Find the absolute maximum and absolute minimum values of $f$ on the given interval.
$ f(t) = t + \cot (t/2) $, $ [\pi /4, 7\pi /4] $
Absolute minimum value $\pi / 2+1$ which occurs at $t=\pi / 2$ ;
Absolute maximum value $3 \pi / 2-1$ which occurs at $t=3 \pi / 2$
let's find the absolute minimum and absolute maximum values of the function F F T equals T plus co tangent of T. Half on the interval by falls seven by fourth. So first we're going to remember the definition of go tangent of T. Have. We know that that's one over tangent of Tr and because tangent of T. Hof is sign of to have over coastline of T. Hof we'll have here T plus go sign of T. Hof over sign of tea health. It means that the function is to find word Whenever sign of Jihad is not zero. But in this case T. is in the interval from P. Pi 4th or through 7.4. And that means that by fourth It is less than or equal to T less than or equal to seven by fourth. That means that by AIDS is less than or equal to T. Half. There's an or equal to seven by eighth. And we know that this value here by eight is positive of course because by its positive number eight number and this less than pie because we have multiple with the scale or 7/8 which is less than one. So we have this inequality that that means that when T. Is in this interval as is the case in this problem, the argument of the sine of T. Half the argument of sign of health. In this denominator here, That argument is between zero and pi but never it's never equal to zero and never equal to buy. So yeah, sign of that argument to have. He's like the graph of sine function but Little bit after zero and a little bit before. But that is your zero here is by let's say. And so the graph of the sine of T. Health would be something like this. It never reaches zero either way Near zero or nearby. So we can say that sign of T have is positive for all. T. In the interval from 5/4 to 7 5/4. Mhm. So that's uh the first thing that is yes. Function is well defined. That is the gold content of jihad never gives an undefined or non finite number. But also we know that because of that the function is continuous on the interval given here. So f he is continuous on five falls seven x 4th. For the same reason that these expression here is continues to his continuous Gazans to have his continues and this expression is continuous and the coalition of the two expression co sign of to health and safety. Half it continues because the denominator sign of the health never is equal to zero. So the expression is a whole function. F. Fifties continues. Um The gloves integral by 47 by fourth. And for that reason being continues to function and close the interval. We know that if attains it's extreme values on the given interval it means there are points or values in this interval for which the image of those funds are equal to the extreme values of the function. We also know that these extreme values are attained at either the end points of the interview or at critical numbers of the function. So we've got to find the critical numbers of F. And for that we calculate first the derivative of F. And this is equal to one plus the derivative of co tangent of tea health. That is negative go second square of T. Hof times the derivative of T hof is one half. So we get one minus one half of of co second square of T. Hof And that's equal to 1 -1 cozy. Can square is one over sine square of tikal. So you get 1/2 signs square of the half. And we know already that this as we said before, this expression sign of T. Hof is never zero for tea, belonging to the internal by 47 5/4. That's what what we see here. And for that same reason this derivative exist for every value of T. In the interval by force. Yeah. 75 Falls. And that's simply because the denominator uh to science square ft have is never zero. And that's because sign of Jihad is not zero 40 in this. Into because sine of T. Hof, we started over here And we saw it's never 040 on the interval by 4th 74th. So uh if this derivative exists for everything value of T in this interval, it means then that the only critical numbers of dysfunction in this interval are those values of T. For which the relative is zero. So we can write that now. So we have this and we had then to solve this equation that is, we start with The narrative of f equals zero. And that's the same thing as is the formula of to the reality, which is this one here, one minus 1/2, signs square of to have a zero is the same as Uh to sign square of Kiev Equal one. And this is the same as signs square of T. F equals one half. And this is even equivalent to sign of jihad equal more or less, one over squared of two, which is equal to more or less. Right? It's creative to over two. Yeah, so this is the question we have to solve and we're not, the sign of some angle is Equal to restorative to over to the annual Gotta be by 4th. But there are two angles that solve these. First of all this, let's put things in perspective. We have two equations here, sign of tie up the T have equal scores of 2/2 and send it to have equal negative skirts of 2/2. But we saw here we said it just here that uh sign of T over two is positive for tea in the interval by four 74th. That's what we saw here up. Okay, because he has 40 in the intervals by force before to have his strictly positive and less than pit on by So we have this situation design. He's always positive. Never questioned the value zero. So, sign of tea health is positive. So this equation Is in reality only one question nor too because the negative sign has no sense for tea in the interval negative by 475/4. Let's say it again here. This is positive for all T in the interval only have to solve the equation. Yeah, sign of Dahaf Equal skirts of two, T. N by fourth, seven by fourth. So now we get to uh look a little bit at the graph of sign of X. That is the same function again we remember is sorry, something like this. Zero. Bye to buy. And as we are here, we know that we only need this part of the graph that is in this part here. And so we stayed right here and we are looking for the angles whose sign is starting to go over to, let's say it's words if you were to something like here. And then we have two solutions even for angles between Syria and by mhm We know exactly that T. Hof is between By eight and 7 x eight. So we're going to solve the equation first and see if we are inside this range, but at least between Syria and pie, which is this graph here. We know that there are evidently two solutions for this equation. These single here, Which we know is by four. That is the argument of the sine function gotta be by 4th and we got to know which is this or the solution is a solution is respect to buy half, which is this out here, we're assigning sequel series equal to one. We know that these distance here is the same as the assistance here. Mhm. So we got to say that we have 12 and three by fourth, so