Problem 20 - Calculus of a Single Variable

Question

In Exercises 5–24, analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results. $$ y=-\frac{1}{3}\left(x^{3}-3 x+2\right) $$

Step-by-Step Solution

Answer

The function \(y=-\frac{1}{3}\left(x^{3}-3 x+2\right)\) has no asymptotes, its extrema are at \(x = -1\) (maxima) and \(x = 1\) (minima), and it has a point of inflection at \(x = 0\). The graph is formed accordingly.

Step 1: Compute the First Derivative

To compute the first derivative of \(y=-\frac{1}{3}\left(x^{3}-3 x+2\right)\), apply the power rule and the chain rule. The first derivative, \(y'\), is \(-\frac{1}{3}(3x^2-3)\), which simplifies to \(-x^2+1\).

Step 2: Find the Critical Points and Extrema

Critical points occur when the first derivative equals 0 or is undefined. Solve \(-x^2+1=0\), which gives two solutions: x=-1, x=1. To classify extrema, use the first derivative test. It can be seen that the function is increasing in \(-\infty, -1\) and \(-1, 1\) and decreasing in \(1, \infty\). Therefore, x=-1 is a relative maxima and x=1 is a relative minima.

Step 3: Compute the Second Derivative and Find Inflection Points

Find the second derivative, \(y''\), by differentiating \(y'\). The second derivative, \(y''\), is \(-2x\). The inflection point is where \(y''\) is zero or undefined, so solve \(-2x=0\) to find x=0. This means there is a point of inflection at \(x=0\).

Step 4: Find the Asymptotes

There are no vertical or horizontal asymptotes as the function \(y\) is a polynomial and is defined for all \(x\) values and \(y\) does not approach a finite number as \(x\) approaches \(\pm \infty\).

Step 5: Draw the Graph

Plot the function's graph by making use of the identified features. Plot the intercepts (\(x = -1, 0, 1\)), extrema (\(x = -1, 1\)), inflection point (\(x = 0\)), and note the absence of asymptotes. The function decreases from \(-\infty\) to -1, increases from \(-1\) to 0, decreases from 0 to 1, and increases from \(1\) to \(\infty\).

Key Concepts

Critical Points

Identifying critical points in graphing polynomial functions is a pivotal step towards understanding the function's behavior. In calculus, critical points occur where a function's first derivative is either zero or undefined. These points are crucial for determining where a function reaches relative maxima or minima.

In our exercise, we computed the first derivative of the given function, set it equal to zero, and solved for the variable. As a result, we found two critical points at \(x = -1\) and \(x = 1\). To understand what happens at these points, we need to analyze the function's behavior around them, which leads us into the first derivative test.

Inflection Points

Inflection points on a graph represent where the curve changes concavity—from being concave up (shaped like a bowl) to concave down (shaped like a cap), or vice versa. These points provide insights into the curvature of the function's graph.

To find the inflection points, we calculate the second derivative and look for where it is equal to zero or undefined. The sign change of the second derivative across a point indicates a change in concavity. In the solution steps, by computing the second derivative as \(y'' = -2x\) and solving \(y'' = 0\), we determined an inflection point at \(x = 0\). At this point, our function changes its curvature, adding to the graph's overall shape and characteristic.

First Derivative Test

The first derivative test is a reliable method to classify critical points as local maxima, minima, or points at which there is no extremum. This test involves analyzing the sign of the first derivative before and after the critical point. If the derivative changes from positive to negative at a critical point, the function has a local maximum there. Conversely, if the derivative changes from negative to positive, there's a local minimum.

In our case, the first derivative \(y' = -x^2 + 1\) changes from positive to negative at \(x = -1\), indicating a relative maximum, and from negative to positive at \(x = 1\), showing a relative minimum. This test complements the identification of critical points and further illustrates the function's behavior.

Second Derivative

The second derivative of a function provides even deeper analysis into the graph's concavity and potential points of inflection. It tells us where the curve of the graph bends and whether the function is accelerating or decelerating.

When computing the second derivative, we look for regions where it's positive (indicating that the graph is concave up) or negative (concave down), as well as points where it equals zero, which are potential inflection points, as we have seen with \(x = 0\) in our example. The second derivative thus plays a critical role in predicting the geometric properties of the function's graph beyond just its slope.