Find the absolute extrema if they exist, as well as all values of x where they occur, for each f

Find the absolute extrema if they exist, as well as all...

Key Concepts

Endpoint Evaluation

Endpoint Evaluation is a necessary step when using the Closed Interval Method. Since the Extreme Value Theorem guarantees that the maximum and minimum occur on the closed interval, analyzing the function’s behavior specifically at the endpoints, in addition to the critical points within the interval, is essential. This evaluation ensures that no potential extrema are overlooked.

Derivative Analysis

Derivative Analysis involves computing the derivative of the function to find critical points. By setting the derivative equal to zero and solving for x, or by identifying where the derivative is not defined, one can find candidates for local extrema. This analysis is a key step in determining where the function might have relative or absolute extrema on a given interval.

Closed Interval Method

The Closed Interval Method is a fundamental concept in calculus used to find absolute extrema of continuous functions on closed intervals. According to the Extreme Value Theorem, a continuous function on a closed interval will always attain both an absolute maximum and an absolute minimum at some points within the interval.

Critical Points and Differentiability

Critical points are those points in the domain of a function where the derivative is either zero or undefined. They are crucial in the search for local extrema because these are the potential candidates where the function's behavior changes. In problems involving functions with fractional exponents or other operations that may cause non-differentiability, identifying and testing these points is essential.

Transcript

00:01 Okay, for 6 .1 number 24, we are again asked to find any absolute extrema for this function f of x that we're given.

00:07 When you're finding extrema, the first thing you need to do is find the derivative.

00:13 So we're going to go ahead and find f prime of x.

00:17 This is going to be equal to 1, the derivative of x, plus you bring down the 2 thirds times 3 x to the negative 1 third.

00:29 And we get the negative 1 3 third because in our original function we're subtracting 3 over 3, which is like subtracting 1.

00:37 So this is what we get.

00:38 You'll notice that the 3s here cancel, which leaves us with a final derivative of 1 plus 2x to the negative 1 third.

00:50 Okay.

00:51 We are going to rewrite this as 1 plus 2 over the cubed root of x.

00:59 And that just gives a little bit of meeting to that negative exponent.

01:02 So with a negative exponent, we're just bringing it down to the bottom.

01:06 So now you have to ask yourself, look, we already know we have to evaluate this function at negative 10 and 1, which are the end points.

01:13 So we already know we have this critical point and this critical point that we need to test.

01:20 But we need to find the other two critical points.

01:22 So we need to check when the derivative is undefined.

01:35 Okay, so we need to figure out when it equals zero.

01:40 So let's just act.

01:41 Let's not say, well, check when is undefined, and we need to set f prime of x equal to zero and solve for x.

01:53 So there's two other points we need to check.

01:55 And i'm going to put a blue star and we're going to do the work in blue.

01:59 So when is f prime of x undefined? well, if we remember f prime of x was equal to.

02:08 1 plus 2 over the cubed root of x and we're going to put that all over a common denominator 3 square root of x plus 2 just rewriting with a common denominator so this is our final derivatives that we're looking at we'll notice it's undefined when x equals 0 f prime of x undefined undefined is that a problem yes, because this number is in the range that we're looking at.

02:47 So we also need to add zero to our list of critical numbers...

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