it's 3 5/4. The other solution And that's the value that must have the argument of sign in order to sign it be equal squirts October two. So looking at that graph, we say that T half, which is the argument of sine function in this equation Get to be equal to buy 4th or T have got to be equal to three x 4th, which is the other solution here. Okay, so from this day you gotta be by 8th, sorry, sorry at eight because we multiply by two, both sides we get by half or do you get to be multiplied by two of size is three by half. And we see that these two values are inside the range given up here between 8th which is smaller than buy house and three by half is smaller than seven By eight. So we are inside the Internet without problem. Um you can also see it in the way that now we have the values of T and T. Is between um Here it is between by force and by 4th and we know that by half the square than by fourth and three by half this less than seven by fourth. So we are good here. So both of these are critical numbers of the function inside the given interval. Then There are two critical numbers of F. Within by fourth. Seven by fourth. They are T equals by half and t equal three by half decided to critical numbers of the function. There are no more critical numbers because these are the only solutions to the equation after relative of F equals zero for pallets of tea inside the interval by fourth and my fourth and the relative exists in all other and all points of that interval, Meaning that the only critical points are told who is the first derivative zero. And then in this to the only solution to that equation of relativity equals here they are then the only critical numbers of the function. Okay, so we got to even late the function at uh the critical numbers In the interval here. And that is to these two values and at the end points of the interval that is by 4th And seven x 4th. Yeah, I had to evaluate F at these two critical numbers and at the and points of the interval. Okay, by fourth, 7.4 that is captive evaluate the function at my 4th and at 7.4. and between those four values will be or will have the extreme values of the function. So the largest value will be the absolute maximum value of the function on the interval by fourth and my fourth. And the smallest of the values will be the absolute minimum value of the function over the given into. Okay. So we start by evaluating the function at the end points Let's say f at by 4th at the left and point. So by definition is 5 4th plus plus mm go tangent of by fourth over To that is by 8th and that's equal to Yeah. Is rather a prison. It'll equal not equal. Okay it's rather a presumably equal To using a calculator. 3.20. Now we evil at f at the um Writing .7 x 4th That is 75 falls. Yeah. Okay this word garden here template false plus go tangent off seven x 8th because remember the coating agent is evaluated at T. Have So here is seven x 4th half is 758 and that's approximately equal to so by fourth let's go content of seven. It is And a calculator would find 3.08. Okay now we calculate F at the critical numbers by half and three by half. Let's start with by half. Okay that's but I have plus go tangent off The half of this value. That is by 4th. you know the code tangent of five fold is one over tangent of by force. There is one. So this is exactly by half plus one. And to have a reference of comparison with the previous values, this is approximately equal to two 17 57. An f at three pi half is by half. So if I have plus co tangent of the half of this, that is +33 by fourth and that's equal to three by half. And we now cover charge and a three by four is in 81. So we get this value here which is approximately equal to 3.71. Mhm. And so the s largest value will be the absolute maximum value of the function over the interval. The largest value here is this one here. So this is the absolute maximum value of the function. That is exactly three by half minus one. And it's attained at the critical number three by half. And now the smallest of the values is this one here, that is exactly by half was juan. Then that will be the absolute minimum value of the function over the given interval and is obtained at the other critical number by health. So we can uh um Right the answer to the problem. So the absolute minimum value of F On the interval by 4th, Seven by fourth is Uh I have plus one properly here. I have plus one Which is about two points. uh 57. And that value of course at the critical number t equal by health. Now we go forward the absolute maximum, so the absolute maximum value of the function f All the closed interval by 4th. Seven by fourth. Yes. Three by half plus minus one. Okay, yeah, Which is approximately equal to 3 71. And that value of course at the other critical number of the function T equal three by half. So this is the final transfer to the given problem and we we call it to here. So first of all we analyze a little bit the function in order to see if it is well defined and also it is continuous on the given close to interval and for that reason we know that their function obtains there, its extreme values and we also know the eastern values are attained either at the end points of the close to interval or at the critical numbers of the function. In that interview we calculate it for conservative and we obtain an expression which is never on defined that it exists for every value of teen given interval. And that's because the denominator is just sign of t have as in the definition of the function. So with uh prove or verified above that these denominator is always positive and not zero. So uh that's why we arrived too. Apparently two different equations we have to solve but in fact it's only one because the sign and the negative sign cannot be considered because the functions sine of the fourth and sign up to half is positive for all t in the given into. So we have to solve only one equation and the solution to the equation are in fact two solutions 40 between spy 1/4 and sent by fourth. And these two solutions aren't equal, I have anti equal through by half. So we have two critical and they are both in the given interval. So there are two only two critical numbers of the functions in the interval by fourth and my fourth and they are by half and three by half. So we got to give away the function of those critical numbers and at the end points of the interval by 4th and 7.4 We did that. And the largest value found was uh the image of three by half. That is the image of one of the critical numbers. And then that largest value of the four calculated is the absolute maximum value of the function over the given interval. And the smallest of the four values was probably have a swan, which is the image of the other critical number by health. And then that's the absolute minimum value of the function over the given interval. And then that's the answer to these problems