Problem 34 - Algebra A Combined Function

Question

Solve each inequality. Write the solution set in interval notation. $$ (3 x-12)(x+5)(2 x-3) \geq 0 $$

Step-by-Step Solution

Answer

The solution set is \([-5, \frac{3}{2}] \cup [4, \infty)\)."

Step 1: Identify Critical Points

We need to find the values of \(x\) that make each factor zero, as these critical points define where the sign of the inequality might change. - For \(3x - 12 = 0\), solving gives \(x = 4\).- For \(x + 5 = 0\), solving gives \(x = -5\).- For \(2x - 3 = 0\), solving gives \(x = \frac{3}{2}\).

Step 2: Determine Intervals

The critical points \(-5\), \(\frac{3}{2}\), and \(4\) divide the number line into intervals: 1. \((-\infty, -5)\)2. \((-5, \frac{3}{2})\)3. \((\frac{3}{2}, 4)\)4. \((4, \infty)\)

Step 3: Test Intervals

Choose a test point from each interval and compute the sign of the expression:- In \((-\infty, -5)\), choose \(x = -6\): \((3(-6)-12)(-6+5)(2(-6)-3) = (-30)(-1)(-15) = -450\), negative.- In \((-5, \frac{3}{2})\), choose \(x = 0\): \((3(0)-12)(0+5)(2(0)-3) = (-12)(5)(-3) = 180\), positive.- In \((\frac{3}{2}, 4)\), choose \(x = 3\): \((3(3)-12)(3+5)(2(3)-3) = (-3)(8)(3) = -72\), negative.- In \((4, \infty)\), choose \(x = 5\): \((3(5)-12)(5+5)(2(5)-3) = (3)(10)(7) = 210\), positive.

Step 4: Include Critical Points

Determine if the critical points satisfy the inequality \((3x-12)(x+5)(2x-3) \geq 0\). Substituting in the critical points:- \(x = -5\): Expression is zero.- \(x = \frac{3}{2}\): Expression is zero.- \(x = 4\): Expression is zero.Since the inequality is \(\geq\), include the points where the expression equals zero.

Step 5: Write the Solution in Interval Notation

Combine the intervals that satisfy \((3x-12)(x+5)(2x-3) \geq 0\):- From Step 3 tests, the intervals \((-5, \frac{3}{2})\) and \((4, \infty)\) are positive.- Including the points where the expression equals zero, the solution in interval notation is \([-5, \frac{3}{2}] \cup [4, \infty)\).

Key Concepts

Interval Notation

Interval notation is a way to describe a range of values on a number line. It uses parentheses and brackets to indicate whether endpoints are included or excluded. It's a compact way to represent the solution sets of inequalities, such as the intervals in which a function is positive or negative.
For example, if an interval includes its endpoints, we use square brackets like \([a, b]\). If the interval does not include an endpoint, we use parentheses, such as \(a, b\).

• Square brackets [ ] mean the endpoint is included, also known as "closed."
• Parentheses ( ) mean the endpoint is not included, also known as "open."

In the solution of the inequality \( (3x-12)(x+5)(2x-3) \geq 0 \), interval notation was used to express the intervals that include critical points where the expression is zero. The solution was written as \([ -5, \frac{3}{2} ] \cup [4, \infty)\), ensuring a clear understanding of the range where the solution is valid.

Critical Points

Critical points are crucial in understanding the behavior of polynomial functions. They are the values of \(x\) that make any factor of the polynomial zero. These points help us determine where the polynomial might change sign.
For a polynomial function, such as \( (3x-12)(x+5)(2x-3) \), critical points are obtained by setting each factor equal to zero:

• \(3x - 12 = 0\), solving gives \(x = 4\).
• \(x + 5 = 0\), solving gives \(x = -5\).
• \(2x - 3 = 0\), solving gives \(x = \frac{3}{2}\).

These values divide the number line into different regions. By identifying these critical points, you help map regions where the expression is positive or negative. It's essential to note where these changes occur as they contribute to defining the solution set of the inequality.

Inequality Sign Analysis

Analyzing the sign of an expression across different intervals is an effective way to solve inequalities. This technique helps us discover which regions of the number line make the expression positive or negative.
Once the critical points have been established, they split the number line into intervals, like so:

• \((-\infty, -5)\)
• \((-5, \frac{3}{2})\)
• \((\frac{3}{2}, 4)\)
• \((4, \infty)\)

For each interval, select a test point and substitute it into the inequality expression. Determine whether the result is positive or negative. Consider the interval \( - \infty, -5 \); testing with \(x = -6\) results in a negative sign. Meanwhile, the interval \((-5, \frac{3}{2})\) with \(x = 0\) yields positive.
By performing this sign analysis, we decide which intervals are solutions to the inequality. Then, these intervals are combined to give the complete solution in interval notation.

Zero of Polynomials

The zeros of a polynomial are the \(x\) values that make the polynomial equal to zero. These are calculated by setting each factor in the polynomial to zero and solving for \(x\). This process is vital because zeros mark the critical points on the graph of the polynomial, showing where it crosses the x-axis.
For the polynomial \((3x-12)(x+5)(2x-3)\), the zeros are:

• For \(3x - 12 = 0\), the zero is \(x = 4\).
• For \(x + 5 = 0\), the zero is \(x = -5\).
• For \(2x - 3 = 0\), the zero is \(x = \frac{3}{2}\).

Determining these zeros is essential in the analysis of inequalities. They help divide the number line into helpful intervals to check sign changes, thereby guiding us towards a complete solution of the inequality.