Chapter 2: Problem 9
If \(S\) is the set of all real \(x\) and such that \(\frac{2 x-1}{2 x^{3}+3 x^{2}+x}\) is positive, then \(S\) contains (A) \(\left(-\infty,-\frac{3}{2}\right)\) (B) \(\left(-\frac{3}{2},-\frac{1}{4}\right)\) (C) \(\left(-\frac{1}{4}, \frac{1}{2}\right)\) (D) \(\left(\frac{1}{2}, 3\right)\)
Short Answer
Expert verified
S contains options A and D.
Step by step solution
01
Identify the Signs of the Factors
To determine where the expression \(\frac{2x-1}{2x^3+3x^2+x}\) is positive, analyze the numerator \(2x-1\) and the denominator \(2x^3+3x^2+x\). Check for values of \(x\) that make the expression positive or negative.
02
Factoring the Denominator
Factor the denominator: \(2x^3 + 3x^2 + x = x(2x^2 + 3x + 1)\). Find the roots of \(2x^2 + 3x + 1\) using the quadratic formula, which gives \(x = -1\) and \(x = -\frac{1}{2}\). Thus, the denominator is zero at \(x = 0, -1, -\frac{1}{2}\), creating potential sign changes.
03
Determine Numerator's Sign
The numerator \(2x-1\) is zero at \(x = \frac{1}{2}\). It is negative when \(x < \frac{1}{2}\) and positive when \(x > \frac{1}{2}\).
04
Analyze Intervals Between Roots
Divide the real number line into intervals based on points where the expression is zero: \((-\infty, -1), (-1, -\frac{1}{2}), (-\frac{1}{2}, 0), (0, \frac{1}{2}), (\frac{1}{2}, \infty)\). Evaluate the sign of the expression within each interval by picking test points.
05
Evaluate Expression's Sign in Each Interval
- In \((-\infty, -1)\), pick \(x = -2\): both factors are negative, the expression is positive.- In \((-1, -\frac{1}{2})\), pick \(x = -\frac{3}{4}\): numerator is negative, denominator is positive, the expression is negative.- In \((-\frac{1}{2}, 0)\), pick \(x = -\frac{1}{4}\): both factors are negative, the expression is positive.- In \((0, \frac{1}{2})\), pick \(x = \frac{1}{4}\): numerator is negative, denominator is positive, the expression is negative.- In \((\frac{1}{2}, \infty)\), pick \(x = 1\): both factors are positive, the expression is positive.
06
Identify Positive Intervals
The expression is positive in the intervals \((-\infty, -1)\), \((-\frac{1}{2}, 0)\) and \((\frac{1}{2}, \infty)\). Match these with the given options to check which intervals are entirely contained.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Real Numbers
Real numbers include every kind of number you can visualize on a number line. This includes integers like -3, 0, or 4, and fractions like -1/2 or 2.5. Essentially, any number that doesn't have an imaginary component (like the square root of a negative number) falls under real numbers. When you work with inequalities, you're often using real numbers as these variables can continuously vary.
Here is a quick breakdown of real numbers:
Here is a quick breakdown of real numbers:
- **Rational Numbers**: Numbers that can be expressed as a fraction, such as 1/2, -3/4, or 5.
- **Irrational Numbers**: Numbers that cannot be expressed as fractions, like \( ext{\(\pi\)}\) or the square root of 2.
- **Integers**: Whole numbers that can be positive, negative, or zero.
Interval Notation
Interval notation is a shorthand used to describe a range of numbers on the real number line. It's a convenient way to communicate precisely where numbers fall within the boundaries. When you're working with inequalities, interval notation helps you express the solution set clearly.
Here's how it breaks down:
Here's how it breaks down:
- **Closed Intervals [a, b]**: Include both endpoints, meaning all numbers between a and b are included.
- **Open Intervals (a, b)**: Do not include the endpoints. They only contain numbers strictly between a and b.
- **Half-Open Intervals [a, b) or (a, b]**: Include one of the endpoints—either a or b—but not both.
Polynomial Factorization
Polynomial factorization involves expressing a polynomial as a product of its simpler components, which are usually polynomials of lower degree. This process helps simplify equations, making it easier to identify solutions, especially in solving inequalities.
When you factor a polynomial, you're looking for terms that can be regrouped into a product of terms:
When you factor a polynomial, you're looking for terms that can be regrouped into a product of terms:
- **Factor out Common Terms**: First, look for common terms that can be factored out, such as pulling out an 'x' in the expression \(2x^3 + 3x^2 + x\).
- **Use Special Factoring Formulas**: Apply formulas for special cases like difference of squares or perfect squares. These make certain expressions easier to factor.
- **Quadratic Formula for Non-Trivial Cases**: When a polynomial cannot be factored easily, use the quadratic formula to find roots. For example, use it on \(2x^2 + 3x + 1\) to find the roots \(x = -1\) and \(x = -\frac{1}{2}\).