Find the absolute extrema of the function on the closed interval. Use a graphing utility to veri

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

Key Concepts

Absolute Extrema

Absolute extrema are the highest and lowest values that a function attains on a given domain. For functions defined on a closed interval, these extrema are identified by comparing function values at critical points—where the derivative is zero or undefined—and at the endpoints of the interval.

Extreme Value Theorem

The Extreme Value Theorem states that if a function is continuous on a closed interval, then it must have both an absolute maximum and an absolute minimum on that interval. This theorem is foundational for problems involving optimization and ensures that evaluating the function at critical points and endpoints will capture the absolute extrema.

Critical Points

Critical points are the values in the domain of a function where the derivative is zero or does not exist. They are important because they represent potential locations for local or absolute maxima and minima within the interval, and must be examined along with the endpoints when determining absolute extrema.

Graphing Utility

Graphing utilities are technological tools used to visually represent functions. They can help verify analytical results by providing a graphical depiction of the function's behavior, which can confirm the locations and values of the absolute extrema found through calculation.

Transcript

00:01 Okay, so let's say given this function, i want to find the absolute extrema on a close interval.

00:07 Absolute extrema, i'm going to have one absolute men and one absolute max, versus the relative extrema are just kind of hills and valleys in between or on my entire function.

00:18 So i could have multiple or no answers for relative extrema, whereas absolute minimum max, i will have at least one answer.

00:25 And i can include endpoints as a possible min or a max.

00:30 But i can also really include critical points.

00:33 So the first thing i'm going to do is i'm going to algebraically calculate my critical points.

00:37 See if i have any.

00:39 This happens by taking the first derivative and setting equal to zero.

00:42 So first derivative, this is just a polynomial.

00:44 So power rule, three times two thirds is going to be six thirds.

00:48 There's going to be two.

00:49 X the two thirds minus one.

00:51 Two thirds minus one, the same thing as negative one third, minus the derivative of negative two x is just negative two.

01:00 So, simplifying this a little more, this could look like 2 over x to the 1 3rd minus 2 to make this exponent positive.

01:13 And i'm going to set this first derivative equal to 0.

01:15 See if i have any values that could plug in.

01:20 I'll get 2 equals 2 over x the 3rd to the 1 3rd.

01:25 Okay.

01:27 What can i do to get x by itself? well, i'm going to multiply by x to 1 3rd to both sides.

01:35 Continuing to simplify 2 x the 1 3rd equals 2, divide by 2 to both sides.

01:42 So i'm going to get x the 1 3rd, 2 divided by 2 equals 1, right, anything divided by itself is 1.

01:48 How do i get rid of this 3? well, i'm going to raise it to the third root.

01:53 Kind of cancel that.

01:56 And since this is dividing by 3, i know this is an odd function, so it's 1 to 1.

02:02 1 to the 3 is just 1.

02:03 So i could have a critical point at x.

02:06 X equals 1.

02:08 Critical points, i could set my first row of equal to zero, or i can look where my function has a denominator of zero, or is undefined.

02:17 So x of the one -third equals zero...

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