Video: Finding the Absolute Maximum and Minimum Values of a Cubic Function on a Closed Interval

Determine the absolute maximum and minimum values of the function 𝑦 = −2𝑥³ on the interval [−1, 2].

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Video Transcript

Determine the absolute maximum and minimum values of the function 𝑦 equals negative two 𝑥 cubed on the interval negative one, two.

I’ve drawn a quick sketch just to help us understand what we’re looking to do and it’s also a quick sketch of our function. And in this question, we actually want to find out what the maximum and minimum values are in the interval from negative one to two. And so to do that, we’re actually gonna use a method. And that method is called the closed interval method. And it’s called that because actually we’re dealing with a very definite interval between negative one and two.

Okay, so what’s the first step? Well, stage one is to actually find values of 𝑓 𝑥 at the critical numbers. And we actually find the critical numbers by differentiating our function and then setting the differential to zero because actually the critical numbers are at the points where the slope is actually zero.

So as I said, first of all, we’re going to differentiate 𝑦 equals negative two 𝑥 cubed. And when we do that, we get negative six 𝑥 squared that’s because we multiplied our exponent three by our coefficient negative two, which gives us negative six. And we’ve reduced our exponent by one. So we get 𝑥 squared.

Okay, great, so we’ve now actually found our differential. As we said before, now what we need to do is actually set this equal to zero because we want to find the critical values which are actually where the slope is equal to zero. So therefore, we get zero is equal to negative six 𝑥 squared. And therefore, we get that 𝑥 is equal to zero. So this is actually 𝑥 is equal to zero — our critical number.

But now, we haven’t quite finished this first step because what we want to do is actually to find the values of 𝑓 𝑥 at this critical number. So this means actually substituting our value 𝑥 equals zero into our original function, which means we’re gonna get function of zero is equal to negative two multiplied by zero cubed which is equal to zero. Okay, so great, we’ve completed our first step.

Now, the second stage is actually to find the values of 𝑓 𝑥 at our end points. And our end points are negative one and two. So first of all, we’re gonna start at negative one. So therefore, if we substitute negative one for 𝑥 in our function, we get negative two multiplied by negative one cubed, which will give us negative two multiplied by negative one. So we get the answer two because if you have negative two multiplied by negative one, you just get two. Negative multiplied by a negative gives us a positive.

Okay, great, we found one of our end points. Let’s find the other one. To do this, we substitute 𝑥 equals two into our function. So we get negative two multiplied by two cubed, which gives us negative two multiplied by eight which is equal to negative 16. So now, what we’ve done is we’ve actually found the three key points on our graph and the values actually associated with them because we found the value when 𝑥 is equal to negative one, when 𝑥 is equal to two, and when 𝑥 is equal to zero.

So then, we move on to our third step, which says that the largest of the values is the absolute maximum and the smallest is the minimum. So if we go and have a look back at the values that we’ve got, when 𝑥 is equal to negative one, we got two, which is our largest value — so therefore, the absolute maximum. And then, when 𝑥 is equal to two, we got negative 16, which is our smallest value — so therefore, the absolute minimum.

So therefore, the absolute minimum value of the function 𝑦 equals negative two 𝑥 cubed on the interval negative one, two is equal to negative 16. And the absolute maximum is equal to two.

And we can check this by taking a look at our sketch where we can actually see by looking at the graph the point that’s at 𝑥 is equal to negative one, well, that would be our maximum value and the point that falls at 𝑥 is equal to two will actually be our minimum value. So yes, our graph actually agrees with the values that we found.

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