In Exercises 21-36, find the absolute maximum and minimum...

In Exercises $21-36,$ find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ g(x)=\sec x, \quad-\frac{\pi}{3} \leq x \leq \frac{\pi}{6} $$

In Exercises $21-36,$ find the absolute maximum and minimum values
of each function on the given interval. Then graph the function. Identify the points
on the graph where the absolute extrema occur, and
include their coordinates.
$$
g(x)=\sec x, \quad-\frac{\pi}{3} \leq x \leq \frac{\pi}{6}
$$

Key Concepts

Absolute Extreme Values

Absolute extreme values are the highest and lowest values that a function attains over a given interval. These values are found by evaluating the function at all critical points and at the boundaries of the interval, ensuring that no potential maximum or minimum is overlooked.

Critical Points

Critical points of a function are those points within its domain where the derivative is either zero or undefined. These points are significant because they indicate where local maxima or minima might occur, and thus must be considered when determining absolute extrema.

Extreme Value Theorem

The Extreme Value Theorem guarantees that if a function is continuous over a closed interval, then it will attain both an absolute maximum and an absolute minimum on that interval. This theorem forms the basis for the process of checking both the interior critical points and the endpoints of the interval.

Function Graphing with Extrema

Graphing a function along with its extrema involves plotting the function over the specified interval and clearly marking the locations where the absolute maximum and minimum occur. Visual representation of these extrema can provide deeper insight into the function's overall behavior and its key characteristics.

Transcript

00:01 All right, here we have g of x is second x defined on negative pi over 3, 2, pi over 6, and this function is continuous.

00:17 Okay, so between, so specifically between negative pi over 2 and pi over 2, it has kind of a parabolic shape.

00:24 It's going to be continuous on, you know, some closed interval contained in between negative pi for 2 and pi over 2.

00:31 So it has an absolute max in men, and it will occur at either a critical point or at one of these endpoints.

00:41 So the derivative is sickenx tanx.

00:45 Well, the only point where sicken x tan x is going to be 0 is where tan x is 0, and that occurs at 0, and that's the only point in this interval.

00:55 So g prime of 0 is 0, so x equals 0 is a critical point.

01:01 Okay, and so let's test the value at our endpoints, pye over three.

01:11 So sikant of negative pi over three, that's the same thing as cosine of pi over three.

01:17 Sorry, one over cosine of pi over three, because cosines even.

01:21 So cosine of pi over three is one half, so this should be two.

01:26 And then g of zero, critical point, is one.

01:30 Seekin of zero is one, because cosine of zero is one.

01:34 And then g of pi over six.

01:38 Well, cosine of pi over six is root 3 over 2, so secant will be the reciprocal if that's a 2 over root 3.

01:50 And 2 over root 3 is bigger than 1, but it's going to be less than 2...