Problem 35 - Beginning and Intermediate Algebra

Question

Solve each inequality. Graph the solution set and write the solution in interval notation. $$(r+2)(r-5)(r-1) \leq 0$$

Step-by-Step Solution

Answer

The solution set for the inequality \((r+2)(r-5)(r-1) \leq 0\) is \([-2, 1] \cup (1, 5]\).

Step 1: Find the critical points of the inequality

To find the critical points of the inequality, we need to find the values of \(r\) that make the polynomial equal to 0: $$(r+2)(r-5)(r-1) = 0$$ We have three factors, so there are three critical points: \(r+2=0 \Rightarrow r=-2\) \(r-5=0 \Rightarrow r=5\) \(r-1=0 \Rightarrow r=1\)

Step 2: Test the intervals between the critical points

Now we will test the intervals between the critical points. We need to find the value of the polynomial at any point within each interval and check the sign of the product. Interval 1: (-∞, -2), let's test \(r=-3\): \(((-3)+2)((-3)-5)((-3)-1) = (-1)(-8)(-4) > 0\) Interval 2: (-2, 1), let's test \(r=0\): \(((0)+2)((0)-5)((0)-1) = (2)(-5)(-1) > 0\) Interval 3: (1, 5), let's test \(r=2\): \(((2)+2)((2)-5)((2)-1) = (4)(-3)(1) < 0\) Interval 4: (5, ∞), let's test \(r=6\): \(((6)+2)((6)-5)((6)-1) = (8)(1)(5) > 0\)

Step 3: Find the solution set based on the sign of the polynomial

Since the inequality is less than or equal to 0, we need to find the intervals where the polynomial is non-positive (negative or equal to 0). The polynomial is non-positive in the interval: - From the analysis above, we can see that the polynomial is negative in Interval 3: (1, 5) We also need to include the critical points since the inequality is less than or equal to 0: - Critical points: -2, 1, and 5 Therefore, the solution set is \([-2, 1] \cup [5, 5] \cup (1, 5)\).

Step 4: Graph the solution set

Plot the critical points on a number line: -2------1-------5 Shade the intervals where the inequality is satisfied: [-2, 1] ∪ [5, 5] ∪ (1, 5) Since there is only one point in the interval [5, 5], we can represent it as {5}. So, we have: [-2, 1] ∪ {5} ∪ (1, 5)

Step 5: Write the solution set in interval notation

The solution set in interval notation will be a combination of all the intervals and critical points: \([-2, 1] \cup (1, 5] \) # Note: I clustered the critical point 5 into the interval (1, 5) -> (1, 5].

Key Concepts

Interval Notation

When solving inequalities, understanding interval notation is important. It helps us visually represent which values satisfy an inequality. For the inequality \[(r+2)(r-5)(r-1) \leq 0\], we find the intervals where the polynomial is less than or equal to zero.

• Intervals are written as pairs, like \((a, b)\) for open intervals, where neither endpoint is included.
• Use square brackets for closed intervals, \([a, b]\), where both endpoints are included.
• A combination of open and closed intervals can express more complex solutions.

For example, in this problem, the solution set was expressed as: \([-2, 1] \cup (1, 5]\). This means:

• All numbers from -2 to 1, including -2 and 1.
• All numbers greater than 1 and up to 5, including 5.

Critical Points

Critical points help identify where a polynomial might change its sign. To find them, we set the polynomial equal to zero. In this problem, solving \((r+2)(r-5)(r-1) = 0\)gave us critical points: \(r=-2, 1, 5\).

• Set each factor to zero: \(r+2=0\), \(r-5=0\), \(r-1=0\)
• Solve for \(r\).
• These points \(-2, 1, 5\) are critical because the sign of the polynomial may change at these points.

By testing intervals between these points, we can determine where the polynomial is positive, negative, or zero.

Polynomial Inequalities

Polynomial inequalities like \((r+2)(r-5)(r-1) \leq 0\) require finding where the expression is less than or equal to zero.

• First, identify the critical points where the polynomial is zero.
• Test the sign of the polynomial between each critical point.
• Select the intervals fulfilling the inequality.

For example, by choosing test points in different intervals and checking the sign of the result:

• \((-\infty, -2)\) is positive.
• \((-2, 1)\) is positive.
• \((1, 5)\) is negative, so included in the solution.
• \((5, \infty)\) is positive.

Ultimately, writing the solution in interval notation guides us to correctly represent where the inequality holds true.