Question

In Exercises \(15-22,\) use analytic methods to find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. $$k(x)=\frac{1}{x^{2}}$$

Short answer

The function \(k(x)=\frac{1}{x^{2}}\) has a local maximum at \(x=0\). It is increasing on the interval \(-\infty < x <0\) and decreasing on the interval \(0< x <\infty\).

Step-by-step solution

Compute the derivative

First, compute the derivative \(k'(x)\) of the function \(k(x)=\frac{1}{x^{2}}\) which results in \( k'(x) = \frac{-2}{x^{3}}.\)

Determine critical points

Next, set the derivative equal to zero and solve for \(x\). However, since \(k'(x)\) is never zero and is only undefined at \(x=0\), therefore, 0 is the only critical point.

Analyze the signs

Using a sign chart or test points, it can be found that on the interval \( -\infty< x<0\), \(k'(x)>0\) and on the interval \(0< x <\infty\), \(k'(x)<0\). Meaning the function is increasing on the interval \(-\infty< x<0\), and decreasing on the interval \(0< x <\infty\).

Find local extrema

Since the function changes from increasing to decreasing at x = 0, that means x = 0 is a local maximum.

Key concepts

Finding Local Extrema

Understanding how to find local extrema is an important skill in calculus, particularly for identifying the highest or lowest points on a graph within a specific interval. These points are known as local maxima and minima. To find local extrema of a continuous function, one typically computes the derivative of the function and then determines where this derivative equals zero. These points are potential candidates for local extrema because they indicate where the function's slope changes direction.

From the provided exercise, we see that the function given is \( k(x)=\frac{1}{x^{2}} \). After computing the derivative, \( k'(x) = \frac{-2}{x^{3}} \), we can observe that the derivative does not actually equal zero for any value of \(x\). Instead, it is undefined at \(x=0\), which also acts as a critical point and possibly a local extremum. Since the sign of the derivative changes from positive to negative as we pass through \(x=0\), this point corresponds to a local maximum. The function does not have any local minima within the domain.

Intervals of Increase and Decrease

To comprehend how a function behaves, figuring out where it is increasing or decreasing is essential. If the derivative of the function, \(k'(x)\), is greater than zero on a particular interval, the function is increasing there; conversely, if the derivative is less than zero, the function is decreasing. By using this method, one can easily determine the behavior of a function on its domain.

In our example function, \( k(x)=\frac{1}{x^{2}} \), we use the derivative \( k'(x) = \frac{-2}{x^{3}} \) to assess where the function rises and falls. Since \(k'(x)>0\) for \(x<0\), the function is increasing on the interval \(-\infty< x<0\). On the other hand, the derivative is negative for \(x>0\), indicating that \( k(x) \) is decreasing on the interval \(0< x <\infty\). Highlighting these intervals allows students to visualize the function's growth and decay, aiding in a deeper understanding of the function's overall behavior.

Derivative Computation

The derivative of a function signifies the rate at which the function's value changes as its input changes. Calculating derivatives is fundamental in calculus as it helps in determining the slope of the function at any given point. For polynomial functions, derivatives are obtained by applying power rules, whereas for non-polynomials, various other differentiation techniques may be employed.

In the exercise, the function \( k(x)=\frac{1}{x^{2}} \) is not a polynomial but rather a rational function. To find its derivative, we apply the power rule, bringing the exponent to the front and reducing the exponent by one, resulting in \( k'(x) = -2x^{-3} \) or \( k'(x) = \frac{-2}{x^{3}} \). Comprehending derivative computation equips students with the ability to analyze functions not just for slopes, but also for rates of change, areas under curves, and solving problems involving motion.