Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Question:**
Explain why the rational inequality \(\frac{1}{x^2 + 2x + 9} < 0\) has no solution.
**Explanation:**
To understand why this inequality has no solution, let's analyze the expression \(\frac{1}{x^2 + 2x + 9}\).
The denominator is \(x^2 + 2x + 9\). This is a quadratic expression, and it needs to be examined to determine the values of \(x\) where it is less than zero, thereby making the whole fraction negative.
1. **Consider the quadratic expression \(x^2 + 2x + 9\):**
- The discriminant of this quadratic expression is given by \(b^2 - 4ac\), where \(a = 1\), \(b = 2\), and \(c = 9\).
- Calculate the discriminant:
\[
b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 9 = 4 - 36 = -32
\]
- Since the discriminant is negative (\(-32\)), the quadratic equation has no real roots. This means the expression \(x^2 + 2x + 9\) is always positive for all real values of \(x\).
2. **Inequality Analysis:**
- Since the denominator \(x^2 + 2x + 9\) is always positive, \(\frac{1}{x^2 + 2x + 9}\) is always positive as well (since the numerator is positive and non-zero).
- Therefore, the expression \(\frac{1}{x^2 + 2x + 9}\) cannot be negative, which implies that there are no values of \(x\) that satisfy \(\frac{1}{x^2 + 2x + 9} < 0\).
Thus, there is no solution to the inequality \(\frac{1}{x^2 + 2x + 9} < 0\) because the expression is always positive for all real numbers \(x\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ad60fb3-d7c7-4aa0-8ac8-5f8ba0f10fe4%2F18007bdd-b115-4ac4-9e40-e2744039c163%2Frm4n97c_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
Explain why the rational inequality \(\frac{1}{x^2 + 2x + 9} < 0\) has no solution.
**Explanation:**
To understand why this inequality has no solution, let's analyze the expression \(\frac{1}{x^2 + 2x + 9}\).
The denominator is \(x^2 + 2x + 9\). This is a quadratic expression, and it needs to be examined to determine the values of \(x\) where it is less than zero, thereby making the whole fraction negative.
1. **Consider the quadratic expression \(x^2 + 2x + 9\):**
- The discriminant of this quadratic expression is given by \(b^2 - 4ac\), where \(a = 1\), \(b = 2\), and \(c = 9\).
- Calculate the discriminant:
\[
b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 9 = 4 - 36 = -32
\]
- Since the discriminant is negative (\(-32\)), the quadratic equation has no real roots. This means the expression \(x^2 + 2x + 9\) is always positive for all real values of \(x\).
2. **Inequality Analysis:**
- Since the denominator \(x^2 + 2x + 9\) is always positive, \(\frac{1}{x^2 + 2x + 9}\) is always positive as well (since the numerator is positive and non-zero).
- Therefore, the expression \(\frac{1}{x^2 + 2x + 9}\) cannot be negative, which implies that there are no values of \(x\) that satisfy \(\frac{1}{x^2 + 2x + 9} < 0\).
Thus, there is no solution to the inequality \(\frac{1}{x^2 + 2x + 9} < 0\) because the expression is always positive for all real numbers \(x\).
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