Problem 4 - 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 function \(f(x) = x^{3} - 3x + 2\) has one critical point in the interval from 0 to 3, which is \(x = 1\). This point is a local minimum. The function attains an absolute maximum value of 20 at \(x = 3\) and an absolute minimum value of 0 at \(x = 1\).

Step 1: Find the derivative

Firstly, determine the derivative of the given function \(f(x)\). The derivative of \(f(x) = x^{3}-3x+2\) is \(f'(x) = 3x^{2}-3\).

Step 2: Find the critical points

The critical points of a function are found where the derivative is equal to zero or does not exist. Here, as a polynomial, \(f'(x)\) exists for all \(x\), so set \(f'(x) = 0\) to find the critical points. Equate \(3x^{2}-3=0\) to find critical points which gives \(x^{2}=1\) and hence, \(x=1, -1\). But -1 is not in our region (0,3). So only \(x=1\) is the critical point in the given interval.

Step 3: Classify the critical points

A critical point is a local minimum if the function changes from decreasing to increasing at that point, a local maximum if it changes from increasing to decreasing. To classify these, use the second derivative test. Take the second derivative of \(f(x)\), which is \(f''(x) = 6x\). Evaluate \(f''(x)\) at the critical point \(x=1\) which gives \(f''(1) = 6\). Here, the second derivative is greater than zero, suggesting that we have a local minimum at \(x=1\).

Step 4: Check for absolute maximum or minimum

To find if the function has an absolute maximum or minimum on the given interval (0,3), evaluate the function at the critical point and at the endpoints of the interval. Evaluate \(f(0)\), \(f(1)\), and \(f(3)\), which gives \(f(0)=2\), \(f(1)=0\), and \(f(3)=20\). So, on the interval from 0 to 3, the function attains a maximum value of 20 at \(x=3\) and a minimum value of 0 at \(x=1\).

Key Concepts

Second Derivative Test

The second derivative test is a powerful tool in calculus that allows us to classify critical points—those points where the derivative of a function is zero or undefined. To use this test, you first need to find the second derivative of the function. For our function, we already found that \( f'(x) = 3x^2 - 3 \). The second derivative, which is needed for the test, is \( f''(x) = 6x \). Once the second derivative is known, evaluate it at the critical points found. A positive result indicates a local minimum, as the curve is concave up at that point. Conversely, a negative result signifies a local maximum, showing the curve is concave down. If the second derivative equals zero, the test is inconclusive. For our function at the critical point \( x = 1 \), we found \( f''(1) = 6 \), which is greater than zero, making \( x=1 \) a local minimum. Thus, always remember:

• If \( f''(c) > 0 \), there's a local minimum at \( x = c \).
• If \( f''(c) < 0 \), there's a local maximum at \( x = c \).
• If \( f''(c) = 0 \), the test is inconclusive.

Local Maximum and Minimum

Local maximum and minimum points are essential features of a function’s graph. They are points where the function attains a maximum or minimum value within an interval surrounding that point. This gives insights into the behavior and shape of the graph. To identify these points, you usually need to find where the derivative changes sign.In our exercise, the critical point was \( x = 1 \), and by using the second derivative test, we identified it as a local minimum. As \( f'(x) \) changes from negative to positive at \( x = 1 \), the function goes from decreasing to increasing. Remember:

• A local maximum is where the function switches from increasing to decreasing, with \( f'(x) \) moving from positive to negative.
• A local minimum, on the other hand, occurs when the function changes from decreasing to increasing, with \( f'(x) \) turning from negative to positive.

Local extrema provide a snapshot of the function's behavior but doesn’t necessarily represent the highest or lowest overall points on the entire domain.

Absolute Maximum and Minimum

Finding the absolute maximum and minimum of a function on a closed interval involves evaluating the function not only at critical points but also at the endpoints of the interval. In our specific example, we evaluated the function \( f(x) = x^3 - 3x + 2 \) over the interval \((0, 3)\).To find the absolute extrema:

• Calculate the function's value at each critical point within the interval; for our exercise, this was \( x = 1 \).
• Evaluate the function at the endpoints of the interval \( x = 0 \) and \( x = 3 \).

By evaluating, we found:

• \( f(0) = 2 \)
• \( f(1) = 0 \) (a local minimum)
• \( f(3) = 20 \)

Hence, the absolute maximum value is 20 at \( x = 3 \) and the absolute minimum value is 0 at \( x = 1 \). Remember, unlike local extrema, absolute extrema are the points where a function reaches its highest and lowest values over a specific interval. They can occur at critical points or endpoints of the interval. This makes them especially significant in optimization and analysis applications.