Chapter 1: Problem 74
Inequalities Involving Quotients Solve the nonlinear inequality. Express the solution using interval notation, and graph the solution set. $$x^{5} > x^{2}$$
Short Answer
Expert verified
The solution is \((-\infty, 0) \cup (1, \infty)\).
Step by step solution
01
Move All Terms to One Side
To solve the inequality \(x^5 > x^2\), we start by moving all terms to one side of the inequality to set it to zero. This gives us \(x^5 - x^2 > 0\).
02
Factor the Expression
The expression \(x^5 - x^2\) is factorable. We can factor out the greatest common factor \(x^2\) to get \(x^2(x^3 - 1) > 0\).
03
Factor the Remaining Polynomial
The cubic expression can be factored further. Recognize \(x^3 - 1\) as the difference of cubes: \((x-1)(x^2 + x + 1)\). Thus, the inequality becomes \(x^2(x-1)(x^2 + x + 1) > 0\).
04
Find Critical Points
The solutions to \(x^2 = 0\) and \(x-1 = 0\) are the potential critical points. These are \(x = 0\) and \(x = 1\). The quadratic \(x^2 + x + 1 = 0\) has no real roots (discriminant \(b^2 - 4ac = -3\)).
05
Test Intervals Around Critical Points
The critical points break the number line into intervals: \((-\infty, 0), (0, 1), (1, \infty)\). Pick test points from each interval and substitute into the factored expression \(x^2(x-1)(x^2+x+1)\):- For \((-\infty, 0)\), pick \(x = -1\); expression is positive.- For \((0, 1)\), pick \(x = 0.5\); expression is negative.- For \((1, \infty)\), pick \(x = 2\); expression is positive.
06
Determine Solution Set
From the testing, the expression is positive on \((-\infty, 0)\) and \((1, \infty)\). Therefore, the solution in interval notation is \((-\infty, 0) \cup (1, \infty)\).
07
Graph the Solution Set
On a number line, represent the solution set \((-\infty, 0) \cup (1, \infty)\) using an open interval notation. Draw open circles at \(x = 0\) and \(x = 1\) with lines extending leftwards from \(x = 0\) and rightwards from \(x = 1\), indicating the intervals.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interval Notation
Interval notation is a method for denoting subsets of the real number line. It is used to represent the solution sets of inequalities in a compact and easy-to-read format.
In our problem, we need to express when the expression \(x^2(x-1)(x^2 + x + 1) > 0\) is positive. The intervals we found where the expression is positive are \((-\infty, 0)\) and \((1, \infty)\).
Here's how interval notation is structured:
In our problem, we need to express when the expression \(x^2(x-1)(x^2 + x + 1) > 0\) is positive. The intervals we found where the expression is positive are \((-\infty, 0)\) and \((1, \infty)\).
Here's how interval notation is structured:
- Parentheses \(( )\) signify that a boundary point is not included, which is also known as an open interval.
- Brackets \([ ]\) indicate that a boundary point is included, and this is a closed interval.
- The union symbol \( \cup \) combines separate intervals.
Factoring Polynomials
Factoring polynomials is a critical skill in solving inequalities, especially nonlinear ones. It involves expressing the polynomial as a product of simpler polynomials, which helps to identify critical points that determine the intervals of the solution.
In this exercise, we start with the inequality \(x^5 - x^2 > 0\). The first step is to factor out the greatest common factor, which is \(x^2\). This gives us \(x^2(x^3 - 1) > 0\).
Further factoring involves recognizing the cubic expression \(x^3 - 1\) as a difference of cubes. The difference of cubes formula is:
This factorization breaks the solution process into evaluating when this product of expressions is greater than zero. Factoring is not just about simplifying; it's about making the problem more manageable by revealing potential critical points for testing.
In this exercise, we start with the inequality \(x^5 - x^2 > 0\). The first step is to factor out the greatest common factor, which is \(x^2\). This gives us \(x^2(x^3 - 1) > 0\).
Further factoring involves recognizing the cubic expression \(x^3 - 1\) as a difference of cubes. The difference of cubes formula is:
- \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)
This factorization breaks the solution process into evaluating when this product of expressions is greater than zero. Factoring is not just about simplifying; it's about making the problem more manageable by revealing potential critical points for testing.
Quadratic Expressions
Quadratic expressions are another concept you'll often encounter, especially while factoring or analyzing polynomial inequalities. A quadratic expression is typically in the form \(ax^2 + bx + c\).
In this exercise, after factoring, we identify \(x^2 + x + 1\) as part of our inequality. This quadratic does not factor further with real numbers because its discriminant (from the formula \(b^2 - 4ac\)) is negative, specifically \(-3\).
When you cannot factor a quadratic expression further due to a negative discriminant:
In this exercise, after factoring, we identify \(x^2 + x + 1\) as part of our inequality. This quadratic does not factor further with real numbers because its discriminant (from the formula \(b^2 - 4ac\)) is negative, specifically \(-3\).
When you cannot factor a quadratic expression further due to a negative discriminant:
- The quadratic expression doesn't cross the x-axis—this means it does not affect the intervals of positivity or negativity directly by adding more critical points.
- Its sign will be determined by either evaluating the expression at a test point or understanding its standard shape (whether it opens upwards or downwards).