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Numerade Educator



Problem 50 Medium Difficulty

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

$ f(x) = x^3 - 6x^2 + 5 $, $ [-3, 5] $


Absolute minimum value $-76$ which occurs at $x=-3$ ;
Absolute maximum value $5$ which occurs at $x=0$


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Video Transcript

we're going to find the absolute minimum and absolute maximum values of the function F of X equals x cube minus six X square plus five. And interval negative +35. First of all, the function is continuous on a closed interval. So we know it attains its extreme values at points inside the interval. But those points where these tribbles are attained can be either the end points of the interval or critical numbers of F. So we start by finding critical numbers of F. For that we need the first derivative of F which is three X square minus 12 X. And this derivative here exist for all values of acts inside the interval 93 5. It means that the only critical Numbers of dysfunction are those wells of X. for its the first relative is equal to zero and he is Mhm. This And then we had to start by solving the equations for conservative of f called zero. So this equation it's equivalent to three X squared minus 12 X equals zero. She's the same as three times x square minus four X. Okay equals zero. And because three is not zero is equivalent to X square -4. x equals zero. Which is equivalent again too X times x minus four equals zero. And because we have factor out people in normal which is a credit to zero, we get that either effect of the factors can be serious. So X can be zero or excuse before And in this case these two solutions of the equations after a difficult zero are both inside the interview -35. So we can say that the two The only two critical numbers of F N. Negative 35, R zero and four. So now we know that the extreme values of the functions on the interval. Netted 345 happen or a core at either at the end points or these two critical numbers. So we get to evaluate the function at those four numbers and shoes. The largest and smallest values. Then what we have here is that F at 93? She's a left and point of the interval is let's remember F is X cubed. That is native three cube- seats negative three square When a 693 square plus five. And if we do this calculation that give us 1976. Now we evaluate at the writing point of the interval That is five we get five cube minus 65 square plus five. And that give us uh negative 20. Now we evaluate at the critical numbers zero and 40 we get five and at four is four Q -6 times four square plus five. And that's equal to -27. So at this results and then The largest value here is five. So that the absolute maximum of f on the interval negative 35, The smallest of the values is -76. This is the absolute minimum of f over 93 5. So we give that answer here, the absolute minimum value of F On the interval negative 35 Is -76. And that value of course at X equal the fucked value. Where these absolute Munich or is the left hand point of the interval. So at the left and point X equals 93. And the other answer is the absolute maximum value of F. On the close interval for -3-5. Yes. Mhm five mm. Which of course At this case zero. That is a critical number at the critical number X equal syria. That's the answer to this problem. And in summary, we can say that first we ensure the existence of the extreme violence, that function over the given interval because the injury was closed and the functions continues there. And we know those uh true values are attained either at the end points of the interval or critical numbers of function. So we find the derivative and because this derivative exists for every value of X. In the interval negative 35 we are sure that the only critical numbers of these functions are those values of the argument X. For which the first derivative is equal to zero. So we solve the equations first serve a difficult zero. We found two solutions, both of which lie lies in the interval negative 35. And for that reason we have to consider both critical numbers and the two critical numbers devaluation of the function give us 50 and negative 27 4. And the evaluation at the end points gave us at negative three. We obtain negative 76 at five negative 20. So the largest of these four values is five attained at zero. So that's the absolute maximum of the function of virgin interval, And the smallest of the values is negative 76 attained at 93. So this is the absolute minimum Value of the function on the interval 93 5. And that's the final answer to this problem.