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Find the absolute maximum and absolute minimum va…

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Problem 47 Hard Difficulty

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

$ f(x) = 12 + 4x - x^2 $, $ [0, 5] $

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Video by Oswaldo Jiménez

Numerade Educator

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October 21, 2021

Excellent way to explain the concept. Gracias Oswaldo, explicas muy bien, gran ayuda.

Top Calculus 2 / BC Educators

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Video Transcript

We are going to find the absolute minimum and absolute maximum values of the function f of x equals 12 plus four X minus x squared on the interval +05. So the first thing we gotta use here is that this function is continuous because the polynomial On the close interval 05. So being continuous. And to find a close interval we are sure to function attained its extreme values. So there are points or values in the interval 05 where the images of those values is equal to the absolute maximum and The exit of minimum of the function over or on the interval serve five. And we also know that those extreme values can be obtained Either at the end points of the interval 05 or critical points. So we gotta find the critical points of F. The first relative of path is four minus sorry four minus good. Yeah, A distributive exists at all points or values eggs on the interval 05. And because that because of that we can ensure that the only critical points of these functions are those values of X. For which the first derivative is zero. Because the first survey delicacies were all points. So the only critical points can be those and the domain of the function is uh 05. So we got to find those critical points inside the interval 05. So this implies that the only critical numbers of F are those values of eggs sewage. The first derivative is here. So we start with that equation for servants because zero which is the same as four minus two, X equals zero which is the same as two x equal four which is the same as X equal to. That's the only critical point Of f. And this point is in the interval 05. So the only critical number of FNC 05 He is x equals two. So we get to evaluate the function at this critical point and and at the end points. And among those three values we will have the absolute minimum and accident absolute maximum of the function. So F zero is equal to we've allied effort zero. We get 12. If we evaluate at five we get 12 plus four times 5 minus X Square. That is five square. That is 12 Plus 20-25 20-25 -5. 12 -5 is seven. And now F at the critical point to is 12 plus four times two minus two square. There is 12-plus 8 minus four. eight metaphor is four plus wealthy, 16. Then the minimum value of this free values is the smallest is seven. This is then the absolute minimum of the function and the greatest value is 16. So this is the absolute maximum of the function over the interval three of five. So the absolute minimum value yeah of F on syria five is seven. Which a core of course at X war to be in Rome for size ad uh the writing point except for five. And the absolute maximum yeah value of f on 05 mm is 16. Which of course had um to which is a critical point. That's critical number X equal to. So this is the answer to this problem. We have then always to you apply the community of the functional gloves interval that ensure the existence of the extreme values of the function on the interval. And we know that those extreme values are attained at either the end points of the interval or at critical points. So we find the critical points. We know the critical numbers in this case are the solutions of the equation after a difficult zero because there are no possibility for the relative not to exist. And that is because the relative exists for all X. Then we solve the equation first, very difficult zero. And we in this case we got only one solution Inside which is inside the internal 05. So this is the only critical number of the function. And then we had the only critical number and the endpoints. We voted at those three numbers. And the largest value is the absolute maximum the function or the interval and the the smallest value is the absolute minimum of the function over to interval. As we said in the answer here

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