In the previous article, we learned about finding critical points using the first derivative test. Now, let's dive deeper into determining whether these critical points are maximum or minimum values.

To do this, we need to analyze the behavior of the function around each critical point. One way to do it is by using the second derivative test. If the second derivative at a critical point is positive, then it is a local minimum. Conversely, if the second derivative is negative, it is a local maximum. If the second derivative is zero, the test is inconclusive.

Let's consider the function f(x) = x^3 - 6x^2 + 9x + 1 as an example. We previously found the critical points to be x = 1 and x = 3. To determine their nature, we calculate the second derivative, f''(x) = 6x - 12. Substituting the critical points, f''(1) = -6 and f''(3) = 6. As f''(1) < 0, x = 1 represents a local maximum, while f''(3) > 0, indicating x = 3 as a local minimum.

Remember, this test is only applicable to the interval where the critical point exists. Additionally, it's crucial to consider the behavior of the function near the endpoints of the interval to determine if there are any absolute maximum or minimum values.

Keep practicing and exploring different functions to strengthen your understanding. You're making great progress!