Problem 9 - Calculus - AP Edition

Question

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$f(x)=(x-2)^{3} ;[1,4]$$

Step-by-Step Solution

Answer

Max: 8 at x=4; Min: -1 at x=1.

Step 1: Find the derivative of f

First, we need to find the derivative of the function. The given function is \( f(x) = (x-2)^3 \). To find the derivative, use the power rule: \( f'(x) = 3(x-2)^2 \).

Step 2: Solve for critical points

The critical points occur where the derivative is zero or undefined. Since \( f'(x) = 3(x-2)^2 \), set \( 3(x-2)^2 = 0 \). Solving this, we get \( x = 2 \), which is a critical point within the interval \([1,4]\).

Step 3: Evaluate f at critical points and endpoints

Now, evaluate the function \( f(x) \) at the critical point \( x = 2 \) and at the endpoints \( x = 1 \) and \( x = 4 \):- \( f(1) = (1-2)^3 = -1 \)- \( f(2) = (2-2)^3 = 0 \)- \( f(4) = (4-2)^3 = 8 \)

Step 4: Determine absolute maximum and minimum

Compare the values obtained: \( f(1) = -1 \), \( f(2) = 0 \), and \( f(4) = 8 \). The greatest value is \( 8 \), and the smallest value is \( -1 \). Hence, the absolute maximum value is \( 8 \) at \( x = 4 \) and the absolute minimum value is \(-1\) at \( x = 1 \).

Key Concepts

Absolute Maximum and Minimum

When you are asked to find the absolute maximum and minimum values of a function on a given interval, you are essentially looking for the highest and lowest points on the graph of the function over that interval. These points are critical because they highlight the function's top and bottom values within the specified range. Here's how you generally find them:

• Identify all critical points by taking the derivative of the function and setting it equal to zero.
• Check the endpoints of the interval because maximum and minimum values can also occur there.
• Evaluate the function at these points to determine the largest and smallest values.

In our original exercise, we compared the function values at the critical point and endpoints to find that the absolute maximum was at point 4, where the value was 8, and the absolute minimum was at point 1, where the value was -1.

Derivatives

The derivative of a function is a fundamental concept in calculus. It represents how a function changes as its input changes. More practically, the derivative measures the slope of the tangent line at any point on the graph. Derivatives are crucial for several reasons:

• They help us find critical points, where the slope is zero.
• They provide insights into the behavior of the function, such as increasing or decreasing trends.

In our exercise, we used the power rule to find the derivative of the function \(f(x) = (x-2)^3\). That gave us \(f'(x) = 3(x-2)^2\), which is crucial for identifying critical points and understanding the function's behavior on the interval.

Critical Points

Critical points are where the derivative of a function is zero or undefined. These points are significant because they indicate where the function might have a maximum or minimum value, or where it changes direction. To find critical points:

• Calculate the derivative of the function.
• Set the derivative equal to zero and solve for the variable.
• Check if these points are within the given interval.

In the exercise, we found that the critical point was \(x = 2\) by setting \(f'(x) = 3(x-2)^2\) to zero. Since this point lies within the interval [1, 4], it was included in our evaluation of the function's maximum and minimum values.

Closed Interval

A closed interval in calculus is marked by two endpoints where both end values are included in the evaluation. We denote it using square brackets, like \([a, b]\).Working within a closed interval has specific implications:

• Both endpoints must be considered when finding absolute maximum and minimum values.
• The function is evaluated not only at the critical points but also at these boundaries.

In our task, the function \(f(x) = (x-2)^3\) was evaluated on the closed interval [1, 4]. This means we had to calculate the function values at both endpoints, x = 1 and x = 4, in addition to any critical points found within the interval.