Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For each function: (a) Find all critical points on the specified interval. (b) Classify each critical point: Is it a local maximum, a local minimum, an absolute maximum, or an absolute minimum? (c) If the function attains an absolute maximum and/or minimum on the specified interval, what is the maximum and/or minimum value? \(f(x)=x^{3}-3 x+2\) on \([0,3]\)

Step-by-Step Solution

Answer

The critical point on the interval \([0,3]\) is \(x=1\) and it is a local minimum and an absolute minimum. The function \(f(x)=x^{3} - 3x + 2\) attains an absolute maximum of 20 at \(x=3\) and an absolute minimum of 0 at \(x=1\).

Step 1: Find the Critical Points

First, differentiate the function \(f(x)\) to find \(f'(x)\). \(f(x)= x^{3} - 3x + 2\) gives \(f'(x) = 3x^{2} - 3\). The critical points of the function are the solutions to the equation \(f'(x) = 0\). Solve \(3x^{2} - 3 = 0\), which gives \(x^{2} = 1\) and then \(x= -1, 1\). Only \(x=1\) lies in the interval \([0,3]\), so this is the only critical point on the interval.

Step 2: Classify the Critical Point

To classify the critical point, take the second derivative of the function. Differentiating \(f'(x)\) gives \(f''(x) = 6x\), and evaluate this at the critical point \(x=1\), to get \(f''(1) = 6\). Since \(f''(1) > 0\), the second derivative test indicates that \(x=1\) is a local and absolute minimum.

Step 3: Find the Absolute Extrema

To find if the function attains an absolute maximum and/or minimum on the interval \([0,3]\), apply the closed interval method, by evaluating the function at its critical points and at the endpoints of the interval. We have \(f(0) = 2\), \(f(1) = 1^{3} - 3*1 + 2 = 0\) and \(f(3) = 3^{3} - 3*3 + 2 = 20\). Thus, the function attains an absolute maximum of 20 at \(x=3\) and an absolute minimum of 0 at \(x=1\).

Key Concepts

Second Derivative Test

Understanding the 'Second Derivative Test' is crucial for determining the concavity of a function at its critical points and classifying the nature of these points. After identifying a critical point—where the first derivative is zero or undefined—we can use the second derivative to determine if the point is a local maximum, local minimum, or neither.

The second derivative of a function, denoted by f''(x), provides us with a concavity check: if f''(x) is positive, the function is concave upwards, resembling a cup, and the critical point is a local minimum. Conversely, if f''(x) is negative, the function is concave downwards, like an arch, and the critical point is a local maximum. If f''(x) equals zero, the test is inconclusive—we must seek other methods to determine the point's nature.

In the example of the function f(x) = x^3 - 3x + 2, we find the critical point at x=1. By applying the second derivative test, we established that f''(1)=6 is greater than zero, which means the function is concave up at x=1, and we have a local (and in this case, also absolute) minimum at x=1.

Absolute Maximum and Minimum

The 'Absolute Maximum and Minimum' are the highest and lowest points, respectively, that a function will attain over a specified interval. Unlike local extrema, which are relative to the vicinity of a critical point, absolute extrema are the supreme values across the entire domain of interest.

Finding these values is a key step in many calculus problems since they often represent optimal solutions in real-world scenarios. To locate these points, one must evaluate the function at critical points and the boundaries of the given interval.

Using the aforementioned function as an example, after identifying the critical point and its nature, we evaluate the function at both the critical point and the interval endpoints. The lowest value obtained is the absolute minimum, and the highest is the absolute maximum. In our case, f(1) yields the absolute minimum of 0, while f(3) gives us the absolute maximum of 20.

Closed Interval Method

When a function is continuous on a closed interval, the 'Closed Interval Method' allows us to find its absolute extrema. It involves evaluating the function at all critical points as well as at the endpoints of the interval. For functions that are continuous and differentiable within the given interval, this method ensures that none of the absolute extrema are overlooked.

To apply this method, we follow these steps:

1. Find all critical points within the interval where the first derivative equals zero or is undefined.
2. Evaluate the function at all critical points found in step 1.
3. Evaluate the function at the endpoints of the interval.
4. The highest and lowest values from steps 2 and 3 are then the absolute maximum and minimum, respectively.

By applying this method to f(x) = x^3 - 3x + 2 on the interval [0,3], we concluded that the function reaches its absolute maximum value of 20 at x=3 and its absolute minimum value of 0 at x=1, adhering to the completed steps of the closed interval method.