Solve: ${ \frac{x 3-x}{x 2+1} \ \textgreater \ 0 }$Find The Intervals Where { X$}$ Satisfies The Inequality.

Introduction

In this article, we will explore the solution to the inequality $\frac{x^3-x}{x^2+1} > 0$. This involves finding the intervals where the expression $\frac{x^3-x}{x^2+1}$ is greater than zero. To solve this inequality, we will use the concept of sign charts and critical points.

Understanding the Inequality

The given inequality is $\frac{x^3-x}{x^2+1} > 0$. We can see that the numerator is a cubic polynomial, while the denominator is a quadratic polynomial. To solve this inequality, we need to find the values of $x$ for which the expression $\frac{x^3-x}{x^2+1}$ is positive.

Finding Critical Points

To find the critical points, we need to find the values of $x$ that make the numerator and denominator equal to zero. The numerator is equal to zero when $x^3-x=0$, which gives us $x(x^2-1)=0$. This gives us three critical points: $x=0$, $x=1$, and $x=-1$. The denominator is equal to zero when $x^2+1=0$, but this equation has no real solutions.

Creating a Sign Chart

To create a sign chart, we need to determine the sign of the expression $\frac{x^3-x}{x^2+1}$ in each interval defined by the critical points. We can do this by choosing a test point in each interval and evaluating the expression at that point.

Interval 1: $(-\infty, -1)$

Let's choose a test point $x=-2$. Evaluating the expression at this point, we get:

$$\frac{(-2)^3-(-2)}{(-2)^2+1} = \frac{-8+2}{4+1} = \frac{-6}{5} < 0$$

So, the expression is negative in this interval.

Interval 2: $(-1, 0)$

Let's choose a test point $x=-0.5$. Evaluating the expression at this point, we get:

$$\frac{(-0.5)^3-(-0.5)}{(-0.5)^2+1} = \frac{-0.125+0.5}{0.25+1} = \frac{0.375}{1.25} > 0$$

So, the expression is positive in this interval.

Interval 3: $(0, 1)$

Let's choose a test point $x=0.5$. Evaluating the expression at this point, we get:

$$\frac{(0.5)^3-0.5}{(0.5)^2+1} = \frac{0.125-0.5}{0.25+1} = \frac{-0.375}{1.25} < 0$$

So, the expression is negative in this interval.

Interval 4: $(1, \infty)$

Let's choose a test point $x=2$. Evaluating the expression at this point, we get:

$$\frac{(2)^3-2}{(2)^2+1} = \frac{8-2}{4+1} = \frac{6}{5} > 0$$

So, the expression is positive in this interval.

Conclusion

Based on the sign chart, we can conclude that the expression $\frac{x^3-x}{x^2+1}$ is positive in the intervals $(-1, 0)$ and $(1, \infty)$. Therefore, the solution to the inequality is $x \in (-1, 0) \cup (1, \infty)$.

Final Answer

The final answer is $\boxed{(-1, 0) \cup (1, \infty)}$.

Introduction

In our previous article, we explored the solution to the inequality $\frac{x^3-x}{x^2+1} > 0$. We found that the expression is positive in the intervals $(-1, 0)$ and $(1, \infty)$. In this article, we will answer some frequently asked questions related to this inequality.

Q&A

Q: What is the significance of the critical points in the inequality?

A: The critical points are the values of $x$ that make the numerator and denominator equal to zero. In this case, the critical points are $x=0$, $x=1$, and $x=-1$. These points divide the number line into intervals, and we need to determine the sign of the expression in each interval.

Q: How do we determine the sign of the expression in each interval?

A: To determine the sign of the expression in each interval, we need to choose a test point in the interval and evaluate the expression at that point. If the expression is positive at the test point, then it is positive in the entire interval. If the expression is negative at the test point, then it is negative in the entire interval.

Q: What is the difference between a sign chart and a graph?

A: A sign chart is a table that shows the sign of the expression in each interval, while a graph is a visual representation of the expression. A sign chart is a useful tool for determining the sign of the expression in each interval, while a graph can provide more information about the behavior of the expression.

Q: Can we use the sign chart to find the solution to the inequality?

A: Yes, we can use the sign chart to find the solution to the inequality. By looking at the sign chart, we can see that the expression is positive in the intervals $(-1, 0)$ and $(1, \infty)$. Therefore, the solution to the inequality is $x \in (-1, 0) \cup (1, \infty)$.

Q: How do we know that the expression is not positive in the intervals $(-\infty, -1)$ and $(0, 1)$?

A: We know that the expression is not positive in the intervals $(-\infty, -1)$ and $(0, 1)$ because we evaluated the expression at test points in these intervals and found that it was negative.

Q: Can we use the same method to solve other inequalities?

A: Yes, we can use the same method to solve other inequalities. The key is to find the critical points, create a sign chart, and determine the sign of the expression in each interval.

Conclusion

In this article, we answered some frequently asked questions related to the inequality $\frac{x^3-x}{x^2+1} > 0$. We discussed the significance of the critical points, how to determine the sign of the expression in each interval, and how to use the sign chart to find the solution to the inequality. We also compared the sign chart to a graph and discussed how to use the same method to solve other inequalities.

Final Answer

The final answer is $\boxed{(-1, 0) \cup (1, \infty)}$.