Find the absolute extrema of the function on the closed interval. g(x)=(lnx)/(x),[1,4]

Name: Problem 40, Chapter 4: Calculus Early Transcendental Functions
Uploaded: 2021-08-18T16:31:27-08:00
Duration: 8 min 18 s
Channel: Diane Koenig
Description: Find the absolute extrema of the function on the closed interval. g(x)=(lnx)/(x),[1,4]

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

Key Concepts

Differentiation

Differentiation is the process of finding the derivative of a function, which provides information about the rate of change of the function. By setting the derivative equal to zero, one can solve for the critical points that may correspond to local or absolute extrema.

Critical Points

Critical points are points within the domain of a function where the derivative is zero or undefined. They are potential candidates for local extrema, and within a closed interval, they must be evaluated along with the endpoints to find the absolute extrema.

Closed-Interval Method

The closed-interval method is a systematic procedure for finding absolute extrema on a closed interval. It involves finding the critical points in the interior of the interval, evaluating the function at these points as well as at the endpoints, and then comparing these values to determine the absolute maximum and minimum.

Extreme Value Theorem

The Extreme Value Theorem states that if a function is continuous on a closed interval, then it must attain both an absolute maximum and an absolute minimum on that interval. This theorem provides the foundational guarantee needed to search for these extrema.

Absolute Extrema

Absolute extrema refer to the global maximum and minimum values that a function attains over a given domain. These are the highest and lowest points on the graph of the function when considering all values in the specified interval.

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 Now, our function here is g of x is equal to the natural log of x over x.

00:17 And the interval we're talking about is the closed interval from 1 to 4.

00:22 Now, the natural log of x over x is only defined for x values that are strictly greater than zero.

00:31 And so we wanted to just make sure it was defined on this closed interval.

00:35 And then there aren't any values of x that would make the denominator zero in that domain, nor in our closed interval that we are even focusing more of our attention on.

00:47 So our g of x for this problem is continuous on that closed interval.

00:52 Now to do this, it's a four -step process.

00:54 The first step we're going to find the critical values of our function on our closed interval.

00:59 So we want to find the, once we find the critical values, we want to make sure they're in the closed interval.

01:04 Next, we are going to evaluate our function at the critical values found in step one.

01:11 Then we are going to evaluate our function at the end point of the closed interval.

01:15 And to be honest with you, this is the step that a lot of times students miss.

01:20 So you want to make sure you don't forget step three in this type of a problem.

01:26 And then finally, you compare the function values from step.

01:29 2 and 3.

01:30 The smallest function value is the absolute min and the x value is where it occurs.

01:36 And then the largest function value is the absolute max.

01:40 So let's get started with this one.

01:42 Define the critical values.

01:44 We want the values in the domain of the function for which the first derivative is zero or undefined.

01:51 And specifically here we want to be in the closed interval from 1 to 4.

01:54 Now our derivative g prime of x we can find by either keeping it written the way it is, or we could rewrite our function g of x as bring that division by x up as a factor of x to the negative one power.

02:13 So x to the negative one times the natural log of x when i'm looking at it.

02:17 And then i can find my derivative using the product rule.

02:19 So g prime of x is the derivative of the first factor, negative 1, x to the negative 2 times the natural log of x, plus the first factor x to the negative 1 times the derivative of the second factor.

02:35 And the derivative of the natural log of x is 1 over x.

02:39 So i have my g prime of x is equal to, and remember this is multiplication.

02:46 So when i think of a fraction bar being here, i'm going to have that x to the negative 2, i'll go down.

02:51 As an x squared in the denominator.

02:54 So i have negative the natural log of x over x squared.

03:00 And then plus i have one over x times one over x.

03:04 So that's one over x square...