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Magazine Chapter 3: Problem 28
Solve the inequality. $$\frac{1}{x}+\frac{1}{x+1}<\frac{2}{x+2}$$
Short Answer
Expert verified
The solution is \((-\infty, -1) \cup (2, \infty)\).
Step by step solution
01
Identify Critical Points
To solve the inequality \(\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+2}\), start by finding the critical points: these are values of \(x\) that make any denominator zero. The critical points for this inequality are at \(x = 0\), \(x = -1\), and \(x = -2\).
02
Find Common Denominator
Rewrite the inequality with a common denominator of \((x)(x+1)(x+2)\):\[\frac{(x+1)(x+2) + x(x+2)}{x(x+1)(x+2)} < \frac{2x(x+1)}{x(x+1)(x+2)}\]. Simplify the numerators.
03
Simplify the Numerators
Simplify each term: \((x+1)(x+2) + x(x+2) = x^2 + 3x + 2\) and \(2x^2 + 2x\). Rewrite the inequality: \[x^2 + 3x + 2 < 2x^2 + 2x\].
04
Rearrange and Simplify Further
Rearrange the terms to clarify the inequality: \[x^2 + 3x + 2 - 2x^2 - 2x < 0\]. Simplify it to \[-x^2 + x + 2 < 0\] or \[x^2 - x - 2 > 0\].
05
Factor the Quadratic
Factor \(x^2 - x - 2\) to find the roots: \((x-2)(x+1)\). So, \((x-2)(x+1) > 0\).
06
Determine Intervals and Test
The inequality \((x-2)(x+1) > 0\) implies the solution set is determined by testing intervals around the roots \(x = 2\) and \(x = -1\): \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\). Use test points to determine which intervals satisfy the inequality.
07
Analyze Intervals for Solutions
Choose test points: For \((-\infty, -1)\) use \(x = -2\), for \((-1, 2)\) use \(x = 0\), and for \((2, \infty)\) use \(x = 3\). Evaluate \((x-2)(x+1) > 0\). The intervals \((-\infty, -1)\) and \((2, \infty)\) satisfy the inequality.
08
Final Solution Set
Combine intervals to obtain the solution to the inequality: \((-\infty, -1) \cup (2, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
When solving inequalities, identifying the critical points is a crucial first step. Critical points are the values where the denominator of a fraction equals zero because division by zero is undefined. In the inequality \(\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+2}\), finding these points helps break the domain into sections where you can test solutions later.
For this inequality:
For this inequality:
- \(x = 0\) makes \(\frac{1}{x}\) undefined.
- \(x = -1\) makes \(\frac{1}{x+1}\) undefined.
- \(x = -2\) makes \(\frac{1}{x+2}\) undefined.
Quadratic Factoring
Factoring a quadratic expression can simplify solving inequalities substantially. Consider the inequality from the solution: \(x^2 - x - 2 > 0\). Factoring quadratics means breaking it down into simpler expressions that multiply to give the original quadratic.
Let's derive this from our simplified inequality:
Let's derive this from our simplified inequality:
- The standard form is \(x^2 - x - 2\).
- Factor it into \((x-2)(x+1)\).
Interval Testing
Once a quadratic is factored, interval testing helps find which portions of the number line satisfy the inequality. For \((x-2)(x+1) > 0\), you can test between and outside the roots found.
First, identify the intervals based on the critical points and roots:
First, identify the intervals based on the critical points and roots:
- \((-\infty, -1)\)
- \((-1, 2)\)
- \((2, \infty)\)
- For \((-\infty, -1)\) choose \(x = -2\): \((x-2)(x+1) = (-2-2)(-2+1) > 0\)
- For \((-1, 2)\) choose \(x = 0\): \((x-2)(x+1) = (0-2)(0+1) < 0\)
- For \((2, \infty)\) choose \(x = 3\): \((x-2)(x+1) = (3-2)(3+1) > 0\)
