Problem

(a) Use a graph to estimate the absolute maximum …

16:23

Problem 65 Medium Difficulty

(a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places.
(b) Use calculus to find the exact maximum and minimum values.

$ f(x) = x^5 - x^3 + 2 $, $ -1 \leqslant x \leqslant 1 $

Answer

(b) $f(x)=x^{5}-x^{3}+2 \Rightarrow f^{\prime}(x)=5 x^{4}-3 x^{2}=x^{2}\left(5 x^{2}-3\right) \cdot$ So $f^{\prime}(x)=0 \Rightarrow x=0, \pm \sqrt{\frac{3}{5}}$
$\begin{aligned} f(-\sqrt{\frac{3}{3}}) &=(-\sqrt{\frac{3}{5}})^{5}-(-\sqrt{\frac{3}{5}})^{3}+2=-\left(\frac{3}{5}\right)^{2} \sqrt{\frac{3}{3}}+\frac{3}{5} \sqrt{\frac{3}{3}}+2 \\ &=\left(\frac{3}{5}-\frac{9}{25}\right) \sqrt{\frac{3}{5}}+2=\frac{6}{25} \sqrt{\frac{3}{5}}+2 \quad(\text { maximum }) \\ \text { and similarly, } f(\sqrt{\frac{3}{5}})=-\frac{6}{25} \sqrt{\frac{3}{5}}+2( \text { minimum) } \end{aligned}$

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Video Transcript

all right, toe. Let's go ahead and answer this question. So the question is we have Quint IQ X to the fifth minus X cubed plus two. And then it is limited between the region. Negative 1 to 1. The question is, graph it and approximate the maximum and the minimum value. And the other part is use calculus and what you've learned to precisely locate the exact value off the maximum and the minimum so as usual, because the region is limited, we need to keep in mind the extreme value theory, Um, saying that if a maximum or minimum occurs, we have to check the minimum and the maximum value, not the minimum and maximum. Excuse me, the endpoints, negative one and one. So we're going to check what f of one in efforts Negative one are going to be okay. If you plug in these numbers, you can see that the negative one minus negative one cancels each other out. So you will get a two. And the same thing happens when you plug in that one. Okay, one. My next one is zero so f of 1.5 of negative one are both to Okay, so it's neither a maximum or minimum right away, because we know that this is not a straight line. Okay, so we're going to now take a look at the critical points. If the derivative is equal to zero, you know that those points are likely to be a maximum and the minimum. And this is a polynomial with, um, odd power. So we know that we're not gonna be dealing with a situation where, um the maximum and the minimum bounces off the the X axis and such. So it's highly likely, but it's not a saddle, but we gotta check it anyways. Okay. So, first, let's take a look at the graph I already grafted for you. X cubed minus x x to the fifth minus X cubed. Plus two is given right there. And to make it look a little bit better, X equals two negative. 1 to 1. I do this negative one positive one. I know that it is this region to that region that I'm focusing on. So the maximum value happen some around here. Let's see if I can help. I can. Maybe around there. So it seems like 2.2186 or so is going to be my maximum. The minimum here is 1.814 So let's see if we can get those two numbers. And as you can see, it's indicating that it's in metric. So possibly the solution is going to be a very, um, plus or minus something kind of number where if we plugged it in, we get these heights. Okay, so that's one thing that we could observe. All right, so let's go ahead and try to do this, um, analytically. So we're going to find the derivative first. This one shouldn't be that big of a deal by now. Have prime off X is five x to the fourth minus three X square to just become zero. We're gonna let this equal to zero. So X equals to zero is, of course, going to be a saddle. As we saw over here, it decreases but decreases again. So it's not a critical point. So it actually did show up my dad before the explanation that I made, um so I factored out on X squared and did that calculation. What's the other part? It is going to be plus ra minus the square root off 3/5. So those are the critical points. If you take a look, you can see that it's going to give you a maximum here, minimum there. If we started off with the analytic process, I could have probably find the second found the second derivative. Plug these numbers to see if it's going to be positive or negative, because if you get a positive, sick and derivative, we know that we will end up having a minimum because it's can't give up and vice versa. If it's can't keep down, it's a maximum. Okay, moving on. So we don't want to end here. We want to actually find the maximum and minimum, not where it's located at. So we're going to plug in these two numbers into F. So I am only going to do the case where it's the square root of 3/5. Just to show you the trick off what we can do to make the process a little bit simpler. Okay, so if we plug this number in, we can do the calculation, of course, but to make it simpler, you can see that I can factor out on X cubed from both of these expressions. So X cubed times X squared, minus one plus two. Why did I do that? Well, X squared is a nice number. It's 3/5. Right? So this expression right here inside the parentheses, it's 3/5 minus one, which is negative to fifth X. Cubed is also not that difficult to calculate because I know that X squared is to fifth. I meant 3/5. So this is 3/5 square root of 35 So if I go through the calculation, it is going to simplify to negative 6/25 the square root of 3/5. I am not going to rationalized the denominator, because to me, this honestly looks simple enough. Plus two. And this is the expression that you will get over here. Okay. If you plug in f off negative, you, uh, the skirt off 3/5. It has to go all the way there. You will get the positive version of this expression plus two. Okay. And then these air going to be the precise location off the maximum and the minimum, and then you can see that if you plug it in a calculator. It will actually be these two numbers that you see right there. Okay. And this is how you solve this problem?

Yuki H.

University of California, Berkeley

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 1

Maximum and Minimum Values

Problem