Problem 43 Solve the polynomial inequality ... [FREE SOLUTION]

Chapter 4: Problem 43

Solve the polynomial inequality (a) symbolically and (b) graphically. $$ x^{3}+x^{2} \geq 2 x $$

Short Answer

Expert verified

The solution is $(-2, 0] \cup [1, \infty)$.

Step by step solution

Rewrite the Inequality

To solve the inequality $x^3 + x^2 \geq 2x$, start by moving all terms to one side to get a single inequality: $x^3 + x^2 - 2x \geq 0$.

Factor the Polynomial

Factor the polynomial $x^3 + x^2 - 2x$. Start by factoring out the common term $x$: $x(x^2 + x - 2) \geq 0$. Then, factor the quadratic $x^2 + x - 2$ to get $(x - 1)(x + 2)$. So, the inequality becomes $x(x - 1)(x + 2) \geq 0$.

Identify Critical Points

Set each factor equal to zero to find the critical points: $x = 0$, $x = 1$, and $x = -2$. These points divide the number line into intervals.

Test Intervals

Test each interval determined by the critical points to see where the product is non-negative. The intervals are: $(-\infty, -2)$, $(-2, 0)$, $(0, 1)$, and $(1, \infty)$. Choose a test point from each interval and substitute it into $x(x - 1)(x + 2)$ to determine the sign: - For $(-\infty, -2)$, test $x = -3$: negative.- For $(-2, 0)$, test $x = -1$: positive.- For $(0, 1)$, test $x = 0.5$: negative.- For $(1, \infty)$, test $x = 2$: positive.

Write the Solution Symbolically

The solution is where the product is non-negative. This occurs in the intervals $(-2, 0] \cup [1, \infty)$, including $x = 0$ and $x = 1$, where the product is zero.

Solve Graphically

To solve graphically, sketch the graph of $y = x^3 + x^2 - 2x$. The graph crosses the x-axis at $x = -2$, $0$, and $1$. Observe where the graph is above (or on) the x-axis, which corresponds to the solution found symbolically. It is above the x-axis in the intervals $(-2, 0] \cup [1, \infty)$.

Verify the Solution

Confirm the solution by checking the critical points and the sign change at each interval. Both the symbolic and graphical solutions should match the intervals: $(-2, 0] \cup [1, \infty)$.

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 91影视!

Key Concepts

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

Factoring Polynomials

Factoring polynomials is a key step in solving polynomial inequalities. It involves breaking down a polynomial into simpler "factor" polynomials that, when multiplied together, produce the original polynomial. In our exercise, we started with the inequality $x^3 + x^2 - 2x \geq 0$.

First, we factored out the common factor $x$ from each term:

The expression $x(x^2 + x - 2)$ was formed.

Next, we factorized the quadratic $x^2 + x - 2$ into

$(x - 1)(x + 2)$.

These factored terms allowed us to rewrite the polynomial as $x(x - 1)(x + 2) \geq 0$. This simplification is crucial as it prepares the inequality for identifying critical points and subsequently testing intervals.

Critical Points

Critical points in a polynomial inequality are values of $x$ where the polynomial equals zero. These points are essential because they mark transitions on the graph, potentially indicating changes in inequality sign.

For $x(x - 1)(x + 2) \geq 0$, we determined the critical points by setting each factor equal to zero:

$x = 0$
$x = 1$
$x = -2$

These critical points divide the number line into distinct intervals where we can test the polynomial's sign. Remember, finding critical points is pivotal as they help identify where the inequality holds true or changes direction.

Graphical Solution

Visualizing the polynomial can be an intuitive way to solve inequalities. When graphing $y = x^3 + x^2 - 2x$, you're observing the curve's interaction with the x-axis, where the value of $y = 0$.

In the graphical solution, the polynomial crosses the x-axis at the critical points:

$x = -2$
$x = 0$
$x = 1$

The solution to the inequality is where the graph is above or touches the x-axis ($(y \geq 0)$). Observing the graph, you see it remains above or on the x-axis within the intervals

$(-2, 0]$
$[1, \infty)$

The graphical method complements symbolic solutions and is especially useful for confirming results.

Interval Testing

Interval testing involves choosing sample points from different intervals, determined by the critical points, to decide where the inequality holds. For the polynomial $x(x - 1)(x + 2)$, we had these intervals:

$(-\infty, -2)$
$(-2, 0)$
$(0, 1)$
$(1, \infty)$

By picking a test point from each interval and substituting it back into the factors:

$(-\infty, -2): \; x = -3$ (negative)
$(-2, 0): \; x = -1$ (positive)
$(0, 1): \; x = 0.5$ (negative)
$(1, \infty): \; x = 2$ (positive)

These tests allowed us to detect where the polynomial's product was non-negative. It's a crucial step to establish the specific regions where the inequality holds true, especially when complemented by the critical points.

91影视

Solve the polynomial inequality (a) symbolically and (b) graphically. $$ x^{3}+x^{2} \geq 2 x $$

Short Answer

Step by step solution

Rewrite the Inequality

Factor the Polynomial

Identify Critical Points

Test Intervals

Write the Solution Symbolically

Solve Graphically

Verify the Solution

Key Concepts

Factoring Polynomials

Critical Points

Graphical Solution

Interval Testing

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Statistics

Probability and Statistics

Logic and Functions

Theoretical and Mathematical Physics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.