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Numerade Educator



Problem 48 Hard Difficulty

Find the absolute maximum and absolute minimum values of $f$ on the given interval.

$ f(x) = 5 + 54x - 2x^3 $, $ [0, 4] $


Absolute minimum value $5$ which occurs at $x=0$ ;
Absolute maximum value $113$ which occurs at $x=3$


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Video Transcript

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.

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