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

$ f(x) = xe^{x/2} $, $ [-3, 1] $

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Absolute minimum value $-\frac{2}{e} \approx -0.7357588823$ which occurs at $x=-2$ ; Absolute maximum value $\sqrt{e} \approx 1.6487212707$ which occurs at $x=1$

01:46

Wen Zheng

01:04

Amrita Bhasin

03:15

Chris Trentman

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 1

Maximum and Minimum Values

Derivatives

Differentiation

Volume

Campbell University

Oregon State University

Baylor University

University of Michigan - Ann Arbor

Lectures

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Find the absolute maximum …

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let's find the absolute minimum and absolute maximum values of the function F of x equals X times exponential of X. Health. On the interval -3 1. First we know that this function is continuous over these closed interval and for that reason it attained its extreme values over that interval and we know more over that the uh after the extreme values of the function are attained at either to create the end points of the interval or political numbers of function. So we start by finding the critical numbers of F. For that we need the first derivative of F which is the relative at this product here, X times exponential of X half. That is the derivative of X which is one times E. To the X half plus X times the derivative of E to the X. Health Is the same exponential function times the derivative of the exponents which is 1/2 and that is E. To the X. Health. Common factor of one plus X over two. And that's the same as to E to the X have times two plus X. And that's the expression of the preservative of F. And we see that this derivative exists for every value of X in the interval From -3 1. Mhm. And for that reason we can say that The only critical numbers of f in the interval from 93 to 1 Are those values of the argument eggs in the interval from 83 to 1, for which the first derivative is equal to zero. So we've got to solve that equation for serve a difficult zero. And that's equivalent to the equation to E. To the X half times two plus X equals zero. Because the first derivative of f human with this expression here. And we know that too. And the exponential of X to the exile is neatness. Always positive. So it's going to say that two plus X is zero and this is the same as X equal narrative to So the only critical number of the function In the interval from -321 clubs interval Is x equals -2. So now we got to evil way to function at this critical number. And at the end points that into that is negative three and one. And from those fighters we have the extreme violence of functions. So first the image of negative three, it's nearly three times E. To the negative three house. And that value it's about if you use a calculator Uh to negative 0.66 9390 4804. Now we evaluate at the right point which is one And we get one times into the 1/2 that is worth of E. Which is about 1.64 8721 2707. And finally, f at the curriculum number negative too Is -2 times e. To the -2/3. Sorry, I made a mistake here. It's not like dad, It's active two times negative to have met. And that is native to E to the -1, that is 82 over E. And that's about zero 7357. Okay, properly 7th 357 5 3 times eight 23. That's the three images we need. And now we see from here that the largest value is 1.6487 etcetera. Because this is creator than the values. And it of course had The Writing .1. The value of the sec values for a spirit of E. And the smallest of these values is native 7357 cetera which is -2 ovary. And that of course a -2 which is a critical number. So now we can say the answer to the problem. Is that the absolute minimum value? Okay. Yeah. Mhm. Okay. Of the function on the interval uh Native 31. Mm. Yes. So here negatives, there are seven which is -2, ovary is native to over 80, which is about negative 0.73575 triple a 23. And that minimum absolute minimum value of course, at the critical number X equal native to that's about the absolute meeting. Now, the absolutely absolute maximum value of the function On the closed intervals from 83 to 1 is uh squares of E. We know it's about 1.64 87 2127 07. And that actually the maximum value of course. Okay, so here worse at uh in this case yeah at the writing point of the interval of the right and point X equal one. And so this is the answer to the problem. And now we get uh remember we call or did here. The first thing is to notice that the function being continuous and defined in a closed interval. We know that it attains its extreme values on that interval. And the points where the values where the function obtains those extreme values, maybe either the end points of the interval or critical numbers has functioned in that into So to find the critical numbers, we find the first serve a tive. We noticed that that derivative is exist for every value of X in the interval. And so the only critical numbers in f in that interval, out those values for is the relative is zero. We stayed that equation preservative of F equals zero. And for that we get that the only critical number of the function of the interval from -321 is native to. Then we evaluate the function at the right end point left and point. And the critical number and we found that The largest of these values is squares of E which is about 1.648 7: 1- 707. And that it is attained at The writing point of the interval that is number one and the smallest of these values which is then the absolute minimum value of the function over a given interval is negative two very, which is about negative zero point 73575, Triple 8- three. And it, of course, at the critical number X equals -2. And so this is the final answer.

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