Problem 52 Hard Difficulty

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

$ f(t) = (t^2 - 4)^3 $, $ [-2, 3] $

Answer

More Answers

01:47

WZ

Wen Z.

01:02

Amrita B.

03:42

Chris T.

Watch More Solved Questions in Chapter 4

Video Transcript

let's find the absolute minimum and absolute maximum values of the function F of T equals yeah, equals d squared minus four value to the third. On the interval negative 23. So first we know this function is a polynomial function and t of the 36. And for that reason it's a continuous function everywhere in the real numbers. But in particular is continuous on the interval native to three because it's continuous and a close interval, we know it has or attain its extreme values. Uh on that interval that is there are points where the images at those points equal the extreme values. Well, we also know that these two brothers are attained either on the end points of the interval or at critical numbers. So we start by calculating the first derivative of T of F meant and this is three times t squared minus four square times derivative of t square minus for respect. And t is to t. So this is 60 times the square minus four square. As we can see that f derivative exists for every T in the changeable narrative to three because it's a polynomial function again. And because of that, because this derivative exists at every point, we can say that the only critical numbers of dysfunction or the values of T for which the first derivative equals zero. So we can write that the only critical numbers of F are those values of T for which first serve a tive of F a T Z equals zero. So we got to start with that equation. First derivative of F T equals zero. And this is equivalent to the equation 60 times the square minus four square equals zero because this is the formula for the preservative of F. And we can factor out this based is where well for us T plus two times t minus two. And that it this square in the civil service. And so this is the same as saying 60 times T plus two square times t minus two square equals zero. And now this completely factor out. So we can say that tea can be zero or t equal to or t equal negative two. So these are the three critical numbers of five. Mhm. And they all lie in the interval native to three. In fact native too easy. And point of the interval, the left hand point. So um we got an evil way to function at this. Uh Three critical numbers. In fact in these 20 on two because a negative two, we had to relate it as uh an end point of the interval. So there are 1234 evaluations and we get F at at negative two is equal to the function is t square minus force cube. So it's needed to square minus four cube. And that is equal to serum. We have also F at three which is the writing point is three square minus four cube. And that give us 125. Yeah then F at the critical number zero is serious. square minus four cube is native for cube um negative 64. And F at the regal number two is to square minus four Q zero. So from these four values we can see that the largest value is 125 uh three And that the absolute maximum of the function on the internet to three. And the lowest value is negative 64 at the critical number zero. So we can say that to max the absolute minimum value of F on the interval negative 23 is 1964. And we say here which of course yeah, add zero which is as we saw here a curriculum number of F at the critical number X equal to. We also have the absolute maximum mhm of F on at 2 to 3 East 125. Yeah. Which of course okay, create and uh three which is the right and point of the interval. And this is the answer to the problem here. And we in summary we did the following steps first we ensure us that we have a function with extreme values obtained in the interval. And that's true because function is continuous and closed interval so it attains its extreme values And we know those extreme values are attained at either the end points of the interval or a typical notes. Critical numbers. So we find the derivative of function and because that derivative exists for every argument t then we know that the only critical numbers of the function are those fellows of the perimeter chief, which the first derivative is zero. So we state that equation first relativity equals zero. And we found three values. In fact one of those critical numbers is exactly the same as the left and point of the interval. So we got to evaluate the function at those critical numbers and at the end points. In fact there are four evaluations because but we said that is one. Critical point is equal to the left and point when we evaluate we found the largest value of the function was 125 at three. That is at the writing point and the lowest value was negative 64 at the critical number zero. So on the 25 is the absolute maximum value of the function attained at. Do you write in 0.3? And native 64 is the absolute minimum value of the function attaining the critical numbers zero. And here I have to repair the fact that here get this moment Christie because it's a variable we were using in this problem. Yeah. Here is the equal to and here is Teagle three. That's it. This is as I said, the final solution to this problem.