Find the absolute maximum and absolute minimum values of $f$ on the given interval.
$ f(x) = 2x^3 - 3x^2 - 12x + 1 $, $ [-2, 3] $
Absolute minimum value $-19$ which occurs at $x=2$ ;
Absolute maximum value $8$ which occurs at $x=-1$
we want to find the absolute minimum and absolute maximum values of the function F of x equal to x cubed minus three X squared minus 12 X plus one on the interval Added to three. Which is a closed interval. So the first thing we got to notice is that F attains that those extreme values that is its absolute minimum and it's absolute maximum values on this interval because interval schools and the function is continuous because simple normal function. So we know that those extreme value exists. But besides that we know that uh those extreme values are images of either the end points of the interval or critical numbers as a function. So we got to start by calculating too Critical numbers of the function for that. We need the first derivative and that is equal to six x Square -6 X -12. And we'll see that this derivative exists for every value X in the interval negative 23. And for that reason the only critical numbers f king half are those files of X. For which the derivative is equal to zero. So yeah. Mhm. And then we get a start by solving the equation After relative equals zero. And that's the same thing as six X squared minus six. X minus 12 equals zero because the relative discretion here and that's the same as six times X square minus x minus two equals zero. And because he six is not no longer we can say this is the same as x square- X -2.0. And we can factor out this polynomial S X minus two times X plus one in that equal zero. And this product series either as a factory zero, so X is two or X is negative one. And these two values in this case are both in the interval negative 23. So they are they're both has to be considered. Then the two critical numbers of F In the interval netted 2, 3 are two and 81. So we got to find the images of these two critical numbers and the images of the end points of the interval. F at let's say negative too is equal to two times negative two cube minus three times Native to square minus 12 times -2 Plus one. And if we calculate all these we get negative three, that's the image of the left hand point negative two. Now the image of the writing 20.3 is to attend three Cube -3 times three square mm hmm minus 12 times three Plus one. And that give us -8. And now we have a late at the critical numbers so F two, he is two times two Q minus three times to square minus 12 times two plus one. There is negative 19. And finally f at the other critical number 91 is two times negative one cubed minus three times negative one square minus 12 times negative one plus one. And that give us and eight. Mhm. So the largest value of these four is eight with the absolute maximum of f over the interval. Native to three and the smallest Of the virus is -19. Okay. So this is -19. And see absolute minimum of the function over the interval. So we got that answer then if has an absolute minimum value 80 negative 19. Which accor's at uh to which is one of the critical numbers at the critical number. Okay, mm It's equal to and f has an absolute maximum value, maximum value we said eight. And that value of course At -1 which in fact is the other critical number at the critical number Ux x equals 91. So this is an example where the the two extreme values of the function happened or a core at the critical numbers inside the close interval. So this is the final answer and then propping all with it here. First thing we noticed that the extreme values of the function exists because the function is continuous and defined on a close interval. And we know those extreme values are images of either the end points of the interval or critical numbers of the function we find the derivative and because that derivative exists at every point or a number in the close interval. So the only critical numbers of the functions are those surveyed for which the relative is zero. We solve that equation derivative of ethical zero. We stick with the solutions that are inside the close interval In this case both of them are so we have a lady function at two critical numbers, we found, and the end points and the largest Value of these four values is the absolute maximum of the function over the interval on the smallest of the values is the absolute minimum value of the function on the into. So this is a final answer to this problem yeah.