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Absolute extrema are the highest and lowest points on a function within a specific interval. When dealing with functions' extrema, there are often two types to consider: absolute maximum and minimum.
For a given function, like in our exercise with \( g(x) = \sec(x) \), the goal is to find these absolute extrema on the specified interval. The interval provided is \(-\frac{\pi}{3} \leq x \leq \frac{\pi}{6}\). To locate the absolute extrema, we need to assess both critical points within the interval and the function's values at the endpoints.
Since no critical points occur due to zero derivatives within the interval (as shown in the solution), we rely on evaluating the function at \( x = -\frac{\pi}{3} \) and \( x = \frac{\pi}{6} \).

• Absolute Maximum: Occurs when the function reaches its highest value. For our function, this is \( 2 \) at \( x = -\frac{\pi}{3} \).
• Absolute Minimum: This is the lowest value the function achieves within the interval. For \( g(x) \), this is \( \frac{2\sqrt{3}}{3} \) at \( x = \frac{\pi}{6} \).

After identifying these points, they are often marked on the graph to visually represent the extrema's location.

The secant function, denoted as \( \sec(x) \), is directly related to the cosine function. It is actually the reciprocal, meaning \( \sec(x) = \frac{1}{\cos(x)} \). Understanding the secant function requires knowing some features of cosine, like where it is zero or undefined.

• Continuity: The secant function is continuous everywhere its friend, the cosine function, is not zero. This is crucial as a discontinuity affects where we can find maxima and minima.
• Graph Behavior: The graph of \( \sec(x) \) tends to have vertical asymptotes where \( \cos(x) = 0 \). In our exercise interval, \( \cos(x) \) remains non-zero, ensuring continuity.

The secant function is known for its steep rising and falling, significantly affecting the investigation of its extrema, especially on intervals that do not include its undefined points.

Critical points of a function are locations where the derivative either equals zero or does not exist. These points are vital for determining possible extrema since they indicate a potential shift in the function's direction.
To find the derivative of \( g(x) = \sec(x) \), we use the formula:

• \( g'(x) = \sec(x) \tan(x) = \frac{\sin(x)}{\cos^2(x)} \).

Critical points occur where \( g'(x) = 0 \) or is undefined.
In our interval \( -\frac{\pi}{3} \leq x \leq \frac{\pi}{6} \), the derivative does not equal zero because \( \sin(x) eq 0 \). The derivative is also defined throughout the interval, leaving no critical points to consider.
With no critical points emerging in the open interval, only the values at the endpoints matter for finding absolute extrema in this case.