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Chapter 2: Problem 43

Solve the inequalities Suggestion: A calculator may be useful for approximating key numbers. $$\frac{x-1}{x+1} \leq 0$$

Short Answer

Expert verified
The solution is \(x \in (-1, 1]\).

Step by step solution

01

Identify Critical Points

Firstly, identify where the expression changes its sign by setting the numerator and denominator to zero. For the numerator, solve \(x-1 = 0\) which gives \(x=1\). For the denominator, solve \(x+1=0\) which gives \(x=-1\). These are the critical points where sign changes potentially occur.
02

Determine Intervals Around Critical Points

Using the critical points, divide the number line into intervals: \((-\infty, -1)\), \((-1, 1)\), and \((1, \infty)\). These will help us test where the expression is less than or equal to zero.
03

Test Each Interval

Select a test point from each interval to determine the sign of the fraction in that interval:- For \((-\infty, -1)\), test \(x = -2\): \(\frac{-2-1}{-2+1} = \frac{-3}{-1} = 3\) (positive).- For \((-1, 1)\), test \(x = 0\): \(\frac{0-1}{0+1} = \frac{-1}{1} = -1\) (negative).- For \((1, \infty)\), test \(x = 2\): \(\frac{2-1}{2+1} = \frac{1}{3}\) (positive).
04

Include Boundary Points

Consider the equality from the inequality \(\leq 0\). At \(x=1\), the function equals zero \(\left( \frac{1-1}{1+1} = 0 \right)\). Since the function is undefined at \(x=-1\), it cannot be included.
05

Write the Solution Set

From the test and boundary inclusion, the solution interval where the inequality holds is \(x \in (-1, 1]\). This is where the expression is either zero or negative.

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Key Concepts

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

Critical Points
Identifying critical points is fundamental when solving inequalities. It involves finding values where the expression might change its behavior, such as switching from positive to negative. In rational inequalities like \( \frac{x-1}{x+1} \leq 0 \), critical points occur where either the numerator or denominator equals zero. The numerator zeroes out when \( x-1 = 0 \), leading to \( x = 1 \). Meanwhile, the denominator is zero at \( x+1 = 0 \), or \( x = -1 \). Critically, these are the potential sign-changing points on the number line.
  • The numerator zero \( (x = 1) \) creates a point where the value of the fraction is exactly zero.
  • The denominator zero \( (x = -1) \) indicates an undefined point, which cannot be part of the solution due to division by zero.
Identifying and understanding these critical points ensures that the subsequent interval testing covers all possible changes in sign behavior.
Interval Testing
Interval testing involves dividing the number line based on critical points and choosing test values from each segment to see where the inequality holds true. With our critical points at \( x = -1 \) and \( x = 1 \), the number line is split into three intervals:
  • \((-\infty, -1)\) 鈥� values smaller than \( x = -1 \)
  • \((-1, 1)\) 鈥� values between \( x = -1 \) and \( x = 1 \)
  • \((1, \infty)\) 鈥� values larger than \( x = 1 \)
By selecting a test point in each interval:
  • For \((-\infty, -1)\), \( x = -2\) results in a positive fraction.
  • For \((-1, 1)\), \( x = 0\) results in a negative fraction.
  • For \((1, \infty)\), \( x = 2\) results in a positive fraction.
This testing method reveals which intervals satisfy the condition given in the inequality.
Inequality Solution
An inequality solution refers to the range of values that satisfy the given inequality condition, \( \frac{x-1}{x+1} \leq 0 \). After testing the intervals, you identify where the inequality is true depending on the desired condition less than or equal to zero. From the tested values, you find:
  • The interval \((-1, 1)\) corresponds to a negative result.
  • At \( x = 1 \), the inequality becomes exactly zero.
Therefore, the solution set includes the interval from \(-1\) to \(1\) but excludes \(-1\) because the expression is undefined there. So, the solution interval is written as \( x \in (-1, 1] \), emphasizing closed inclusion at \( x = 1 \) where it is equal to zero.
Rational Inequalities
Rational inequalities involve expressions where a polynomial is divided by another polynomial and typically require careful analysis due to potential undefined points. They often include both equality and inequality conditions that restrict the solution set further.
These inequalities, like \( \frac{x-1}{x+1} \leq 0 \), necessitate:
  • Identifying where the rational expression isn't defined (critical points).
  • Finding where the numerator equals zero, contributing valid solutions.
  • Testing intervals to determine signs across different ranges.
Solving these inequalities not only requires logical steps in determining sign changes but also demands understanding the behavior of functions around those key points. Importantly, undefined points due to zero denominators are excluded from the solution set, ensuring valid, actionable conclusions.

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Most popular questions from this chapter

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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