Problem 9 - Algebra A Combined Function

Question

Write the solution set in interval notation. $$ (x-6)(x-4)(x-2)>0 $$

Step-by-Step Solution

Answer

The solution set is \((2, 4) \cup (6, +\infty)\).

Step 1: Find the Zero Points

Set each factor of the inequality \((x-6)(x-4)(x-2) > 0\)to zero to find the critical points.- \(x-6 = 0\) gives \(x = 6\)- \(x-4 = 0\) gives \(x = 4\)- \(x-2 = 0\) gives \(x = 2\)These are the points where the expression changes sign.

Step 2: Determine Intervals Between Critical Points

The critical points divide the number line into the intervals:- \((-\infty, 2)\)- \((2, 4)\)- \((4, 6)\)- \((6, +\infty)\).

Step 3: Test Sign in Each Interval

Choose a test point from each interval and evaluate the sign of the expression \((x-6)(x-4)(x-2)\):- For \((-\infty, 2)\), test \(x = 0\): \((0-6)(0-4)(0-2) < 0\) (negative)- For \((2, 4)\), test \(x = 3\): \((3-6)(3-4)(3-2) > 0\) (positive)- For \((4, 6)\), test \(x = 5\): \((5-6)(5-4)(5-2) < 0\) (negative)- For \((6, +\infty)\), test \(x = 7\): \((7-6)(7-4)(7-2) > 0\) (positive).

Step 4: Write the Solution Set

The inequality is satisfied where the expression is positive.Thus, the solution set in interval notation is:\((2, 4) \cup (6, +\infty)\).

Key Concepts

Interval Notation

When dealing with inequalities in algebra, interval notation is a concise way of writing the set of solutions. This is especially useful when the solution involves a range of values. Instead of listing every single value, interval notation captures all values within a start and end point, using parentheses or brackets.

• Parentheses, like \(a, b\), indicate that the endpoints \('a' and 'b'\) are not included in the set.
• Brackets, like \[a, b\], indicate inclusion of the endpoints.

A union symbol \(\cup\) is used when there are multiple intervals forming the solution. For instance, \(2, 4\) \cup \(6, +\infty\) indicates that values between 2 and 4, as well as those greater than 6, are solutions. This approach is not only efficient but also eliminates ambiguity.

Zero Points

Zero points, or roots, of an expression occur where each factor in the inequality equals zero. Solving for these zero points is critical because they are potential places where signs can change. These are sometimes referred to as the 'critical points.'
Consider the inequality \(x-6)(x-4)(x-2) > 0\). Setting each factor to zero helps us find the transition points:

• Set \(x - 6 = 0\), you find \('x = 6'\).
• Set \(x - 4 = 0\), you find \('x = 4'\).
• Set \(x - 2 = 0\), you find \('x = 2'\).

These points, 2, 4, and 6, act as boundaries that split the number line into intervals. Understanding these points helps in determining where the inequality holds true.

Critical Points

Critical points are where the function changes behavior, like crossing from positive to negative or vice versa. These are determined by solving \((x-a)\) for zero in each factor of the inequality. In our example, the critical points were found to be at \(x = 2\), \(x = 4\), and \(x = 6\).
Each critical point divides the number line into different intervals. These divisions are essential for testing the sign of the function in each interval, which is the next step in solving the inequality. Knowing the position and effect of these points paints a clearer picture of where solutions may lie, based on positive or negative intervals.

Sign Testing in Intervals

After identifying intervals created by critical points, the next step is to test the sign of the expression within each interval. This process helps us determine where the original inequality holds true. Select a test point from each interval and substitute it into the expression.
For the inequality \(x-6)(x-4)(x-2) > 0\), choose:

• For \((-\infty, 2)\), test \('x = 0'\); the result is negative.
• For \(2, 4\), test \('x = 3'\); the result is positive.
• For \(4, 6\), test \('x = 5'\); the result is negative.
• For \(6, +\infty\), test \('x = 7'\); the result is positive.

The positive intervals, \(2, 4\) and \(6, +\infty\), satisfy the inequality, showing where solutions exist. This systematic approach ensures every interval is checked for consistency with the original condition.