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\(33-66\) me solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ (x-5)(x+4) \geq 0 $$

Short Answer

Expert verified
The solution is \((-\infty, -4] \cup [5, \infty)\).

Step by step solution

01

Identify the Critical Points

To find the critical points of the inequality \((x-5)(x+4) \geq 0\), set the expression equal to zero: \((x-5)(x+4) = 0\). Solve for \(x\) to get the critical points: \(x-5=0\) and \(x+4=0\). Therefore, the critical points are \(x=5\) and \(x=-4\).
02

Test Intervals

Use the critical points to divide the number line into intervals: \((-\infty, -4)\), \((-4, 5)\), and \((5, \infty)\). Choose a test point from each interval to determine where the inequality is non-negative \((\geq 0)\).
03

Test the Interval (-∞, -4)

Choose a test point for the interval \((-\infty, -4)\), such as \(x=-5\). Substitute \(x = -5\) into \((x-5)(x+4)\):\((-5-5)(-5+4) = (-10)(-1) = 10\). Since 10 is positive, the inequality holds in this interval.
04

Test the Interval (-4, 5)

Choose a test point for the interval \((-4, 5)\), such as \(x=0\). Substitute \(x = 0\) into \((x-5)(x+4)\): \((0-5)(0+4) = (-5)(4) = -20\). Since -20 is negative, the inequality does not hold in this interval.
05

Test the Interval (5, ∞)

Choose a test point for the interval \((5, \infty)\), such as \(x=6\). Substitute \(x = 6\) into \((x-5)(x+4)\):\((6-5)(6+4) = (1)(10) = 10\). Since 10 is positive, the inequality holds in this interval.
06

Analyze the Critical Points

At the critical points \(x = -4\) and \(x = 5\):Substitute \(x = -4\): \((-4-5)(-4+4) = (-9)(0) = 0\)Substitute \(x = 5\): \((5-5)(5+4) = (0)(9) = 0\)Since the inequality is \(\geq 0\), the solution includes \(x = -4\) and \(x = 5\).
07

Express Solution in Interval Notation

Combine the results from each interval and critical points. The inequality holds for the intervals \((-\infty, -4] \cup [5, \infty)\).
08

Graph the Solution Set

Draw a number line and shade the intervals where the inequality is true: \((-\infty, -4]\) and \([5, \infty)\). Use a closed dot at \(x = -4\) and \(x = 5\) to indicate they are included in the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
To solve nonlinear inequalities like \[(x-5)(x+4) \geq 0\] we start by identifying the critical points. These are the values of \(x\) that make the expression inside the inequality equal to zero.
To find them, set the expression \((x-5)(x+4) = 0\). This equation gives us the critical points by setting each factor equal to zero separately: - \(x-5 = 0\) leads to \(x = 5\)- \(x+4 = 0\) leads to \(x = -4\).The critical points, \(-4\) and \(5\), help divide the number line into separate intervals, each of which we'll test to see if they satisfy the inequality.
Interval Notation
Once critical points are determined, they are used to divide the number line into specific sections, or intervals. For the inequality \((x-5)(x+4) \geq 0\), the critical points \(-4\) and \(5\) split the number line into the following intervals:
  • \((-\infty, -4)\)
  • \((-4, 5)\)
  • \((5, \infty)\)
Interval notation is a compact way to express these sections and their boundaries. In our solution, we determined that the inequality holds in \((-\infty, -4]\) and \([5, \infty)\), which is expressed in interval notation as \((-\infty, -4] \cup [5, \infty)\). A closed bracket "]" or "[" indicates the endpoint is included, as the critical points in this case satisfy the inequality exactly, meaning the inequality is true even when it's zero.
Inequality Solution
The solution to an inequality is the set of values which make the inequality true. For our inequality \((x-5)(x+4) \geq 0\), we found the solution in several stages. We determined the critical points, which define the boundary of our test intervals. After testing each interval, we found that the solution set includes the intervals where the inequality is positive or zero:
  • The interval \((-\infty, -4]\) is valid because the expression is positive there.
  • The interval \([5, \infty)\) is also valid for the same reason.
The intervals are combined using the union symbol "\(\cup\)" in interval notation, forming our complete solution: \((-\infty, -4] \cup [5, \infty)\). It reflects where the original inequality holds true across the number line.
Test Intervals
The concept of test intervals is crucial in finding inequality solutions. Once critical points are established—\(-4\) and \(5\) in this case—they create boundaries on the number line, dividing it into intervals. For \((x-5)(x+4) \geq 0\), these intervals are:
  • \((-\infty, -4)\)
  • \((-4, 5)\)
  • \((5, \infty)\)
We select a test point from within each interval to substitute back into the inequality, checking whether it returns a positive or negative result:
  • For \((-\infty, -4)\), \(x = -5\) yields a positive result, so it holds.
  • For \((-4, 5)\), \(x = 0\) results in a negative outcome, so it doesn't satisfy the inequality.
  • For \((5, \infty)\), \(x = 6\) again gives a positive value, so it holds.
Test intervals effectively help determine where the inequality is true. This enables us to confidently define our solution set in interval notation.

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