Find the absolute extrema of the function on the closed interval. h(s)=(1)/(s-2),[0,1]

Name: Problem 29, Chapter 4: Calculus Early Transcendental Functions
Uploaded: 2021-08-17T17:53:54-08:00
Duration: 4 min 8 s
Channel: Diane Koenig
Description: Find the absolute extrema of the function on the closed interval. h(s)=(1)/(s-2),[0,1]

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

Key Concepts

Endpoint Evaluation

In problems involving closed intervals, evaluating the function at the endpoints is crucial because, according to the Extreme Value Theorem, absolute extrema might occur at these boundary points even when the derivative does not indicate any critical points in the interior of the interval.

Critical Points

Critical points are values in the domain of a function where the first derivative is zero or fails to exist. These points are potential candidates for local extrema, and in the context of a closed interval, must be considered along with the endpoints when searching for absolute extrema.

Extreme Value Theorem

This theorem guarantees that if a function is continuous on a closed interval, it must attain both an absolute maximum and an absolute minimum somewhere on that interval. It is foundational in analysis and optimization, ensuring that the search for extrema is meaningful and complete when working with closed and bounded intervals.

Absolute Extrema

Absolute extrema refer to the highest or lowest values a function takes on its domain. When applied to closed intervals, these extrema can occur at critical points within the interval or at the endpoints, and determining them is key in optimization problems.

Transcript

00:02 In this video, we're going to look at how to find the absolute max and absolute min of a function that's continuous on a closed interval.

00:10 Now, our function is h of s is equal to 1 over s minus 2.

00:15 And while this function is undefined at s equal 2, 2 is not in our closed interval from 0 to 1.

00:23 So h of s equaling 1 over s minus 2 is continuous on the closed interval from 0 to 1, and it's a type of question that this applies to.

00:32 Now to do this process, you start with finding the critical numbers of the function that are in the interval, evaluate the function at the critical numbers, evaluate the function at the end points of the interval, and then compare the outputs in steps two and three.

00:51 The smallest is the absolute min, the largest is the absolute max.

00:56 So when we come up looking for the critical numbers, we're going to rewrite our h of s to be the quantity s minus 2 to the negative 1 power to get ready to differentiate it differentiating it as h prime of s is equal to and we're going to use the chain rule so negative 1 times s minus 2 now to the negative 2 power and then times the derivative of s minus 2 which is 1 so my h prime of s is equal to negative 1 over s minus 2 quantity squared now that that is undefined at 2, but 2 is not in the domain of the original function, so that's not a critical value, nor is it in our interval that we're interested in.

01:43 When is it 0? well, the output of the derivative would be 0 when i have 0 is equal to negative 1 over s minus 2 quantity squared.

01:54 But that is never 0 because a fraction of 0 only when the numerator is 0 and the denominator is not...

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