Question

Express the solution set of the given inequality in interval notation and sketch its graph. $$ (x+2)(x-1)(x-3)>0 $$

Short answer

The solution set is \((-2, 1) \cup (3, \infty)\).

Step-by-step solution

Identify Critical Points

To solve \[(x+2)(x-1)(x-3) > 0\]we first find the critical points by setting \[(x+2)(x-1)(x-3) = 0\].This gives us the critical points: \[x = -2, \quad x = 1, \quad x = 3.\] These are the points where the expression can change its sign.

Determine Test Intervals

The critical points \(x = -2\), \(x = 1\), and \(x = 3\) divide the number line into four intervals:1. \((-\infty, -2)\)2. \((-2, 1)\)3. \((1, 3)\)4. \((3, \infty)\).

Test each Interval

Choose a test point from each interval to determine the sign of the expression:- For \((-\infty, -2)\), choose \(x = -3\): \(((-3)+2)((-3)-1)((-3)-3) < 0\)- For \((-2, 1)\), choose \(x = 0\): \((0+2)(0-1)(0-3) > 0\)- For \((1, 3)\), choose \(x = 2\): \((2+2)(2-1)(2-3) < 0\)- For \((3, \infty)\), choose \(x = 4\): \((4+2)(4-1)(4-3) > 0\).

Form the Solution Set

From Step 3, the expression is positive in intervals \((-2, 1)\) and \((3, \infty)\). Therefore, the solution set in interval notation is:\[(-2, 1) \cup (3, \infty)\].

Sketch the Graph

On a number line:- Place open circles at \(x = -2\), \(x = 1\), and \(x = 3\) because the inequality is strict (greater than).- Shade the region between \(-2\) and \(1\) and the region from \(3\) to \(\infty\) to indicate where the solution set is.

Key concepts

Critical points in inequalities

When solving polynomial inequalities, identifying critical points is a crucial first step. Critical points occur where the polynomial is equal to zero. This is because these points are where the expression can potentially change its sign from positive to negative, or vice versa. To find the critical points, you set the polynomial equal to zero and solve for the variable.

In our exercise, we have the inequality \((x+2)(x-1)(x-3) > 0\). By setting \((x+2)(x-1)(x-3) = 0\), we solve for the critical points:

• \(x = -2\)
• \(x = 1\)
• \(x = 3\)

Each of these points is where the inequality could change its sign. They partition the number line into intervals that we can test for signs.

Interval notation

Interval notation is a succinct way to express sets of numbers, commonly used to describe solution sets of inequalities. It uses brackets and parentheses to define the included and excluded endpoints of intervals.

Parentheses \((\) and \()\) are used to indicate that an endpoint is not included in the interval. This is often the case with strict inequalities like \(>\) or \(<\). Whereas brackets \([\) and \()]\) are used when endpoints are included in the interval, usually with \(\geq\) or \(\leq\) inequalities.

For the inequality \((x+2)(x-1)(x-3) > 0\), the solution in interval notation, as derived from sign testing, is \((-2, 1) \cup (3, \infty)\). Open intervals \((-2, 1)\) and \((3, \infty)\) show where the inequality is greater than zero, excluding the critical points themselves.

Sign testing in polynomial inequalities

Sign testing is a systematic approach for determining the sign of a polynomial over intervals defined by critical points. After identifying critical points, the number line is divided into regions or intervals to test.

You then choose a test point from each interval, substitute it back into the polynomial, and observe the sign of the result.

For example, if we take the polynomial \((x+2)(x-1)(x-3)\), it is divided into four intervals by the critical points:

• \((-\infty, -2)\)
• \((-2, 1)\)
• \((1, 3)\)
• \((3, \infty)\)

We test a number from each interval:

• For \((-\infty, -2)\), test \(x = -3\): expression is negative.
• For \((-2, 1)\), test \(x = 0\): expression is positive.
• For \((1, 3)\), test \(x = 2\): expression is negative.
• For \((3, \infty)\), test \(x = 4\): expression is positive.

This allows us to identify which intervals satisfy the inequality \(> 0\). In this case, it is positive in \((-2, 1)\) and \((3, \infty)\).