 3)\), \((3, 15)\), and \((15, fty)\) to test the inequality. A critical point where the numerator is zero (like \( x = 15\) in our example) is part of the solution if it satisfies the inequality \(rac{-x + 15}{x - 3} \geq 0\).

Rational Inequalities

Rational inequalities involve expressions that are ratios of polynomial functions. Solving them requires careful manipulation to avoid errors, as they often have undefined points that influence where the expression changes sign.
To solve rational inequalities:

• Move all terms to one side of the inequality, setting the expression \(rac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials.
• Simplify the expression, and then find the values that make either the numerator or the denominator zero. These are your critical points.
• Identify the intervals between and beyond these critical points to test.

In our example, we arrived at \(rac{-x + 15}{x - 3} \geq 0\), introducing the need to identify when this fraction changes from positive to negative as \(x\) passes certain values.

Test Intervals

After determining the critical points, create test intervals. These intervals help us identify where the expression satisfies the inequality. Test intervals allow us to evaluate the sign of the expression within each segment of interest.
Follow these steps to work with test intervals effectively:

• Divide the number line into sections using your critical points. For our example, these intervals are \((-fty, 3)\), \((3, 15)\), and \((15, fty)\).
• Choose a simple test point from each interval to substitute into the simplified inequality. Often, choosing integers makes calculations easier. In our exercise, points like \(x = 0\), \(x = 4\), and \(x = 16\) were tested.
• Determine whether the result is positive or negative for each test point and identify which intervals satisfy the inequality condition \(rac{-x + 15}{x - 3} \geq 0\).

This testing process helps us confirm that only \((3, 15]\) meets the inequality condition, guiding us toward the correct solution set.