Consider The Function F ( X ) = − X 5 X 2 + 1 F(x) = \frac{-x}{5x^2 + 1} F ( X ) = 5 X 2 + 1 − X ​ On The Interval 0 ≤ X ≤ 2 0 \leq X \leq 2 0 ≤ X ≤ 2 . For Each Of The Following, Enter DNE If A Value Does Not Exist.This Function Has An Absolute Minimum Value Equal To □ \square □ Which Is

Introduction

In this article, we will be analyzing the function $f(x) = \frac{-x}{5x^2 + 1}$ on the interval $0 \leq x \leq 2$. Our primary goal is to determine the absolute minimum value of this function within the given interval. To achieve this, we will employ various mathematical techniques, including differentiation and the first derivative test.

Understanding the Function

The given function is $f(x) = \frac{-x}{5x^2 + 1}$. This is a rational function, where the numerator is a linear function and the denominator is a quadratic function. The function is defined for all real values of $x$, except when the denominator is equal to zero.

Finding the Critical Points

To find the critical points of the function, we need to find the values of $x$ for which the first derivative is equal to zero or undefined. We will start by finding the first derivative of the function using the quotient rule.

First Derivative

The first derivative of the function $f(x)$ is given by:

$$f'(x) = \frac{(5x^2 + 1)(-1) - (-x)(10x)}{(5x^2 + 1)^2}$$

Simplifying the expression, we get:

$$f'(x) = \frac{-5x^2 - 1 + 10x^2}{(5x^2 + 1)^2}$$

$$f'(x) = \frac{5x^2 - 1}{(5x^2 + 1)^2}$$

Finding the Critical Points

To find the critical points, we need to set the first derivative equal to zero and solve for $x$.

$$\frac{5x^2 - 1}{(5x^2 + 1)^2} = 0$$

Since the denominator is always positive, we can set the numerator equal to zero and solve for $x$.

$$5x^2 - 1 = 0$$

$$5x^2 = 1$$

$$x^2 = \frac{1}{5}$$

$$x = \pm \sqrt{\frac{1}{5}}$$

However, since we are only interested in the interval $0 \leq x \leq 2$, we will only consider the positive value of $x$.

$$x = \sqrt{\frac{1}{5}}$$

Checking for Undefined Points

We also need to check if the first derivative is undefined at any point in the interval. The first derivative is undefined when the denominator is equal to zero.

$$(5x^2 + 1)^2 = 0$$

$$5x^2 + 1 = 0$$

$$5x^2 = -1$$

$$x^2 = -\frac{1}{5}$$

Since $x^2$ cannot be negative, the first derivative is never undefined in the interval.

Applying the First Derivative Test

Now that we have found the critical points, we can apply the first derivative test to determine the nature of the critical points.

First Derivative Test

The first derivative test states that if the first derivative changes sign at a critical point, then the critical point is a local extremum.

We can evaluate the first derivative at the critical point to determine the sign of the first derivative.

$$f'(\sqrt{\frac{1}{5}}) = \frac{5(\frac{1}{5}) - 1}{(5(\frac{1}{5}) + 1)^2}$$

$$f'(\sqrt{\frac{1}{5}}) = \frac{\frac{1}{5} - 1}{(\frac{1}{5} + 1)^2}$$

$$f'(\sqrt{\frac{1}{5}}) = \frac{-\frac{4}{5}}{(\frac{6}{5})^2}$$

$$f'(\sqrt{\frac{1}{5}}) = \frac{-\frac{4}{5}}{\frac{36}{25}}$$

$$f'(\sqrt{\frac{1}{5}}) = -\frac{4}{5} \times \frac{25}{36}$$

$$f'(\sqrt{\frac{1}{5}}) = -\frac{20}{36}$$

$$f'(\sqrt{\frac{1}{5}}) = -\frac{5}{9}$$

Since the first derivative is negative at the critical point, we can conclude that the critical point is a local maximum.

Evaluating the Function at the Critical Points

Now that we have determined the nature of the critical points, we can evaluate the function at the critical points to determine the absolute minimum value.

