 In this video, we provide the solution to question number 11 for practice exam number four for math 1210 We're asked to find the absolute maximum and minimum values of the function f of x equals one half sine of 2x on the interval from 0 To 2 pi thirds and so we to find these absolute maximum and minimum values We have to first investigate the critical number. So let's calculate the derivative of f to find these critical numbers So f prime of x the derivative is Is going to be one half times two times cosine of 2x by the usual chain rule That is the derivatives equal to cosine of 2x and we have to investigate when that's equal to zero Well, what is cosine equal to zero? Well cosine So we get that 2x is going to equal cosine is equal to zero at the top of the unit circle in the bottom of the unit circle So we're going to get pi halves plus, you know, a multiple of well Let's think about that for a second get pi halves. You also get three pi halves. So if you add pi to it or subtract pi We get that the solutions are gonna be pi halves plus two or plus pi k That's what two that will give us 2x. We have to divide by two and get that x is equal to pi force plus pi halves Okay, so for example, we have solutions at pi force at pi At three pi force would be the next possible one, right? And then add another pi halves to that We're going to get five pi force. We can stop once we get past two pi thirds Which would notice that five pi force is too big But four pi force would be in there. Even three pi force is too big, right? So this is our only critical number now This doesn't mean that the extremum happens at the critical number But we have to make sure that it fits inside of the domain that's we've constructed here The next thing we're going to do is we're going to build a sign T chart, excuse me for all of your x coordinates We're going to list the boundaries of the interval. So zero and two pi thirds And we also need to include the critical numbers if there any so we get the critical number Pi force and then we're going to put these into the function f of x the original function right here So if we take one half sign Of two times zero, right two times zero zero sign of zero is zero one half times zero is zero So we're going to get zero as the value right there For the next one, we're going to get one half sign of two times pi force is pi halves Pi halves well if you take sign of pi halves, that's going to give you one So we actually get this value of one half And then for the last one we're going to do two pi thirds there put that into our function one half sign Of two times two pi thirds. It's going to give us four pi thirds Which be aware that four pi thirds, where is that in terms of the unit circle? That actually will be in the Excuse me, that'll be in the third quadrant sign is going to be negative that situation And so we're going to end up with negative one half times the sign of pi thirds, which is root three over two So we end up with a negative root three over four As this last value so the largest value present is going to be the maximum which we see is right here This is where the absolute maximum is for our function the absolute maximum value will be y equals one half It's obtained at pi force and then the absolute minimum value is going to be this one right here negative Three negative square root of three over four. This is going to be our absolute minimum And this is obtained at the at the number x equals two pi thirds