Question

In Exercises, find all relative extrema of the function. Use the Second- Derivative Test when applicable. $$ f(x)=\frac{x}{x^{2}-1} $$

Short answer

The function \( f(x)=\frac{x}{x^{2}-1} \) has a local maximum at \( x=1 \) and a local minimum at \( x=-1 \).

Step-by-step solution

Find the first derivative

The first derivative of the function will be determined using the quotient rule for derivatives, which states that the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. Here, the numerator is \( x \) and the denominator is \( x^{2}-1 \). So, the first derivative \( f'(x) \) will be \( f'(x) = \frac{(x^{2}-1)-(2x^{2})}{(x^{2}-1)^{2}} = -\frac{x^{2}+1}{(x^{2}-1)^{2}} \)

Find the critical points

The critical points of the function are the values of \( x \) that either make the derivative \( 0 \) or undefined. Setting \( f'(x) = 0 \) and solving for \( x \) gives no solutions, our critical points come where \( f'(x) \) is undefined, which is \( x = \pm1 \).

Find the second derivative

The second derivative of the function is needed for the Second-Derivative Test. Using the quotient rule again, we get the second derivative \( f''(x) = \frac{2x(x^{2}-3)}{(x^{2}-1)^{3}} \)

Use the Second-Derivative Test

The Second-Derivative Test will classify the nature of each critical point. Evaluate the second derivative at each critical point. If the value of the second derivative at a point is positive, then the function is concave up at that point, and the point is a local minimum. If the value is negative, then the function is concave down, and the point is a local maximum. We get \( f''(1) = -2 < 0 \) and \( f''(-1) = 2 > 0 \). So, \( x=1 \) is a local maximum and \( x=-1 \) is a local minimum.

Key concepts

Second-Derivative Test

The Second-Derivative Test is a useful method to determine the nature of critical points of a function, specifically whether these points are relative maxima, minima, or neither. To use this test, we first find the second derivative of the function.

• If the second derivative evaluated at a critical point is positive, the function is concave up at that point, indicating a local minimum.
• If it is negative, the function is concave down, indicating a local maximum.
• If the second derivative is zero, the test is inconclusive, meaning other methods must be used to classify the point.

In our exercise, we found that at the critical point \(x = 1\), the second derivative is negative \(f''(1) = -2\), indicating a local maximum. Conversely, at \(x = -1\), the second derivative is positive \(f''(-1) = 2\), revealing a local minimum. This classification helps us understand the behavior of the function around these points.

Critical Points

Critical points are the points on a graph where the derivative of a function is either zero or undefined. These are often points where the function reaches local extrema — local maxima or minima. Finding critical points involves calculating the first derivative of the function and determining where this derivative equals zero or does not exist.
In our example, the first derivative of the function \(f(x) = \frac{x}{x^2 - 1}\) is \(f'(x) = -\frac{x^2 + 1}{(x^2 - 1)^2}\). When we set \(f'(x) = 0\), we find no solutions, as the numerator \(-x^2 - 1\) never equals zero. However, critical points arise where the derivative is undefined, which happens at \(x = 1\) and \(x = -1\) due to the zero denominator. These critical points are significant, as they guide us in examining the function's extremity behavior through the Second-Derivative Test.

Concavity

Concavity refers to the curvature direction of the graph of a function. It can visually indicate where a function is bending upwards or downwards.

• A function is concave up when its graph looks like an upward-opening bowl, meaning the second derivative is positive in that interval.
• It is concave down when the graph resembles a downward-opening bowl, with the second derivative being negative.

Understanding concavity is crucial when analyzing functions because it tells us about the rate at which the slope of the tangent line is changing. It also helps in predicting turning points or inflection points where concavity changes occur. In the given exercise, concavity helps us classify critical points and identify whether they are local maxima or minima through the Second-Derivative Test.

Quotient Rule

The Quotient Rule is crucial for differentiating functions that are the quotient of two other functions. If you have a function \(f(x) = \frac{u(x)}{v(x)}\), the derivative can be calculated using the formula: \[f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}\]
This rule is invaluable when dealing with rational functions, such as our given function \(f(x) = \frac{x}{x^2 - 1}\).
By applying the quotient rule:

• The derivative of the numerator \(u(x) = x\) is \(1\)
• The derivative of the denominator \(v(x) = x^2 - 1\) is \(2x\)

Substituting these into the quotient rule gives \(f'(x) = -\frac{x^2 + 1}{(x^2 - 1)^2}\).
This result allows us to identify critical points and further analyze the function through the Second-Derivative Test.