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Chapter 2: Problem 29

Solve the inequalities. Suggestion: A calculator may be useful for approximating key numbers. $$(x-1)(x+3)(x+4) \geq 0$$

Short Answer

Expert verified
The solution is \([-4, -3]\cup[1, \infty)\).

Step by step solution

01

Identify Critical Points

To solve the inequality \((x-1)(x+3)(x+4) \geq 0\), begin by finding the critical points where each factor equals zero. These points are \(x-1=0\), \(x+3=0\), and \(x+4=0\). Solving these gives the critical points: \(x=1\), \(x=-3\), and \(x=-4\).
02

Analyze Intervals

Divide the real number line into intervals using the critical points. The intervals are: \((-\infty, -4)\), \((-4, -3)\), \((-3, 1)\), and \((1, \infty)\).
03

Test Intervals

Choose a test point in each interval and determine the sign of the product \((x-1)(x+3)(x+4)\). For \((-fty, -4)\), use \(x = -5\). For \((-4, -3)\), use \(x = -3.5\). For \((-3, 1)\), use \(x = 0\). For \((1, fty)\), use \(x = 2\). Substitute each into the product and note the sign of the result.
04

Interpret Results

Evaluate the test point results: - For \((-\infty, -4)\), \((-5-1)(-5+3)(-5+4) < 0\). - For \((-4, -3)\), \((-3.5-1)(-3.5+3)(-3.5+4) > 0\). - For \((-3, 1)\), \((0-1)(0+3)(0+4) < 0\). - For \((1, \infty)\), \((2-1)(2+3)(2+4) > 0\). The expression is non-negative in the intervals \((-4, -3)\cup(1, fty)\).
05

Incorporate Critical Points

Since the inequality is \(\geq 0\), include the points where the product equals zero: \(x=-4, -3,\) and \(1\). Therefore, the solution to the inequality is \([-4, -3]\cup[1, \infty)\).

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Key Concepts

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

Critical Points
In the context of solving inequalities, critical points are the values of the variable that make each factor in the expression equal to zero. They are crucial because they indicate where the expression changes its behavior, such as switching from positive to negative. In the given inequality \((x-1)(x+3)(x+4) \geq 0\), we set each factor equal to zero to find the critical points:
  • \(x - 1 = 0\) gives \(x = 1\)
  • \(x + 3 = 0\) gives \(x = -3\)
  • \(x + 4 = 0\) gives \(x = -4\)
These critical points \(x = 1, -3, -4\) are the boundaries that help us determine where the inequality holds true. They also help us understand which intervals on the number line to evaluate further.
Number Line Intervals
Once you have the critical points from the inequality, the next step is to use them to divide the number line into intervals. Each interval is defined by the space between these points. For the inequality \((x-1)(x+3)(x+4) \geq 0\), the critical points \(x = -4, -3,\) and \(1\) produce the following intervals:
  • \((-fty, -4)\)
  • \((-4, -3)\)
  • \((-3, 1)\)
  • \((1, fty)\)
Understanding these intervals is key to testing parts of the number line to see where the inequality is valid. Each interval will be analyzed separately to determine if the values within it satisfy the original inequality.
Sign Test
The sign test is a technique deployed on the intervals defined by the critical points to determine the sign of the expression in each interval. You pick a test point from each interval and substitute it back into the expression \((x-1)(x+3)(x+4)\) to check whether it produces a positive or negative result.For example, in each interval:
  • In \((-fty, -4)\), choosing \(x = -5\), results in a negative sign.
  • In \((-4, -3)\), choosing \(x = -3.5\), results in a positive sign.
  • In \((-3, 1)\), choosing \(x = 0\), results in a negative sign.
  • In \((1, fty)\), choosing \(x = 2\), results in a positive sign.
These signs tell us where the inequality \(\geq 0\) holds true, indicating the intervals that require further attention during final solution construction.
Inequality Solution
After performing the sign test, interpret these results to determine where the original inequality \((x-1)(x+3)(x+4) \geq 0\) holds:
  • In \((-4, -3)\), the expression is positive.
  • In \((1, fty)\), the expression is also positive.
To attain the final solution, include the critical points where the expression equals zero because our inequality is\(\geq 0\) rather than simply \(> 0\). This way, the solution consists of closed intervals:
  • \([-4, -3]\)
  • \([1, fty)\)
Thus, unions of these intervals \([-4, -3] \cup [1, fty)\) provide the final answer to the inequality, as these intervals satisfy the non-negativity condition.

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Most popular questions from this chapter

Solve the inequalities Suggestion: A calculator may be useful for approximating key numbers. $$x^{2}\left(3 x^{2}+11\right) \geq 4$$

Solve the inequalities Suggestion: A calculator may be useful for approximating key numbers. $$(x-3)^{2}(x+1)^{4}(2 x+1)^{4}(3 x+2) \leq 0$$

(a) Use a graph to estimate the solution set for each inequality. Zoom in far enough so that you can estimate the relevant endpoints to the nearest thousandth. (b) Exercises \(61-70\) can be solved algebraically using the techniques presented in this section. Carry out the algebra to obtain exact expressions for the endpoints that you estimated in part (a). Then use a calculator to check that your results are consistent with the previous estimates. $$x^{4}-2 x^{2}-1>0$$

Determine all of the real-number solutions for each equation. (Remember to check for extraneous solutions.) $$\sqrt{2-x}-10=x$$

In this exercise we investigate the effect of the constant \(c\) upon the roots of the quadratic equation \(x^{2}-6 x+c=0\) We do this by looking at the \(x\) -intercepts of the graphs of the corresponding equations \(y=x^{2}-6 x+c\) (a) Set a viewing rectangle that extends from 0 to 5 in the \(x\) -direction and from -2 to 3 in the \(y\) -direction. Then (on the same set of axes) graph the equations \(y=x^{2}-6 x+c\) with \(c\) running from 8 to 10 at increments of \(0.25 .\) In other words, graph the equations \(y=x^{2}-6 x+8, y=x^{2}-6 x+8.25, y=x^{2}-6 x+8.50\) and so on, up through \(y=x^{2}-6 x+10\) (b) Note from the graphs in part (a) that, initially, as \(c\) increases, the \(x\) -intercepts draw closer and closer together. For which value of \(c\) do the two \(x\) -intercepts seem to merge into one? (c) Use algebra as follows to check your observation in part (b). Using that value of \(c\) for which there appears to be only one intercept, solve the quadratic equation \(x^{2}-6 x+c=0 .\) How many roots do you obtain? (d) Some of the graphs in part (a) have no \(x\) -intercepts. What are the corresponding values of \(c\) in these cases? Pick any one of these values of \(c\) and use the quadratic formula to solve the equation \(x^{2}-6 x+c=0 .\) What happens?
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