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Problem 56 Hard Difficulty

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

$ f(t) = \frac{\sqrt{t}}{1 + t^2} $, $ [0, 2] $

Answer

Absolute minimum value $0$ which occurs at $t=0$ ;
Absolute maximum value $\frac{3}{4 \sqrt[4]{3}} \approx 0.569876764238694$ which occurs at $t=\frac{1}{\sqrt{3}} \approx 0.577350269$

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Video Transcript

we're going to find the absolute minimum and absolute maximum values of the function F. F. T equals square root of tea or over one Plus T sq On the closed interval. 02. We know this function attains its extreme values on that interval because the interval disclose and the function is continued there. Moreover, we know uh those extreme values are obtained either at the end points of the interval or at critical numbers of F. So we start by calculating the critical numbers of F. For that. We need the first derivative of F. We have a cogent here. So we have to eat derivative respect to t of the numerator sort of tea times. The denominator one plus T square minus square root of T. Times the derivative respected T. Of the denominator want lusty square and all that over one plus T square square. And that's the expression for the fish or the derivative of F. And that is equal to derivative of squadron 30 is the same as the derivative of teaching one health. And that's one half T to the negative one half. That is 1/2 square. It 50 Times one Plus T sq minus the derivative of one plastic squares to T. It's always to tea time squirt of T over one plus T square. And that square now we have in the numerator we have yeah, different fractions. So we get Denominators, two Squares of T. And on that we get one Place city square minus uh two skirts of tee times to t skirts of T. In that over one plus T square in that square and that's equal to And then this denominator gets here down here to skirts of two times one plus t square square. And in the numerator we get one plus T square-. Then we have four four squirts of tea time squirts of Tst tst tst square That's it. Then this is equal to 1 -1 three T square. And that over two squares of tee times one plus t square square. That's the first order derivative of F. And the conservative is not defined at zero. So after relative zero does not exist. and zero is into the main of the function because in fact F is defined at all points in this interval 02 and all points in real numbers in fact, but the given domain 02 55 0. So series in the domain of F. And they're the relatives that does not exist so serious. A critical number of he's critical number mm of F In the Interval 02. Now to find the other critical numbers, we equate the first serve a tiv 20. And that's equivalent to one minus three T square over two squares of tee times one plus t square square equal zero. Here we are assuming that these different from zero all the way through these ambulances here In order to ride in the Denominator scored 50. And this fractions. Here's the numerator is zero. And from here we get three t square is one which is the same as T square is one third. And so t can be once one over square 23 or negative one over square to three. So we have two possible solutions to this equation first. Very difficult zero but T equal negative one over scores of three is not only given interval here. So up the equal negative one over square 23 Is not in the interval 02. Then T equal square one over score 23 is a critical number of F in the industry to in fact one over score to three is about 0.57735. So this number here is in the interval and is a number where the first survey, the other number as we saw here, His native 1/2, 3 is not indeed given integral. So F has two critical numbers On or in the interval 02 and they are zero and 1 over skirts of three. So we have to evaluate the function At those critical numbers. And at the end zero and two we get the following. But let's say we start by the end points at zero F is equal to squares of zero over one plus series square that zero equals zero F. Two. These creative to over 1-plus 2 square That is a creative to over five. Okay, and that's about I see your point 28 28 4271 247 4619. Now, we either like f at the critical numbers first at zero but zero is already here. So let's see that This particular case, zero is a critical number of F. Yeah, in the given interval, but it happens to be also the left point. So We have already evaluated at zero as the left point. And now we got to even late at the other critical number one of our score to three. And we get from here squared of one over square 23. And that over one plus square one over square 23 square. And with a little bit of math here and we are going to get finally 3/4 times 4th Root of three. Here is the that's for again clearer. Four times fourth word of three Or it's the same. 3 to the 1 4th. And that's about you know, more or less what value it is. 0.56 98 987 6764 23, 94. That's the image of one over square 23. In fact, one over square 23 we said here is about 157735. So from these values we have that the absolute maximum value of the function over a given interval interval 02 is to your point 56. There is 3/4 times four ft three And the minimum value is zero attained 0. With that we cannot give the answer. Then the absolute minimum value of F In the interval 02 is zero which of course at uh the left and point. And also a critical number T equals zero. And the absolute maximum value of the function In the interval 02, Yes. Uh 3/4 times four through 2 three Which is about 0.56 and nine 87 6764 238 694. Which of course at t equals one over square 23, Which is in fact about 0577 350 269. So this is a final answer to the problem. And we remember here dad, we started back by saying that this function attains its extreme values at the given interval because if this continues and the interval is closed, we already know that uh the stream values of the functions are attaining either at the end points of the interval or at critical numbers of the function in that interval. So we start by calculating the first derivative of F. And with that we calculate the critical numbers in this case zero is a value in the domain of the function where the preservative does not exist for that reason, the equal seriously critical number of F. And the equation first derivative of ethical, zero has 2 solutions. One over scores of three and negative one over scores of three. And for those the solution negative one over scored 23 is not in the given intervals here too. And for that reason the other critical number besides zero is T. Equals one over scooter through three, Which is about 0.577350- 69. We now calculate the image of the function at the end 0.0 and two and at the critical number one over square 23. And we obtained that the largest value, which is then the absolute maximum value of the function over the interval is 3/4 times Fourth Fruit three, which is about 3.56986764, 694. And it's a it of course at one over square to three which is a critical number. And the absolute minimum value of the function corresponds to the smallest Of these three values 0. And of course at the left hand point also, which is also a critical number T. Equals syria. So this is the final answer.

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