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Find the absolute maximum and minimum, if either exists, for each function on the indicated intervals.$f(x)=x^{4}-4 x^{3}+5$(A) [-1,2](B) [0,4](C) [-1,1]...

Question

Find the absolute maximum and minimum, if either exists, for each function on the indicated intervals.$f(x)=x^{4}-4 x^{3}+5$(A) [-1,2](B) [0,4](C) [-1,1]

Find the absolute maximum and minimum, if either exists, for each function on the indicated intervals. $f(x)=x^{4}-4 x^{3}+5$ (A) [-1,2] (B) [0,4] (C) [-1,1]



Answers

Find the absolute maximum and minimum, if either exists, for each function on the indicated intervals. $f(x)=x^{4}-4 x^{3}+5$ (A) [-1,2] (B) [0,4] (C) [-1,1]

We are going to find the absolute minimum an absolute maximum values of the function F of X equal 5 54 X minus two. X cube on the interval 04. So the first thing we're gonna say is that these functions phenomenal Yuri Degree three. So it's continuous everywhere. In particular in this interval 04. And being this function continues on a closed interval, we know it attains its extreme values. That is the absolute maximum and absolute minimum values of this function on this interval exists and they are images of some points in this interval. And we know more than that. We know that the point where the extreme values can be attained are either the end points of the interval or critical points of critical numbers of death. So we get a fine All the critical numbers of F which lie in side the interval 04. And with that with those critical numbers and the endpoints we evaluated at all those points. And the largest value would be the absolute maximum value of the function of our dangerous. And the smallest of the values of those values would be the absolute minimum of the function on the glass into. So that's all it is. It's happening here. So we find the derivative of this function In order to find a critical numbers and this is 54 -6 x square. This function exists for every X. In and the Intercultural four. No, it means that the only critical numbers for dysfunction are those values of experts. The first relativity about zero. Yeah. So when we solve this equation we are finding all the critical numbers of F. So we start with this equation after relative of a very difficult zero. And that's the same as 54 -6 X Square equal to zero. Then it's the same as six X squared equals 54. And this is the same as um X square Equal to 54 over six. And this is the same as X Square Equal nine. And this is same as X Equal three or X equal near to three. So we have to uh solutions to the equation to this equation here. But remember we are looking at critical numbers In the interval 04. That is why we only Are interested in the solution x equals three. So say that seems 83 is not in the usual three or 4. We know that the only mhm. Critical number of F. In the intel ceo four is three. And so we get to evaluative function at three at the curriculum number At zero and 4 which is which are the endpoints of this interview. F zero yes five F at four which is the point is five plus 54 times five. And the sort of thing before I meant which is uh the argument times four minus two x cube. That is for cute doing all these calculations here. We get 93 And finally f at the critical number three, five plus 54 times three -2 times three. Cute. And this give us 113. So we can see here that this is two largest value of those three values. So this is the absolute maximum of the function over dangerous here. four and The smallest of these three values is five. So this is the absolute minimum of the function over it into a three or 4. So if that answer now so d absolute minimum value of death On the closed interval. 04 yes five. Which of course at Uh zero which is the left point at the left and poin mhm X equals zero. And the absolute maximum value of F Only until three or 4 Is 113. And these of course yeah, add um The critical number three and that's the answer to this problem. It's very important to it remark then that we use first the fact that a continuous function on a close interval attain its extreme values. That is it has an absolute maximum value and an absolute minimum value over that interval. And in fact and those extreme values are obtained at either day in points of the interval or at critical numbers. So we find the critical numbers inside given interval. Then we relate the function of those critical numbers and the endpoints and the largest of those civil relations is the absolute maximum value of the function of our day interval on the smallest of the images we found this way are or is the absolute minimum of the function of related and that's the finances.

So we're wanting to establish absolute maximum and minimum values on the interval from 0 to 5. Um, so we're gonna restricted just within this. We see that the maximum is going to be located right here at 16. When X equals two and the minimum will be located right when access equal to five. So that's going to be seven.

Minimum values of function. Given it a go, the first thing we need to do is find the critical points and this critical points occur where the derivative of the function or the slope of the tangent line to the curve equals zero. So first thing we wanna differentiate dysfunction that comes out ast four minus two x. So now that we have the derivative, if we want to set that equal to zero and so for X Lipes when we find the X equals two. So this is one of our critical points. But we also have to consider the endpoints of Are given Interval so zero and five also are critical points. Now we need to figure out what the value of the function is that each of these critical points so richest plug each of these into original function. So this comes out is 12. Both thes terms are zero at X equals two. We find that the value of our Foshan is 16 and finally X equals five. We see our function gets a value that 25 seven no, no the the absolute maximum is the value of 16 and it occurs at X equals two and our absolute minimum is seven at X equals five

So we have the function here. F of X is equal to two X plus four on the closed interval from negative 1 to 1. Okay, so, um Well, then what we can do is to find the absolute next admitted we're gonna first take the derivative of our function. So the derivative of, um Well, we just have a linear function here. So the derivative of affects that's gonna be f prime of X. That's what's gonna be equal to to right? So then to find the critical values would then take our derivative and we would set it equal to zero. But right, this is never equal to zero right to If we set to equal to zero What values of X make to equal to 02 is never equal to zero. So therefore there are no critical values. Eso then what we do is we list out all critical values and endpoints. But since there are no critical values since the functions is constant, um, well, we only have our endpoint story about which would be negative one and one. So then we go and evaluate our function at those end points on DWI would have Well, First we'd have f of negative one so f of negative one that's just equal to two times negative one plus four, which is two. And then we evaluate at one so f off one. Well, that's just equal to two times one plus four or two plus four, which is six so f of negative one is to f of one is six. So therefore defined the what? The absolute Max and men. But we only have one minimum and one maximum So therefore those would be our absolute max and men. So our absolute maximum is our only maximum in this case is, um, equal to six. So absolute max is six and that would occur when X is equal to one. So the actual taxes six and then occurs at acts being equal toe one and then the absolute minimum is again the Onley minimum in this case. So the absolute minimum that would be, too. And that would occur when X is negative one at our other endpoints. So the absolute men is to and that would occur, um, at X being equal to negative ones


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