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Critical points of a function are crucial for identifying where a function reaches its highest or lowest values—known as extrema. To find them, we look where the derivative of the function equals zero or is undefined. These are the points where the slope of the tangent is zero, which could indicate a peak, a valley, or a flat section.
For example, in the function given in the exercise, we found these critical points by solving the equation where the derivative equals zero. After factoring and solving the equation, we determined the critical points to be at \(x = 0\), \(x = 2\), and \(x = -1\). Recognizing these critical points lays the groundwork for determining the absolute maximum and minimum of a function.

The derivative of a function expresses how the function's output value changes as the input changes. It's a fundamental tool in calculus that's commonly used to find slopes and rates of change.
To find the derivative of a function like \(f(x) = 2x^4 - \frac{8}{3}x^3 - 8x^2 + 12\), we differentiate each term with respect to \(x\). This process transforms each term using rules of differentiation, such as the sum and difference rules, constants rule, and most importantly, the power rule.
Derivatives are key to identifying critical points because they highlight where changes occur, specifically where slopes are zero, allowing us to determine where a function's relative maxima and minima may occur.

The power rule is a straightforward guideline for differentiating expressions of the form \(x^n\). According to the power rule, the derivative of \(x^n\) is \(nx^{n-1}\).
This rule simplifies the differentiation process greatly and is especially useful for polynomial functions. For instance, in the exercise function, \(2x^4\) becomes \(8x^3\) when differentiated, using this rule. Similarly, each term of the polynomial is adjusted accordingly by subtracting one from the exponent and multiplying the term by the original exponent.
Understanding and applying the power rule allows us to effectively manage polynomial functions and is critical for reaching solutions swiftly and accurately.

Endpoints refer to the boundary values within a given interval of a function. When finding absolute extrema (maximum or minimum values) of a function over a closed interval, we must examine these endpoints directly, as they are potential locations of extrema.
In the exercise, our interval is \([-2, 3]\). This means we must check the function value not just at critical points but also at these two endpoints: \(x = -2\) and \(x = 3\). Both endpoints and critical points must be evaluated to determine where the absolute maximum or minimum function values lie within the interval.
By calculating the function's value at \(x = -2\) and \(x = 3\), alongside the critical points, we confirmed that the absolute maximum was at \(x = 0\) and the absolute minimum was at the endpoint \(x = 3\). This comprehensive evaluation ensures the overall solution covers every possible contender for these extreme values.