Question

In Exercises \(23-28\) , find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. $$k(x)=\frac{x}{x^{2}-4}$$

Short answer

The function \(k(x)=\frac{x}{{x^2}-4}\) has no local extrema. The function is increasing on intervals \((-\infty, -2)\) and \((2, \infty)\) and decreasing on interval \((-2, 2)\)

Step-by-step solution

Find the derivative

The first step is to find the derivative of the function \(k(x)=\frac{x}{{x^2}-4}\) using the quotient rule. The quotient rule 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 over the square of the denominator. So, the derivative \(k'(x)\) is given by \(k'(x)=\frac{{(x^{2}-4)-x(2x)}}{{(x^{2}-4)^{2}}}\).

Simplify the derivative

Simplify the expression for the derivative to \(k'(x)=\frac{{-4x^{2}+4}}{{(x^{2}-4)^{2}}}\).

Find critical points

Next, find the critical points by setting the simplified derivative equal to zero and solving for \(x\), which gives \(x = 0\) and \(x = 2, -2\) (the points where the denominator is equal to zero).

Determine intervals and local extrema

Then, determine on which intervals the function is increasing or decreasing and find the local extrema. To do this, substitute a number from each interval (formed by the critical points) into the derivative. If the resulting value is positive, then the function is increasing on that interval. If it's negative, the function is decreasing. The points \(x = 2\) and \(x = -2\) are not included in the intervals because the function is not defined at these points.

Write final answers

Finally, write the final answers based on the signs of the derivative on the intervals. The function has no local extrema because it's not defined at \(x = 2\) and \(x = -2\). The function is increasing on the intervals \((-\infty, -2)\) and \((2, \infty)\) and decreasing on the interval \((-2, 2)\).

Key concepts

Quotient Rule Differentiation

Understanding the quotient rule for differentiation is a cornerstone in calculus, especially when tackling complex rational functions. It essentially provides a method to differentiate a function that is the quotient of two differentiable functions. For a function represented as \( g(x)/h(x) \), the quotient rule states that its derivative is \[ \frac{d}{dx}\left(\frac{g(x)}{h(x)}\right) = \frac{h(x)g'(x) - g(x)h'(x)}{[h(x)]^2} \]. In our exercise, we applied the quotient rule to the function \(k(x)=\frac{x}{{x^2}-4}\).

It's important to remember when using the quotient rule that you must first differentiate the numerator and denominator separately. Then, construct the derivative following the formula, paying special attention to maintaining the correct sign between the terms. This rule is particularly useful when both the numerator and the denominator are relatively simple functions, as in the exercise provided.

Finding Local Extrema

Local extrema refer to the peaks and valleys of a function. These are the points where the function shifts from increasing to decreasing or vice versa, and they are identified where the derivative of the function changes sign. To find local extrema, we first derive the function. If the function's derivative equals zero or is undefined at a certain point, this location may be a potential local maximum or minimum.

In the given exercise, after differentiating the function, we look for values of \(x\) that make the derivative zero or undefined, which can reveal the local extrema. However, simply finding where the derivative equals zero is not enough. We must also check the behavior of the derivative around these points to confirm if they are indeed local maxima or minima.

Critical Points Analysis

Critical points of a function are the values of \(x\) at which the derivative is either zero or does not exist. These points are vital for understanding the function's overall behavior, including where it may have local extrema. Identifying critical points involves calculating the derivative, as we did using the quotient rule, and then solving for where this derivative is equal to zero or when the denominator is zero (indicating the derivative does not exist).

In our case, the critical points were found at \(x = 0\) and where the derivative is undefined at \(x = \pm 2\). However, since the function itself does not exist at \( x= \pm 2 \) because of a zero denominator in the original function, these are not considered points of the function but rather points to consider alongside the function's domain.

Increasing and Decreasing Intervals

A function's intervals of increase and decrease reveal much about its shape. After determining the critical points, we analyze the sign of the derivative on either side of these points to find these intervals. If the derivative is positive in an interval, the function is increasing there. Conversely, if the derivative is negative, the function is decreasing within that interval.

Our exercise demonstrates this analysis. By testing points in the intervals around the critical points (\(x = 0\) and the undefined points at \(x = \pm 2\)), we can determine the behavior on each interval. For \(k(x)\), we discovered that it increases on \( (-\infty, -2) \) and \( (2, \infty) \) and decreases on \( (-2, 2) \), with no local extrema within the domain of the function.