91Ó°ÊÓ

To understand calculus optimization problems, critical points are essential and serve as a starting point for determining other factors like maximums and minimums. Critical points occur at values of x where the derivative of a function equals zero or where the derivative does not exist. However, in most problems, we look for values where the derivative becomes zero as a key indicator of potential peaks or troughs in the graph of the function.
For the function \( f(x) = (x^2 - 1)^3 \), finding critical points involves:

• Taking its derivative, which involves applying the chain rule effectively.
• Setting the derivative equal to zero to solve for x.

When we do this for \( 6x(x^2 - 1)^2 = 0 \), the key values of x to check are: -1, 0, and 1. These points help us move forward in determining the behavior of the function on a given interval.

Once critical points are located, we can identify the absolute maximum of a function on a given interval. The absolute maximum value refers to the highest point a function reaches over an interval. After obtaining the critical points and endpoints of the interval, evaluate the function at these points.
In the exercise, the critical points are x = -1, 0, and 1, with the interval being [-1, 2]. Additionally, we check the endpoint x = 2. Evaluating the function, we find:

• At x = 2, \( f(2) = 27 \)
• Other points have lesser values compared to this.

Thus, the absolute maximum is 27 at x = 2. This example reflects how checking critical points helps determine the highest value of a function in an interval efficiently.

The absolute minimum of a function is the smallest value it attains on an interval. This concept is crucial in optimization problems to identify how low a function can go on a given range. Just like finding the maximum, involve evaluating the function at critical points and endpoints.
In our example, we found:

• At x = 0, \(f(0) = -1\)
• This is smaller than values at any other examined points like -1, 1, and 2.

Hence, the absolute minimum is -1 at x = 0. Such evaluations reinforce the importance of assessing both critical points and endpoints effectively when determining minimum values in optimization problems.

Interval evaluation is a method that helps determine both absolute maximum and minimum values by examining function behavior within a specific range, or interval. This involves not only considering critical points but also examining endpoints, giving a comprehensive view of the function’s behavior over that interval.
For \([−1, 2]\), the critical strategy is:

• Check and calculate function at endpoints and each critical point within the interval.
• Compare these values systematically.

In our problem, the interval evaluation approach confirms absolute maximum at x = 2 and minimum at x = 0 by evaluating among x = -1, 0, 1, and 2. This demonstrates the power and efficiency of interval evaluation in finding extrema within specific bounds.