Question

In Exercise, use a graphing utility to estimate graphically all relative extrema of the function. $$ f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2} $$

Short answer

By graphically representing the provided function and marking the points where derivative equals to zero, one can estimate the relative extrema points and determine their nature using the second derivative test.

Step-by-step solution

Graph the Function

Use a graphing utility to sketch the graph of the function \(f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2}\). Observe the graph for visible high and low points.

Identify Critical Points

Critical points occur where the derivative of the function equals zero or does not exist. In this case, calculate the derivative of the function \(f'(x)=2x^{3}-x^{2}-x\). Set the derivative equal to zero and solve for x to find the critical points.

Determine Relative Maximum and Minimum

Use the second derivative test to determine the nature of the critical points. If the second derivative at a critical point is positive, it is a relative minimum; if it is negative, it is a relative maximum. Note that relative extrema can also be seen from the graph.

Key concepts

Graphing Utility

In the digital age, graphing utilities are essential tools for students and mathematicians alike. These tools, which can be software programs, websites, or even dedicated graphing calculators, help visualize functions by plotting their graphs. When dealing with complex functions, as in our original problem, graphing utilities make it easy to see where a function has highs and lows, known as relative extrema.

To estimate graphically all relative extrema of a function like \(f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2}\), you simply input the function into the tool. Whether it's Desmos, GeoGebra, or another program, the graph appears, revealing the function's shape.

With the graph, we can clearly observe peaks (high points) and valleys (low points), which are critical in finding relative maxima and minima.

Relative Extrema

Relative extrema refer to the high and low points on a graph of a function. These are the points where a function changes direction from increasing to decreasing or vice versa.

Mathematically, relative maximum is where the function reaches a peak, while a relative minimum is a valley. These points are significant because they indicate the most and least value the function will locally take, which has practical applications in optimization.

In the context of our original exercise, identifying these points through the graph revealed by a graphing utility can guide a deeper analytical process, like taking derivatives, to confirm the accuracy and location of these extrema.

Derivative

The derivative of a function is pivotal in determining the nature and position of its extrema. It reveals the function's rate of change at any point, which indicates where the function is increasing or decreasing.

To find the derivative of the function \(f(x)=\frac{1}{2} x^{4}-\frac{1}{3} x^{3}-\frac{1}{2} x^{2}\), we apply basic differentiation rules, yielding \(f'(x)=2x^{3}-x^{2}-x\). This derivative helps locate critical points where the function may have relative extrema.

Once we set the derivative equal to zero, \(f'(x)=0\), the solutions to this equation give the critical points—the potential candidates for relative maximum and minimum. Moreover, points where the derivative does not exist must also be considered, though they are less common in polynomial functions.

Second Derivative Test

The second derivative test is a method to determine the nature of critical points found by the first derivative. Once we have these points from \(f'(x)=0\), the second derivative can help us identify whether each point is a minima, maxima, or neither. To do this, compute the second derivative, \(f''(x)\).

In the original problem, if \(f''(x)\) is positive at a critical point, the point is a relative minimum because the curve is concave up there. Conversely, if \(f''(x)\) is negative, the point is a relative maximum because the curve is concave down.

By applying the second derivative test, we can confirm our graphical observations and ensure we accurately classify the nature of each critical point. This method completes the process of analyzing a function's extrema with precision.