Question

Solve the inequality. Then graph the solution set on the real number line. \(3(x-1)(x+1)>0\)

Short answer

The solution to the inequality \(3(x-1)(x+1)>0\) is \(x < -1\) and \(x > 1\). The graph of the solution set includes the points \(x = -1\) and \(x = 1\) and all points to the left of -1 and to the right of 1.

Step-by-step solution

Expansion of the inequality

First, let's expand the left hand side of the inequality: \(3(x-1)(x+1)>0\). Expanding it will give us \(3x^2-3>0\).

Critical Points

The critical points of an inequality are the points where the function equals to zero. Let's find out the critical points of this function. Setting \(3x^2 - 3 = 0\), we get \(x^2 - 1 = 0\). This gives us two roots x = -1 and x = 1.

Interval Analysis and Solution

The inequality will change its sign whenever it passes through a critical point. Now we will test 3 intervals: x < -1, -1 < x < 1 and x > 1. For x < -1, the value of the function would be negative since \(x^2\) is always positive and we have -3 in the function, For -1 < x < 1, the value of the function would be negative as well. For x > 1, the function value would be positive. Hence the values of x which satisfy the inequality are x< -1 and x > 1.

Graphing the solution

The solution on a number line would be a filled circle at x=-1 and x=1 and all points to the left of -1 and to the right of 1.

Key concepts

Graphing Solution Sets

Graphing solution sets of inequalities involves representing the solutions of an inequality visually on a number line. Let's take the inequality \(3(x-1)(x+1) > 0\) as an example. Once we have determined which intervals satisfy the inequality, graphing helps us see these intervals clearly.
To graph this solution set:

• We first identify our critical points where the inequality changes direction. In our case, these points are \(x = -1\) and \(x = 1\).
• Next, we visually represent these points on the number line. Since the inequality is strict ('>'), we use open circles at \(x = -1\) and \(x = 1\) to indicate these points are not included in the solution.
• Lastly, we shade the sections of the number line corresponding to \(x < -1\) and \(x > 1\) since these intervals satisfy the inequality. This shaded region represents all the possible values of \(x\) for which the inequality holds true.

Adding open or filled circles and shading the number line makes it much easier to convey the solution at a glance. Graphing solution sets is a valuable tool that helps in thoroughly understanding the range of solutions for an inequality.

Interval Analysis

Interval analysis is crucial for solving inequalities and involves examining the function's behavior across different ranges. Once the critical points are identified, we need to understand how the function behaves on the intervals formed by these points.
For the inequality \(3(x-1)(x+1) > 0\), the critical points are \(x = -1\) and \(x = 1\). These critical points divide the number line into three intervals:

• \(x < -1\)
• \(-1 < x < 1\)
• \(x > 1\)

Analyzing each interval involves selecting a test point from each interval and substituting it into the inequality or function. For:

• \(x < -1\): Testing with \(x = -2\), which results in \(3(-2-1)(-2+1)\). The product of negative terms remains negative.
• \(-1 < x < 1\): Testing with \(x = 0\), producing \(3(0-1)(0+1)\), a negative result again.
• \(x > 1\): Testing with \(x = 2\) yields \(3(2-1)(2+1)\), a positive result.

Thus, the function is positive in the interval \(x > 1\) and negative in both \(x < -1\) and \(-1 < x < 1\). Understanding these intervals and signs helps establish which parts of the number line contribute to the solution of the inequality.

Critical Points

Critical points of a function are crucial for solving inequalities, as they are the points at which the value of the function is zero. These points divide the number line into intervals, which we analyze to determine where the function satisfies the inequality.
Given the inequality \(3(x-1)(x+1) > 0\), the process involves:

• Setting the inequality equation to zero: Set \(3(x-1)(x+1) = 0\).
• Finding roots: Expand or solve \((x-1)(x+1) = 0\), which results in \(x = -1\) and \(x = 1\) as roots.
• These roots, \(x = -1\) and \(x = 1\), are our critical points.

Critical points are vital in inequality solving because they often indicate changes in the sign of the inequality. This sign change is essential for determining where the inequality is true or false. Once these points are identified, we can proceed with interval testing using interval analysis to pinpoint the solution set of the inequality accurately.