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Chapter 10: Problem 72

Find the relative extrema, if any, of each function. Use the second derivative test, if applicable. $$ f(x)=\frac{2 x}{x^{2}+1} $$

Short Answer

Expert verified
There is a relative maximum at the point \((-1, -1)\) and a relative minimum at the point \((1, 1)\).

Step by step solution

01

Find the first derivative of the function

The given function is \(f(x) = \frac{2x}{x^2 + 1}\). Using the quotient rule to differentiate, we get: \[f'(x) = \frac{(x^2 + 1)(2) - 2x(2x)}{(x^2 + 1)^2}.\] Simplify the expression to get: \[f'(x) = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} = \frac{-2x^2 + 2}{(x^2 + 1)^2}.\]
02

Determine the critical points

Set the first derivative equal to zero and solve for x: \[\frac{-2x^2 + 2}{(x^2 + 1)^2} = 0.\] To find where the function is not defined, we need to check the denominator. However, since the denominator is never equal to zero, we can ignore this case. Now solve the equation \(-2x^2 + 2 = 0\): \[-2x^2 + 2 = 0 \implies x^2 = 1\] This gives us two critical points, \(x = \pm 1\).
03

Find the second derivative of the function

Differentiate the first derivative to get the second derivative: \[f''(x) = \frac{d}{dx} \left(\frac{-2x^2 + 2}{(x^2 + 1)^2}\right).\] We will apply the quotient rule again: \[f''(x) = \frac{(-4x)(x^2 + 1)^2 - 2(-2x^2 + 2)(2x)(x^2 + 1)}{(x^2 + 1)^4}.\]
04

Apply the second derivative test

Evaluate the second derivative at the critical points: \[f''(1) = \frac{-4(1^2 + 1)^2 - 2(-2(1)^2 + 2)(2(1))(1^2 + 1)}{(1^2 + 1)^4} = 12.\] Since the second derivative at x = 1 is positive, it indicates a relative minimum at this point. \[f''(-1) = \frac{-4(-1^2 + 1)^2 - 2(-2(-1)^2 + 2)(-2(-1))(-1^2 + 1)}{(-1^2 + 1)^4} = -12.\] Since the second derivative at x = -1 is negative, it indicates a relative maximum at this point.
05

Determine the relative extrema

Plug the critical points into the original function to find the function values (relative extrema): For x = 1, \(f(1) = \frac{2(1)}{1^2 + 1} = 1\). For x = -1, \(f(-1) = \frac{2(-1)}{(-1)^2 + 1} = -1\). So, there is a relative maximum at the point \((-1, -1)\) and a relative minimum at the point \((1, 1)\).

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Key Concepts

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

Critical Points
Critical points are specific values in the domain of a function where its derivative is zero or undefined. These points are essential when analyzing the behavior of the function because they potentially indicate where the function has a relative maximum, minimum, or an inflection point.

To find the critical points of the function \(f(x) = \frac{2x}{x^2 + 1}\), we calculate the first derivative \(f'(x)\) and set it equal to zero. The resulting critical points are \(x = \text{\textpm}1\). It's also important to check where the first derivative may not be defined; however, in this example, the derivative is defined for all real numbers since the denominator \((x^2 + 1)^2\) is never zero.

By identifying the critical points, we can then proceed to further analyze the function, especially using the second derivative test to determine the nature of these critical points.
First Derivative
The first derivative of a function gives us the slope of the tangent line at any point along the curve. It also reveals the function's increasing or decreasing nature in intervals around critical points. In calculus, the first derivative is used to determine the location of potential relative extrema and to sketch the graph of the function.

For the given function \(f(x) = \frac{2x}{x^2 + 1}\), the first derivative \(f'(x)\) is calculated using the quotient rule, yielding \(f'(x) = \frac{-2x^2 + 2}{(x^2 + 1)^2}\). The points at which this derivative is zero are the critical points we've found, namely \(x = 1\) and \(x = -1\).

Understanding how to find and interpret the first derivative is critical for analyzing the behavior of functions and is foundational in calculus.
Relative Extrema
Relative extrema are the peaks and troughs of a function, where it reaches local maximums or minimums. To determine relative extrema using the second derivative test, follow these steps:
  • Identify critical points by finding where the first derivative is zero.
  • Compute the second derivative of the function.
  • Plug in the critical points into the second derivative:
If the second derivative is positive at a critical point, that point is a relative minimum; if negative, it is a relative maximum.

Applying these steps to \(f(x) = \frac{2x}{x^2 + 1}\), we find that \(f''(1)\) is positive, indicating a relative minimum at \((1, 1)\), and \(f''(-1)\) is negative, indicating a relative maximum at \((-1, -1)\). This analysis helps to form a complete picture of the function鈥檚 shape and leads to understanding its behavior throughout its domain.

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Most popular questions from this chapter

The total worldwide box-office receipts for a long-running movie are approximated by the function $$ T(x)=\frac{120 x^{2}}{x^{2}+4} $$ where \(T(x)\) is measured in millions of dollars and \(x\) is the number of years since the movie's release. Sketch the graph of the function \(T\) and interpret your results.

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ g(x)=x^{2}+2 x^{2 / 3} \text { on }[-2,2] $$

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One condition that must be satisfied before Theorem 3 (page 713 ) is applicable is that the interval on which \(f\) is defined must be a closed interval \([a, b] .\) Define a function \(f\) on the open interval \((-1,1)\) by \(f(x)=x\). Show that \(f\) does not attain an absolute maximum or an absolute minimum on the interval \((-1,1)\). Hint: What happens to \(f(x)\) if \(x\) is close to but not equal to \(x=\) \(-1 ?\) If \(x\) is close to but not equal to \(x=1 ?\)
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