Question

A function \(f(x)\) is given. Find the critical points of \(f\) and use the Second Derivative Test, when possible, to determine the relative extrema. \(f(x)=\frac{1}{x^{2}+1}\)

Short answer

The critical point is at \(x = 0\) with a local maximum.

Step-by-step solution

Find the First Derivative

To find the critical points of the function, we must first find the derivative of the function. The function is given by \[ f(x) = \frac{1}{x^2 + 1}. \]Using the quotient rule, where if \( u(x) = 1 \) and \( v(x) = x^2 + 1 \), the derivative is\[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} = \frac{(x^2 + 1) \cdot 0 - 1 \cdot (2x)}{(x^2 + 1)^2} = \frac{-2x}{(x^2 + 1)^2}. \]

Solve for Critical Points

Critical points occur where the first derivative is zero or undefined. The derivative found is \[ f'(x) = \frac{-2x}{(x^2 + 1)^2}. \]This is zero when the numerator is zero. Therefore, \[-2x = 0. \]Solving for \(x\), we get \(x = 0\) as the critical point. The denominator \((x^2 + 1)^2\) is never zero since it is a sum of positive terms.

Apply the Second Derivative Test

To determine the nature of the critical point, we find the second derivative. Differentiating the first derivative \[ f'(x) = \frac{-2x}{(x^2+1)^2}, \]using the quotient rule again, we get\[ f''(x) = \frac{(-2)[(x^2+1)^2] - (-2x)[4x(x^2+1)]}{(x^2+1)^4}, \]simplifying to\[ f''(x) = \frac{-2(x^2+1)^2 + 8x^2(x^2+1)}{(x^2+1)^4}. \]Evaluating at the critical point \(x = 0\),\[ f''(0) = \frac{-2(1)^2 + 0}{1^4} = -2. \]Since \(f''(0) < 0\), the point is a local maximum.

Key concepts

First Derivative

Understanding the first derivative is crucial for finding critical points of any function. When given a function like \[ f(x) = \frac{1}{x^2 + 1}, \] we can determine its critical points by first finding its derivative. The derivative, denoted as \( f'(x) \), represents the rate of change or the slope of the tangent line to the function's curve at any point \( x \).For \( f(x) = \frac{1}{x^2 + 1} \), we use the quotient rule to differentiate. The quotient rule states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( f'(x) \) is:\[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}. \]In this case, \( u(x) = 1 \) and \( v(x) = x^2 + 1 \). Differentiating these functions, \( u'(x) = 0 \) and \( v'(x) = 2x \). Applying the quotient rule, we find:\[ f'(x) = \frac{0 - 2x}{(x^2 + 1)^2} = \frac{-2x}{(x^2 + 1)^2}. \] Critical points occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined. Since the denominator \( (x^2 + 1)^2 \) is never zero for any real \( x \), we focus on solving when the numerator \(-2x = 0 \). Thus, the critical point is at \( x = 0 \).

Second Derivative Test

The second derivative test helps determine whether a critical point is a relative maximum, a relative minimum, or neither. After identifying the critical points using the first derivative, we calculate the second derivative, \( f''(x) \). This reveals the concavity of the function at those points.Given \( f'(x) = \frac{-2x}{(x^2+1)^2} \), we find \( f''(x) \) using the quotient rule. The second derivative measures how the rate of change itself changes:\[ f''(x) = \frac{(-2)(x^2+1)^2 - (-2x)(4x)(x^2+1)}{(x^2+1)^4}. \]Simplifying, we have:\[ f''(x) = \frac{-2(x^2+1)^2 + 8x^2(x^2+1)}{(x^2+1)^4}. \]To apply the second derivative test at the critical point \( x = 0 \), substitute \( x = 0 \) in \( f''(x) \):\[ f''(0) = \frac{-2(1)^2}{1^4} = -2. \]The value is negative, indicating that the function is concave down at \( x = 0 \). This means that \( x = 0 \) is a relative maximum. When \( f''(x) < 0 \) at a critical point, it suggests a local maximum, while \( f''(x) > 0 \) would indicate a local minimum.

Relative Extrema

In calculus, relative extrema refer to the points in a function where it reaches a local maximum or minimum within some neighborhood. These extrema are key for understanding the behavior of functions, as they represent points where the function changes direction.Identifying these points starts with finding the critical points using the first derivative. In our case, the critical point found at \( x = 0 \) is analyzed using the second derivative test. The test centered at \( x = 0 \) indicates:\- Since \( f''(0) = -2 \) and is negative, the function reaches a local maximum at \( x = 0 \).Thus, this tells us that the function \( f(x) = \frac{1}{x^2 + 1} \) has a peak at \( x = 0 \).Having knowledge of relative extrema offers insights into a function's graph. It helps predict behaviors in optimization problems, where maximization or minimization is essential.