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Find the absolute maximum and absolute minimum va…

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Problem 59 Hard Difficulty

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

$ f(x) = x^{-2} \ln x $, $ [\frac{1}{2}, 4] $

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Video Transcript

let's find the absolute minimum and absolute maximum values of the function F of X equals extra negative two times natural algorithm of X On the closed interval from 1/2 24. The first we know that this function is continuous and then given interval because the point example series not there. And in fact the natural algorithms are defined for positive values of X. But here we have an interval which is completely on the positive values of X. So uh the functions continue there. And because the internet is closed, we know that the function obtains its extreme values on that interval and the points where the extreme values of F. R attain it can be either the end points of the interval or the critical numbers of F. So we start by calculating the first derivative of F. And that's a product here we have X squared X. To the negative two times Natural algorithm of X. We can see it also as steady as a fraction as coaching last like natural rhythm of X over X squared. But let's see it like a product. And then we apply the product rule for differentiation. So so we get the derivative of extra negative two is negative two X. To the negative three times. And natural rhythm of X plus X to the negative two times the derivative of natural rhythm of X which is one over X. So we get negative two. Natural rhythm of X over X. Q. It's the first term here plus one over X. Q. And that becomes one minus two. Natural rhythm of X over excuse. So that's the derivative and preservative is not defined at zero that does not exist at zero. And in fact it's also defined only for positive values of X. So zero is not a critical number of F. Because even though the derivative does not exist there and that Value zero is not in the domain of the function. So We can say that after evocative at zero does not exist but zero is not in the domain of F. So it is not critical number of F. So we've got to solve the equation. First derivative F equals zero. And that's the same as the equation one minus two. Natural rhythm effects over excuse. And we are supposing on this equivalence is here that X is positive. So that we can take the natural algorithm. So this is equivalent to the equation. One runs to natural algorithm of X equals zero. Which is equivalent to natural rhythm of X equal to one half. Because the exponential function is injected function. We get for the positive numbers. Yet this equivalent to E. To the natural algorithm of Mexico to eat to the one half. And this girl into because the natural algorithm and the exponential function are inverse to each other. We get x equal into the 1/2 which is quite a bit of E. And if we calculate that it's about um Squares of E. is about 1.64 1.64 87 87 1 2- 1- seven. Mm. Oh seven. Okay. This is uh so this is the only uh x equal spirit of E is the only critical number. Let's see in fact that this number is in the interval one has four because one has 0.5. So between 0.5 and four is the number 1.6487. So they see a critical number of F In the interval 1/2 4. In fact, this is the only critical number of functions in its extended domain which is a positive numbers. Okay, so we have this and now we can find the extreme value of the function by evaluating the function at this critical number and at the end points of the interval That is 1/2 and four. So what we have here is f at 1/2 is equal to let's see, Excellent. Native to times. Natural algorithm of eggs. That's the same as natural rhythm of 1/2 over one have square. And remember the natural rhythm of fraction is the natural rhythm of the numerator minus. And natural rhythm of the denominator. There's a property of the logarithms And 1/2 square is 1/4. Natural rhythm of 10. So we get negative. Natural algorithm of two over 1/4 in that negative. Yeah, that's negative. Uh for natural rhythm of two, which is about negative value is about negative. 2.772 five. Yeah. 887. Let's put a little bit too left. That's about 2.7725887 2224 three times two. Yeah. Okay. Now f at the other end point that is at the writing point is for And that is natural rhythm of four over for square. Remember four here is naturally almost two square over uh 16. And these applying the properties of the law Rhythm, we get to natural rhythm of to over 16. And that's the same as Natural algorithm of 2/8. And that's about 0.08 664. Okay. And that's the end points. Now the critical number is scores of E. And the images. Natural rhythm of this word of E. Over mm. Spirits of E. Square. And here we get one half over mm That is one over to eight which is about To your .18 39397 206. So between these values is the absolute minimum and the actual um and similar values of the function. And we recognize Isley that The absolute maximum value of the function is 0.1839397206. Which is one over to E. And it is the image of the critical number, scores of E. And the minimum value is as a minimum value of the function is negative four. Natural rhythm of two which is about 92 7725887 to 2 to four Which is the image of the left hand .1 half. So we can write the answer to the problem. You can say that the absolute minimum value of F on the interval one half four. Yes We said -4 naturally the most to which is about negative 27725 sorry two point 7725 887 2224. And that minimum value of course at um The left hand .1 half. And for the on the other hand the absolute maximum value of F On the close into one have four is 1/2, one over to E. Which is about zero 18 3939 7-06 72. It's your six which of course at um the critical number X equal squares of E. Which in fact is more or less equal to 1.64 8721 2707. So this is if I know the answer to the problem and we remember what we did here. We first recognized that this function is continuous Over the close intervals so it attains its extreme values at that interval. And knowing that those uh stream values are obtained either on the end points of the interval or critical numbers of the function, we calculate the first derivative of F and see if there are values in the domain where the derivative is not defined. It's not the case here. There is even though zero for example is evaluate the preservative does not see exist that value serious not in the domain of function. So it is not a critical number. And we saw that in fact the function and its derivative are defined for positive value of X serious not included. So but the given interval is completely in the interval or is contained in the interval of positive numbers. So the derivative exists for all values of vaccine that into this case. So the only critical numbers here are those for which differ serve a tip zero. We state that equation and then we solve it to find that the only critical number of the function in the interval one half four is X equals scores of E, which is about 1.6487 to 1 to 7 37. Then we evaluated the function at the end points of the interval and at the critical number squared E. And we found that the Largest value is 0. 1 over to me which is about 0.28 0.1839397- six. And it occurs at the critical number of records of E. And the smallest of the values is negative four times natural rhythm of two, which is about negative two .7725887- 2- four. And it accords at the left end point of the interval one half. And so we have found the absolute minimum and excellent maximum value, the function over to give an interview.

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