Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

k(x) = x³ + 3x² + 3x + 1, −∞ < x ≤ 0

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Identify the local extreme values of H of Z is equal to Z 3 minus 3 Z 2 + 3 Z minus 7. Within the domain of negative infinity is less than Z and Z is less than or equal to 2. Awesome. So it appears for this particular problem we're trying to identify the local extreme values of this function of H of Z within this specific specified domain. So now that we know that we're trying to solve for the local minimum and local maximum as our final two answers we're ultimately trying to solve for, let's read off our multiple choice answers to see what our final answer pair might be, noting we'll read the local minimum value first and then our local maximum value second. So A is none and none. B is H1 is equal to -6 and none. C is none and H1 is equal to -6, and finally D is H1 is equal to 6 and none. Awesome. So first off, we need to, in order to solve for this problem, we need to take the first derivative of H of Z, so we need to solve for H of Z, where this prime symbol denotes the first derivative. So awesome. So when we take the first derivative, we will find that H of Z is going to be equal to D by DZ because we're taking the derivative with respect to Z. Of Z to the power of 3 minus 3 to 2, so minus 3 to 2, plus 3 Z minus 7, and when we perform that derivative, we will find that it's equal to 3 z squared minus 6 Z plus 3, fantastic. So now our second step is we need to find the critical points. So we need to take H of Z and set it equal to 0, so we need to take 3 Z square minus 6 Z. Plus 3 is equal to 0, and what we're ultimately trying to do is we're trying to rearrange this expression to isolate and solve for Z. So I'm going to skip a few steps, but ultimately you're going to need to simplify a little bit and rearrange by factoring, and when we factor, we will find that Z minus 1 squared is equal to 0. So that will mean that there's only 1 critical point, and that critical point will be Z is equal to 1. So once again, there's only one critical point and it's Z is equal to 1. So now our third step is we now need to perform the first derivative test. So with that in mind, we need to recall the note that we now need to use the first derivative test to determine the behavior of the function at this critical point of Z equals 1, and at the boundary, let's make a note, and at the boundary, Z equals 2. So once again, we're trying to determine the behavior of H Z. Around our points, so we're trying to find the behavior of H Z around Z is equal to 1, so around Z is equal to 1. So with that said, we need to note that for drum rolls, so for Z is less than 1, that means, for example, Z is equal to 0, we will find that H of 0 is going to be equal to 3 multiplied by 0, squared minus 6 multiplied by 0, plus 3, which will be equal to 3, which 3 is a positive value. So this will mean that the function is increasing to the left of Z is equal to 1, and for the condition where Z is greater than 1, let's say for example, when Z is equal to 2, that will mean that H of 2 is going to be equal to 3, multiplied by 2 to the power of 2, so it's going to be. So it's gonna be 3 multiplied by 2 to 2, minus 6 multiplied by 2 + 3, which will be equal to, once again, a value of 3, which is also a positive value. So that will mean that the function is increasing to the right of Z is equal to 1. So therefore, without a doubt, as we should recall and note, since H Z is positive on both sides of Z is equal to 1, that will mean that the function is increasing on both sides of Z is equal to 1, which will indicate that Z is equal to 1 is a point of inflection. Rather than a local or maximum, so it's gonna be, so it's an inflection value rather than this local maximum or minimum value. So, because it's an inflection point, that means that it's neither a local maximum or a minimum value. So with that said, our final answer without a doubt has to be multiple choice answer, letter A, which once again is local minimum none, local maximum none. And this is because it's, we're dealing with an inflection point. So let's make a note of that. So we're dealing with an inflection point. Fantastic. And that's it. We've officially solved for this problem. So thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

k(x) = x³ + 3x² + 3x + 1, −∞ < x ≤ 0

Key Concepts

Critical Points

First Derivative Test

Domain Consideration

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