# Video: MATH-DIFF-INT-2018-S1-Q18B

Find the absolute extrema values of the function π(π₯) = π₯(π₯Β² β 12) over the interval [β1, 4].

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### Video Transcript

Find the absolute extrema values of the function π of π₯ is equal to π₯ times π₯ squared minus 12 over the interval between negative one and four.

Now, the absolute extrema values over an interval are the maximum and minimum values of that function over the interval. In order to find the absolute extrema values, there are a few things we need to do. Firstly, we need to find the critical points of π of π₯, which are in the interval between negative one and four. Next, we need to evaluate the value of π of π₯ at these critical points and at the end points of the interval. Then finally, we need to identify the maximum and minimum from these values of π of π₯. So starting with point one, we find the critical points of π of π₯, which are in the interval between negative one and four.

Now, critical points occur where the π₯-values of the differential of π of π₯ or thatβs π dash of π₯ are zero. And so we need to find π dash of π₯ and then set it equal to zero. Now, π dash of π₯ is equal to the differential of our function with respect to π₯. So thatβs d by dπ₯ of π₯ times π₯ squared minus 12. Now, we could find this differential by using the product rule since we have a product of π₯ multiplied by π₯ squared minus 12. However, it would be easier to find this differential if we simply expand the brackets and then differentiate the polynomial, which is left. Expanding the brackets gives us π₯ cubed minus 12π₯.

In order to differentiate this polynomial, we simply take each term and multiply it by the power of π₯ and then decrease the power of π₯ by one. So for the first term, which is π₯ cubed, we multiply it by the power, which is three, and then decrease the power by one. So the power goes from three to two. And this gives us π₯ squared. So we found that π₯ cubed differentiates to three π₯ squared.

The next term is negative 12π₯. The power of π₯ here is one since π₯ to the power of one is equal to π₯. So you multiply it by the power, which is just one. And then one times negative 12 is negative 12. And then, we decrease the power of π₯ by one to get π₯ to the power of zero. And π₯ to the power of zero is simply one. So we get negative 12 times one, which is just negative 12. Therefore, we found that π dash of π₯ is equal to three π₯ squared minus 12.

In order to find our critical points, we simply set this equal to zero. And now, we solve three π₯ squared minus 12 is equal to zero for π₯. We isolate the π₯ terms, divide by three, and then take the square root of both sides, remembering the plus or minus since a square root gives both a positive and negative solution. So now, we found the critical points of π of π₯, which are negative two and positive two. However, weβre only interested in critical point in the interval between negative one and four.

Our critical points are π₯ equals negative two and π₯ equals two. Now, π₯ equals negative two does not lie in the interval between negative one and four. And so we can ignore this critical point. However, the critical point of π₯ equals two does lie in the interval between negative one and four. And so we found the critical point of π of π₯, which is in the interval between negative one and four. And weβve completed the first step of this process.

Now we can move on to step two of this process. We need to evaluate the value of π of π₯ at the critical point and at the end point of the interval. So weβre evaluating π of π₯ at two, negative one, and four. Remember that our π of π₯ is equal to π₯ times by π₯ squared minus 12. So when we substitute π₯ equals two into π of π₯, we obtain that π of two is equal to negative 16. Next, we can substitute π₯ equals negative one into π of π₯. And we obtain that π of negative one is equal to 11. And finally, we substitute π₯ equals four into π of π₯. And we obtain that π of four is equal to 16.

Now, we have evaluated the value of π of π₯ at the critical point and at the end points of the interval. So weβve completed step two. Our final step is to identify the maximum and minimum from these values of π of π₯. We can clearly see that the minimum value here is negative 16. And the maximum value here is 16. And so weβve identified the maximum and minimum, completing the third step.

Therefore, we found the solution to this question. The minimum point over negative one to four is at π₯ equals two, where π of π₯ is equal to negative 16. And the maximum point over negative one to four is at π₯ equals four, where π of π₯ is equal to 16.