Problem 72
Question
Find the relative extrema, if any, of each function. Use the second derivative test, if applicable. $$ f(x)=\frac{2 x}{x^{2}+1} $$
Step-by-Step Solution
Verified Answer
There is a relative maximum at the point \((-1, -1)\) and a relative minimum at the point \((1, 1)\).
1Step 1: 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:
\[f'(x) = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} = \frac{-2x^2 + 2}{(x^2 + 1)^2}.\]
2Step 2: 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, \(x = \pm 1\).
3Step 3: 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:
\[f''(x) = \frac{(-4x)(x^2 + 1)^2 - 2(-2x^2 + 2)(2x)(x^2 + 1)}{(x^2 + 1)^4}.\]
4Step 4: 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.
5Step 5: 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 \((1, 1)\).
Key Concepts
Critical PointsFirst DerivativeRelative Extrema
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.
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.
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:
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’s shape and leads to understanding its behavior throughout its domain.
- 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:
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’s shape and leads to understanding its behavior throughout its domain.
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