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The concept of a derivative is fundamental in calculus. It represents the rate at which a function changes at any given point. In simpler terms, it's the function's slope at a particular point. The derivative tells us how the function's output changes as its input changes.
To find the derivative of a function, such as \[g(x) = (x - 3)^{\frac{5}{3}}(x + 2)\], we typically use rules like the Power Rule or the Product Rule, which are designed to make the process easier. Think of the derivative as a mathematical tool that helps us analyze the behavior of functions. It's crucial in understanding how changes in one variable affect another.
In this problem, the derivative \( g'(x) \) was found using the Product Rule, which simplifies the differentiation process when dealing with products of functions. Knowing how to calculate and apply derivatives is essential in discovering more about the function's local behavior, such as identifying critical points where maximums or minimums may occur.

The Product Rule is a valuable technique in calculus for finding the derivative of a product of two functions. It states that if you have a function \( g(x) \) expressed as a product of two functions, each with their own variable dependence, you can find the derivative by differentiating each function separately.
To apply the Product Rule:

• Differentiate the first function while keeping the second function constant.
• Switch roles by keeping the first function constant while differentiating the second function.
• Add the results together.

In our exercise, we have \( g(x) = (x - 3)^{\frac{5}{3}}(x + 2) \). Using the Product Rule: \[ g'(x) = \frac{d}{dx}[(x - 3)^{\frac{5}{3}}] \cdot (x + 2) + (x - 3)^{\frac{5}{3}} \cdot \frac{d}{dx}[(x + 2)] \].
By applying these steps, you can accurately compute the derivative, allowing for further analysis, like finding critical points. Understanding and applying the Product Rule is instrumental in solving many problems involving derivatives of products.

Interval analysis allows us to take a broader view of the behavior of functions within a specific range of x-values. It involves examining the function on an interval, which helps in understanding how the function behaves between specific values.
In the presented problem, interval analysis is applied to find critical points and determine whether there are any maximum or minimum values within the interval In this problem, the provided interval is \([-4, 4]\), and we aim to find critical points where the derivative equals zero and check the function's behavior at these points. When analyzing our function over this interval, we consider these aspects:

• The critical point where the derivative equals zero, providing potential local extrema.
• The endpoints of the interval to ensure potential maximum or minimum values aren't missing.

This method gives us complete insight into how the function behaves over our range of interest. It helps us find where the function reaches its highest and lowest values, both crucial for comprehensive function analysis.

Finding the maximum and minimum values of a function involves pinpointing where the function achieves its highest and lowest outputs. These values are significant as they provide crucial information about the function's range and behavior.
The process typically involves examining critical points—where the function's derivative is zero or undefined—and also evaluating the function at the boundaries or endpoints of our interval. For our function \( g(x) = (x-3)^{\frac{5}{3}}(x+2) \) on \([-4, 4]\):

• We found the critical point at \( x=3 \) by setting the derivative \( g'(x) = 0 \).
• The function's value is calculated at endpoints \( x=-4 \) and \( x=4 \), as well as at the critical point.

Through this, we confirmed that an absolute minimum occurs at \( x=3 \) with \( g(3) = 0 \), and an absolute maximum occurs at \( x=4 \) with \( g(4) = 6 \).
The identification of these values teaches us not only about the behavior of this specific function but showcases methods applicable to a wide range of calculus problems.