Find the absolute extrema of the function on the closed interval. g(x)=secx,[-(π)/(6), (π)/(3)]

Name: Problem 36, Chapter 4: Calculus Early Transcendental Functions
Uploaded: 2021-08-18T16:27:10-08:00
Duration: 7 min 13 s
Channel: Diane Koenig
Description: Find the absolute extrema of the function on the closed interval. g(x)=secx,[-(π)/(6), (π)/(3)]

Find the absolute extrema of the function on the closed...

Key Concepts

Differentiation of Trigonometric Functions

Differentiation of trigonometric functions involves applying rules such as the chain rule and the derivatives of basic trigonometric functions. For example, knowing the derivative of the secant function is essential when finding where the function's rate of change is zero or undefined, which helps in identifying critical points.

Trigonometric Functions

Trigonometric functions, like the secant function, are functions related to angles and periodic phenomena. Understanding the properties and behavior of these functions, including their domains, ranges, and periodicity, is crucial when analyzing and solving optimization problems involving them.

Critical Points

Critical points are the values of the independent variable where the derivative of the function is zero or undefined. These points are significant in determining potential local extrema because they indicate where the function's slope changes or where the function might have a cusp or vertical tangent.

Absolute Extrema

Absolute extrema refer to the highest and lowest values that a function attains on a given domain. In the context of optimization, finding the absolute maximum and minimum involves comparing values at critical points and endpoints of the interval, ensuring that the global peak and valley are correctly identified.

Closed Interval Method

When optimizing functions over a closed interval, the Closed Interval Method is used. This involves evaluating the function at both the critical points within the interval and at the endpoints of the interval, as the absolute extrema are guaranteed to exist on a continuous function over a closed interval, according to the Extreme Value Theorem.

Transcript

00:03 In this video, we are going to look at how to find the absolute min and the absolute max of a continuous function on a closed interval.

00:11 So here we want to find the absolute min and absolute max of g of x is equal to secant of x on the closed interval from negative pi over 6 to pi over 3.

00:21 Now, we know that secant of x has infinitely many different points where it is undefined, and it is undefined where the cosine of x is equal to zero.

00:33 So at like negative pi over two, pi over two, three pi over two, et cetera.

00:38 But our interval, our closed interval, is from negative pi over six to pi over three.

00:44 So there are none of those values where the vertical asymptotes for secant of x occur, where we're focusing our attention.

00:52 So our g of x is equal to secan of x is continuous on our closed interval from negative pi over six to pi over three.

00:59 Now to do this, we are going to do four steps.

01:01 First, we're going to find the critical values of our function g of x is equal to the secant of x on our closed interval from negative pi over six to pi over three.

01:14 So only reporting the critical values of that function that are in that interval.

01:19 Then we are going to evaluate the function at the critical values found in step one.

01:26 The third step, we are going to evaluate the function at the endpoints of the closed interval.

01:31 And then in the fourth step, we're going to compare the function values from steps two and three.

01:37 The smallest is the absolute min.

01:39 The largest is the absolute max.

01:43 So coming back up here to step one, to find the critical values of a function, you take the first derivative and find the values in the domain of the function for which the first derivative is zero or undefined.

01:57 And specifically, we want them in the closed interval from negative pi over six.

02:02 To pi over three.

02:04 So taking the derivative g prime of x, well, the derivative of secant of x is secant x tangent x.

02:14 Now secan x tangent x is also undefined at all of those like negative pi over two, pi over two, etc.

02:21 But those values aren't in the domain of the original function.

02:24 So they aren't critical values, nor are they in our interval.

02:28 Even if we had something, we wouldn't be able to use it, it's out of the interval.

02:34 How about when g prime of x is zero? what values of x make zero come out for the derivative? so zero is equal to sicken of x times tangent of x.

02:44 Well, the product of two variable expressions are zero if one or the other is zero.

02:49 So secant of x equals zero or tangent of x equals zero.

02:55 And i can do that right away because i did have a zero on this left -hand side.

03:00 Well, there aren't any angles for which the secant of the angle is zero...

Please give Ace some feedback