Evaluating the Function at the Critical Point

We will evaluate the function at the critical point $x = \sqrt{\frac{1}{5}}$.

$$f(\sqrt{\frac{1}{5}}) = \frac{-\sqrt{\frac{1}{5}}}{5(\sqrt{\frac{1}{5}})^2 + 1}$$

$$f(\sqrt{\frac{1}{5}}) = \frac{-\sqrt{\frac{1}{5}}}{\frac{5}{5} + 1}$$

$$f(\sqrt{\frac{1}{5}}) = \frac{-\sqrt{\frac{1}{5}}}{1 + 1}$$

$$f(\sqrt{\frac{1}{5}}) = \frac{-\sqrt{\frac{1}{5}}}{2}$$

$$f(\sqrt{\frac{1}{5}}) = -\frac{\sqrt{\frac{1}{5}}}{2}$$

$$f(\sqrt{\frac{1}{5}}) = -\frac{\sqrt{5}}{10}$$

Evaluating the Function at the Endpoints

We also need to evaluate the function at the endpoints of the interval to determine the absolute minimum value.

Evaluating the Function at $x = 0$

$$f(0) = \frac{-0}{5(0)^2 + 1}$$

$$f(0) = \frac{0}{1}$$

$$f(0) = 0$$

Evaluating the Function at $x = 2$

$$f(2) = \frac{-2}{5(2)^2 + 1}$$

$$f(2) = \frac{-2}{20 + 1}$$

$$f(2) = \frac{-2}{21}$$

Comparing the Values

We can compare the values of the function at the critical points and the endpoints to determine the absolute minimum value.

The value of the function at the critical point is $-\frac{\sqrt{5}}{10}$.

The value of the function at the endpoint $x = 0$ is $0$.

The value of the function at the endpoint $x = 2$ is $-\frac{2}{21}$.

Since $-\frac{2}{21}$ is less than $-\frac{\sqrt{5}}{10}$, we can conclude that the absolute minimum value of the function is $-\frac{2}{21}$.

Conclusion

$$$$$$

Introduction

In our previous article, we analyzed the function $f(x) = \frac{-x}{5x^2 + 1}$ on the interval $0 \leq x \leq 2$ and determined the absolute minimum value of the function. In this article, we will answer some frequently asked questions related to the analysis of the function.

Q: What is the absolute minimum value of the function?

A: The absolute minimum value of the function is $-\frac{2}{21}$.

Q: How did you find the critical points of the function?

A: We found the critical points of the function by setting the first derivative equal to zero and solving for $x$. We also checked for undefined points by evaluating the denominator of the first derivative.

Q: What is the first derivative test?

A: The first derivative test is a method used to determine the nature of a critical point. If the first derivative changes sign at a critical point, then the critical point is a local extremum.

Q: How did you apply the first derivative test to the function?

A: We evaluated the first derivative at the critical point to determine the sign of the first derivative. Since the first derivative was negative at the critical point, we concluded that the critical point was a local maximum.

Q: Why did you evaluate the function at the endpoints of the interval?

A: We evaluated the function at the endpoints of the interval to determine the absolute minimum value of the function. Since the function is a continuous function, the absolute minimum value must occur at a critical point or an endpoint.

Q: What is the significance of the absolute minimum value of the function?

A: The absolute minimum value of the function is significant because it represents the lowest value that the function can attain on the given interval. This value is important in various applications, such as optimization problems and data analysis.

Q: Can you provide more examples of functions that can be analyzed using the same techniques?

A: Yes, there are many functions that can be analyzed using the same techniques. Some examples include:

• $f(x) = \frac{x}{x^2 + 1}$ on the interval $0 \leq x \leq 2$
• $f(x) = \frac{x^2}{x^2 + 1}$ on the interval $0 \leq x \leq 2$
• $f(x) = \frac{x^3}{x^2 + 1}$ on the interval $0 \leq x \leq 2$

These functions can be analyzed using the same techniques as the original function, including finding the critical points, applying the first derivative test, and evaluating the function at the endpoints.

Conclusion

In this article, we answered some frequently asked questions related to the analysis of the function $f(x) = \frac{-x}{5x^2 + 1}$ on the interval $0 \leq x \leq 2$. We provided explanations and examples to help clarify the concepts and techniques used in the analysis. We hope that this article has been helpful in understanding the analysis of the function and its applications.