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# Find the absolute maximum and absolute minimum values of $f$ on the given interval.$f(x) = 3x^4 - 4x^3 - 12x^2 + 1$, $[-2, 3]$

## Absolute minimum value $-31$ which occurs at $x=2$ ; Absolute maximum value $33$ which occurs at $x=-2$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

Yeah. Here we want to find the absolute minimum and absolute maximum values of the function F. Of x equals three X. To the fourth minus four X cubed minus 12 X square plus one on the interval negative +23. So first we note is that this function is a polynomial functions so it's continues everywhere. In particular on the closed interval native to three and being continuous identified in a closed interval. We know that this function attains it's extreme values that is it absolute maximum and it absolute minimum values inside that interval. That is there are numbers inside the interval which image is equal to. Yes. So the minimum and other number between is the absolute maximum value. Yeah but we also know that those stream valleys are attained either at the end points of the interval or at the critical numbers of the functions. So we start by Calculating the 1st derivative of F. We get 12 x cube -12 X square -24 x. And this polynomial of degree three. So this derivative which exists for every value of X in the interval negative 23. For that reason we can say that mhm. The only critical numbers of F are those values of X. For which the first serve a ticket zero. So we got to start by solving this equation. First derivative of F. He called zero. This equation is equivalent to 12 X cubed minus 12 X squared minus 24 X equals zero. Because the first serve a tiv is given by this expression here. And now we can factor out this polynomial here like this to f X times X square minus eggs minus six mm minus two. And that is equal to zero. And this is equivalent again. We can factor out this original also so we can get 12 X times X. Um Plus one time 6 -2 equals zero. So now we have 12 is a constant different from zero and we have three factors dependent on X. There they are X. X plus one and x minus two and any of them can be zero in order to the product be zero. So this is equivalent to saying that X is equal zero or X is negative one or X is too. So these three numbers are critical numbers of F. But moreover that they all are in the interval negative 23. So we have to consider them all. Yeah, That is we have to evaluate the function at the three critical notes numbers and at the end points of the sub interval that is -2 and three. So let's see here the critical numbers of F In -2, 3 are x equals zero, X Equal 91 and x equal two. No F at the point native to for example is three X To the 4th 33, 2 To the 4th -4 Times -2 Cubes minus 12 X square. There is minus 12 negative two square plus one. And if we calculate that we get 33 Now it will wait at the writing .3. So is three times 3 to the 4th -4 times three Q minus 12 3 square plus one. And the result is 28. Now we go for the critical numbers efforts zero Yes one. Now f at negative one is three times nearly one to the falls -4 times 91 Cube -1291 sq Last one. And that give us 94 and finally effort to the last critical number is three times 2 to the 4th -4 times to Cube -12 times to Square plus one. The result of this uh Operation is -31. We have been these five values And the largest of these values is 33. So this is the absolute maximum value of F on the interval negative 23. And it of course at the left hand point negative too. The largest of these values as already the smallest of these values is negative 31 and it is attained at X equal to that is a critical number. So this final answer, we are going to right here. The absolute minimum value of f Owned interval negative 2 3 is negative 31 and eat a course at a critical point or the critical number. The critical number X equals two. And the other is the absolute maximum value of F On the interval narrative to three is We said is 33, 33. And that value is obtained or of course at um the left and point of the interval the left and point X equals negative tomb. And is, is the answer to the problem. And as a summary, we always start by uh ensuring that the extreme values of the function exists in the given interval in this case. That is true because the function is continuous because it's a polynomial function and the interval is closed. So we know the extreme values of the function exists in that interval. And we know more than that that those extreme values are attained either at the end points of the interval or at critical numbers of the function. We find the derivative and we verify that this derivative exists for every argument X in the interval. And that implies that the only critical numbers of the function are those vials of the argument X. For which the first derivative is zero. So we state that equation for conservative of F equals zero. We do some algebra and factories ations of polynomial appearing in this case. And we obtain three solutions this example and all of them all the three solutions we got our in the interval netted 23. So all of them have to be considered in the search for the string value. So okay, we have three critical numbers inside the interval negative 23 and the two end points. So we got to evaluate the function at those five arguments. We did that. And the largest Of the values we found was 33 at the at the left and .8 of two. And that the absolute maximum value of the function over the interval And the smallest of the virus is -31. And it of course at The critical number two. And that's the absolute minimum value of the function on the closed interval negative 23, which is in this case attained, had a critical number X equal to. And that's the solution to these problems.

Universidad Central de Venezuela

#### Topics

Derivatives

Differentiation

Volume

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp