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Problem 57 Hard Difficulty

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

$ f(t) = 2 \cos t + \sin 2t $, $ [0, \pi /2] $

Answer

Absolute minimum value $0$ which occurs at $t=\frac{\pi}{2}$ ;
Absolute maximum value $\frac{3 \sqrt{3}}{2} \approx 2.598076211353316$ which occurs at $t=\frac{\pi}{6}$

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Video Transcript

we will find the absolute minimum and absolute maximum values as a function F of T equal to co sign of people. A sign of two times T on the closed the interval from syria to my health. Without this function is continuous in this close interval. And for that reason we know F attains its extreme values at points on that interval. In fact we know those points were F. Attendance. Extreme values may be either the end points of the interval or critical numbers of the function. For that reason we need to calculate the critical numbers of F. For that we calculate first the first order derivative of F. And that is negative to sign of T plus two co sign of duty. And now deserve a tiv is defined everywhere in the real number. So F derivative of T exists four every T in the interval zero by health. And for that reason we can say that The only critical numbers of f are those values of T. four which The first derivative is equal to zero. Yeah. So that reason we get to solve this equation for preservative of F equals zero. And this is equivalent to the equation negative to sign of T. Plus to co sign of duty Equals zero. Which is the same as um go sign of duty equal. Or in fact we can say instead of that All 10 of TTINA son of T Equal to zero. And now we can use the formula of the co sign of the double of an angle which we know is equal to Coulson square of t minus Saint square of team. Okay And that minus sign of T equals zero. and now cause square of tea is one minus sine square of tea. That minus Since we're t minus sine of T equals zero. And then this is equivalent to the equation 1 -2. Sine Square of T minus sine of T equals Iraq. Which can be written as to sign square of T plus sign of team -1 Equals Syrup. And now we can make a change of viable here we named alpha good sign of T. Or maybe they let's not use alphabet and other letters that matter like em Then we get to um square plus M -1 equals zero. And we can solve this equation using the formula we know for the solutions of this equation which is equal to 91, more or less square root of one square which is one minus four times two times negative 1/2 times two that he was negative one more or less were off one plus 98 over four That give us 91 more or less periods of nine. Okay skirts at nine Over four, that is 91. More or less. 3/4. So we get em one is nearly one plus 3/4 which is three minus one is two of the four is one half An M two is 91 month for over four. And that is -3. I meant sorry here. Mhm minus three And that is negative for over four is -1. So remember a sign of tea. So we get sine of T equals one half. Or sign of T Equal 91 4. T. In the interval. Remember the domain 0x2? Okay so we got to remember the sine function a little bit here. Over the interval syrup by half we have Sign of Serious zero. So you get increases let me throw this smaller here. So let's say get this by half. We know sign is equal to one and at 00. So here we can say the value is one. So this is sign of T for this interview. And at that interval this solution has no equation has no solution to this equation has no solution because we have sign a loose positive or zero. So this cannot be and so sign of T equal one. Have you see angle for which we have This value one else here and we know sign of um by six is one half. And The reason we can say that tea is 56 and it's the only solution. In fact because graphically we can see that there are no the points in the interval zero from zero by half where a sign of tea is one half. In fact functions include increasing. So it's 1 to 1 or there. So there's only one value of T between Syrian health for which the image is going to one half And that number is 5, 6 belonging to this interval syrup by half. We have this critical number and that's the only critical number as we said here both the only solution in the given interval of the equation for the difficult zero is the only critical number of F. So the only critical number of F Ian zero by Health S. T. Equals 5 6. So we got to evaluate the function at that critical number and at the end points of the interval let's start by evaluating F0 and my health efforts zero is to go sign of zero plus sign of two times 0. And that is because in a serious one we get to send a serious serious now at the writing point by by health we get to go sign up by half blast sign of two times by half. So this value here is equal to zero and here this value is equal to one. So we get to let's see because I'm serious. One authority I made a mistake here Try to say that this is zero here But this is zero also here because co sign and zero is 1 and then by half as we see here zero. So we got zero here's here here so we get zero sorry for that. Okay. Yeah the image or the images of the endpoints and now we find the image of The curriculum number by six. And we get to go sign of by six blood sign of two times by six. That's the same to co sign up by six plus sign of five third. That is to square 2 3/2 plus square to three over to. And that is Once for 2 3 plus scores of 3/2 is three squirts of 3/2. And that's about the value here. It's about 2.59 8076 21135 3316. Okay so we have this value here and these are all the values we get to consider. So the maximum the largest of the values, the absolute maximum value of the function over a given interval and that is 2.598 etc. Which is the the image at the critical number 56. And the minimum the absolute minimum value of the function which is the smallest of the three values Is zero attained at the writing point by half. So we write that answer here then. D uh absolute maximum that's right. First the minimum to the minimum value of f In the interval zero by half is zero. Which of course okay. At the writing point as we saw here, yeah at the right in point t equal by uh my health and the absolute maximum value of the function in the interval zero by half is three scored 3/2 which is about Which is about 2.59807621135 3316. Which of course diverse and By six which is sick. The critical number also function T equal 56. So exist the answer to the problem. And we remember then procedure here was to verify first that the functions continues over close the interval. And for that reason we know we detains its extreme values on that interval and knowing that the extreme values may be attained either at the end points of the interval or at critical numbers of the function. We start by calculating the first derivative of F. And then we identified the preservative exists for everybody of team the given interval. And so the only critical numbers of those parts of the argument. T. For which the functions first toward the derivative of physical zero, stating that equation for sort of derivative of F. At any T equals zero. We solve that equation and we got two possible solutions to equal uh The equation is equivalent to sign up to equal to 1/2 and sign up to equal nearly one. The equation center Chico native one is impossible for tea on the interval syrup by half because over there The functions sine of T. is always positive or zero. So the question we have to solve finally is sign of equal 1/2. And the only solution on the intervals by half of that equation is 56. So the only vehicle number of f in the intervals by half by six. And now we get Then to the function at the end zero I have And at a critical number by six we did that we found that the largest value of the function on those three points is 2.59, et cetera. With That is three score to three over to just the image of the critical number 56. So that's the absolute maximum value of the function over the given interval and the smallest of the three values is zero. So it's the absolute minimum of the function over the interval. And that of course at the writing point of the intervals uh syrup I have which is by half. And that's the end the solution to the given problem.

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