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. \(f(x)=\frac{x}{x^{2}-1}\)

Short answer

The function has vertical asymptotes at \( x = 1 \) and \( x = -1 \) and no relative extrema in its domain.

Step-by-step solution

Find the First Derivative

To find the critical points, we first need the first derivative of the function. Given the function \( f(x) = \frac{x}{x^2 - 1} \), it is a quotient function. We use the Quotient Rule: \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \). Here, \( u = x \) and \( v = x^2 - 1 \). The derivatives are \( u' = 1 \) and \( v' = 2x \). Thus, the first derivative is: \[ f'(x) = \frac{1 \cdot (x^2 - 1) - x \cdot 2x}{(x^2 - 1)^2} = \frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2} = \frac{-x^2 - 1}{(x^2 - 1)^2}. \] Simplifying further, \[ f'(x) = \frac{-x^2 - 1}{(x^2 - 1)^2}. \]

Find Critical Points

Critical points are found where \( f'(x) = 0 \) or \( f'(x) \) is undefined. Since \( f'(x) = \frac{-x^2 - 1}{(x^2 - 1)^2} \), the derivative does not equal zero because the numerator \(-x^2 - 1\) is never zero. Determine where the derivative is undefined by setting the denominator zero: \( (x^2 - 1)^2 = 0 \). Solving \( x^2 - 1 = 0 \), we find critical points at \( x = 1 \) and \( x = -1 \). However, \( f(x) \) is undefined at these points since they zero the denominator of \( f(x) \), so they're not in the domain of \( f \).

Apply the Second Derivative Test

Since there are no critical points in the domain of \( f(x) \), we do not apply the Second Derivative Test. The function's critical points occur at places where the function itself is undefined, meaning potential asymptotes exist at \( x = 1 \) and \( x = -1 \), not relative extrema.

Key concepts

Second Derivative Test

The Second Derivative Test is a helpful tool in calculus for analyzing the nature of critical points in differentiable functions. It determines whether these points are relative maxima, minima, or points of inflection. To use this test, first find the second derivative of the function, denoted as \( f''(x) \).

• If \( f''(c) > 0 \), the function is concave up at \( x = c \), indicating a potential relative minimum.
• If \( f''(c) < 0 \), the function is concave down at \( x = c \), suggesting a possible relative maximum.
• If \( f''(c) = 0 \), the test is inconclusive.

In our exercise, we weren't able to use the Second Derivative Test directly since there were no valid critical points within the domain of \( f(x) \). The critical points coincided with the asymptotes, so the test was not applicable. Nonetheless, understanding this test is still crucial for other functions where critical points exist within the domain.

Quotient Rule

When faced with a function expressed as a quotient of two differentiable functions, the Quotient Rule comes to the rescue. It provides a systematic way to find the derivative of such functions, and is essential in calculus for solving a wide range of problems.The Quotient Rule states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative is given by:\[(\frac{u}{v})' = \frac{u'v - uv'}{v^2}\]
Here, \( u \) is the numerator and \( v \) is the denominator. Their derivatives are denoted as \( u' \) and \( v' \). In our specific problem, we applied this rule to find that \( f'(x) = \frac{-x^2 - 1}{(x^2 - 1)^2} \), which helped in identifying the non-existing critical points within the domain.

Asymptotes

Asymptotes are lines that a graph of a function approaches but never touches. They are crucial in understanding the overall behavior and limits of functions, particularly rational functions, like the one in our exercise.

• Vertical Asymptotes occur at values of \( x \) that make the denominator zero, provided that these values don't also zero the numerator. In our function \( f(x) = \frac{x}{x^2 - 1} \), \( x = 1 \) and \( x = -1 \) are vertical asymptotes since they make the denominator zero while \( f(x) \) is undefined.
• Horizontal Asymptotes appear as \( x \rightarrow \pm \infty \), describing the end behavior of the function. In this case, \( f(x) \rightarrow 0 \) as \( x \rightarrow \pm \infty \).

Understanding these aspects is vital for sketching accurate graphs of functions and predicting their behavior far into the domain.

Relative Extrema

Relative extrema are peaks or valleys in a function's graph, commonly referred to as relative maximum or minimum points. These points represent local high or low values of the function.To find these points, we often need to first determine the critical points by solving \( f'(x) = 0 \) or finding where the derivative is undefined. However, the critical points should lie within the domain of the function to consider them for relative extrema. In our function, although derivative discontinuities indicated potential extrema, these values actually defined the points where the function was undefined (asymptotes), thus there were no valid relative extrema.Grasping the concept of relative extrema helps in analyzing and optimizing real-world phenomena where determining peak values is essential for decision-making.