# Optimization Using the Closed Interval Method - Problem 1

###### Explanation

To find the absolute maximum and minimum values of a function on an interval, you must do three things: find the relative maxima and minima of the function, determine which lie in the interval, and then check the boundary of the interval.

To find the relative maxima and minima, find the zeros of the first derivative of the function, as we have done before. Then, ignore the zeros that are not within the given interval (since we are only looking at one segment of the graph, we are only interested in maxima/minima that lie between these points). Finally, plug in the boundary values into the function (for example, if you are looking at the bounded interval from [-8, 5] for a function f(x), find the value of f(-8) and f(5)). Out of the boundary values, relative maxima and minima values, the largest value is the absolute maximum and the lowest value is the absolute minimum of the function on the interval.

###### Transcript

Let’s do an optimization problem. Here we are asked to find the absolute maximum and minimum values of the function f(x) equals x³ minus 6x² minus 36x plus 41. So nice cubic function on the interval on the closed boundered interval from -4 to 10.

So we are really only considering the part of this function that goes between x equals -4 and x equals 10. When you are looking for the absolute max and min, on a closed boundered interval like this one, you want to use the closed interval method.

Remember a closed boundered interval, is an interval of finite length where the endpoints are contained by the interval. So first step, is going to be to find the critical points. So I’m going to need to look at the derivative of f. And then I’m going to make a table of values, that examines the value of f at each of the critical points, and at the endpoints. The biggest of these will be the absolute max. The smallest is the absolute mean.

So let’s start with the derivative. F'. So it's just a polynomial. So it’s going to be 3x² minus 12x minus 36. Now right away I can see that, there's a factor of 3 I can pull out. So let me so that,3 times x² minus 4 x minus 12.

Now hopefully this will factor. Let’s take a look. It looks like if I have 2 and 6, 6 minus 2 is 4, so that might work. X minus 6 and x plus 2. I always use the sort of guess and check method for factoring, so let’s try this.

First of all it does give me a -12, when you multiply -6 times 2 so that works. Then -6x plus 2x, does give me a -4. So that works. And that tells me that my critical points are x equals 6 and x equals -2. So critical points; x equals 6 x equals -2.

Now that I have the critical points, I’m ready to make a table of values. So let’s go over here and do that. I want my x values, -2 and 6 and I need my endpoint values of -4 and 10. And I took the liberty of calculating this before. At -4, f(x) is 25, at -2 it's 81, at 6 it's -175 and at 10 it's 81 again. Now what this means is, wherever the value is biggest, that’s the absolute maximum. This is the absolute maximum value here and also here. It’s achieved twice; once at x equals -2 and once at x equals 10.

But a really important point is, the absolute maximum value is actually 81. So that’s the absolute maximum value. Where it occurs, is x equals -2 or 10. So it occurs at x equals -2 and x equals 10.

The absolute minimum value is the smallest of this and it's -175. The absolute min is -175. And it occurs at x equals 6. So that’s the important distinction to keep in mind. The difference between the absolute maximum value with the absolute minimum value, and the x coordinate where it occurs.

So pay a close attention to the instructions of the problem it might ask you where does the absolute maximum occur? It might ask you what the absolute maximum value is.