Problem 33 - College Algebra

Question

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ (2-x)^{2}\left(x-\frac{7}{2}\right)<0 $$

Step-by-Step Solution

Answer

The solution to the polynomial inequality \( (2-x)^2*(x-\frac{7}{2}) < 0 \) is \( (1,2)\cup(2,\frac{7}{2}) \)

Step 1: Expression Simplification

First, simplify the expression \( (2-x)^2*(x-\frac{7}{2}) < 0 \) to standard polynomial form by expanding the squared binomial and distributing the result across \( (x-\frac{7}{2}) \). This simplifies the expression to \( -x^{3}+\frac{7}{2}x^{2}+\frac{1}{2}x-\frac{7}{2}<0 \)

Step 2: Finding Critical Points

The next step involves determining when the inequality is zero, which will yield critical points. Critical points are potential endpoints for the intervals in the solution set. These points are obtained by setting the inequality to zero and solving for x. Doing so gives \( -x^{3}+\frac{7}{2}x^{2}+\frac{1}{2}x-\frac{7}{2}=0 \), which yields the solutions \(x=2\), \(x=1\), and \(x=\frac{7}{2}\) . These values are the critical points that divide the real number line into intervals.

Step 3: Intervals Testing

The critical points divide the real number line into four intervals: \(-\infty,1\), \(1,2\), \(2,\frac{7}{2}\), and \(\frac{7}{2},\infty\). To determine which intervals satisfy the original inequality, select a test point from each interval and substitute it into the simplified inequality. If the result is less than zero, then that interval is part of the solution set. It's found that the intervals \(1,2\) and \(2,\frac{7}{2}\) are part of the solution set.

Step 4: Solution Expression in Interval Notation

The last step is to express the solution in interval notation. The solution is the union of the intervals \(1,2\) and \(2,\frac{7}{2}\), which can be written in interval notation as \( (1,2)\cup(2,\frac{7}{2}) \).

Key Concepts

Inequality Solving

Solving polynomial inequalities involves finding the values of the variable that make the inequality true. Unlike equations, where we're looking for where two expressions are equal, inequalities deal with values that are greater or less than another. Here's a simple process to solve them:

• Simplify the expression: Begin by expanding and combining like terms if necessary.
• Find critical points: Set the inequality equal to zero, solving for values that make it zero.
• Test intervals: Use the critical points to determine intervals on the number line and test points to see where the inequality holds true.
• Express the solution: Write your solution using interval notation.

Interval Notation

Interval notation provides a concise way to describe sets of numbers on the real number line. It's especially useful for inequalities. Here's how we use it:

• Round brackets \(( , )\): Indicate that endpoints are not included (open interval).
• Square brackets \([ , ]\): Indicate that endpoints are included (closed interval).
• Union: Use the union symbol \(\cup\) to combine disjoint intervals.

In our example, the solution was found to be the union of the intervals \((1, 2)\) and \((2, \frac{7}{2})\). This tells us the exact values of \(x\) that satisfy the original inequality.

Critical Points

Critical points are values that make the expression equal to zero, marking potential boundaries for the solution intervals. Finding these involves:

• Set the expression to zero: Solve the polynomial equation.
• List solutions: These are your critical points.

In the example, solving \(-x^3 + \frac{7}{2}x^2 + \frac{1}{2}x - \frac{7}{2} = 0\) gives the critical points \(x = 2\), \(x = 1\), and \(x = \frac{7}{2}\). These divide the real number line into different intervals.

Real Number Line

The real number line is an essential concept for visualizing solutions to inequalities. It's a straight line that represents all real numbers, where each point corresponds to a real number. It helps us:

• Visualize intervals: Mark critical points and identify regions that satisfy the inequality.
• Test intervals: Select test points from these regions to verify where the inequality holds true.

In our inequality, the critical points divided the real number line into intervals, and we discovered that \( (1, 2) \) and \( (2, \frac{7}{2} ) \) satisfied the inequality. The number line helps us see these intervals clearly, making it easier to understand the solution.