In the [url=https://www.geogebra.org/m/x39ys4d7#material/fd7edtmx]previous activities[/url] we discussed the topic of maximums and minimums, and how to find them by way of critical points and the Second Derivative Test.[br][br]In this activity we will discuss how to find the [i]absolute[/i] maximum and absolute minimum values of a function when it is constrained to a closed finite interval. FYI, a "closed finite interval" is an interval of numbers i[i]ncluding the end points[/i]. For example, you might consider the set of numbers between -3 and 6 including the endpoints. This would be every number between -3 and 6, including -3 and 6. The notation for this closed finite interval is -3≤x≤6 or [-3,6]. In other calculus courses you might also study intervals that don't include the endpoints, called "open intervals," but we won't be discussing those here, and will only consider closed intervals. [br][br]Moving on. Check out the applet below where a function is plotted which has been restricted to the closed interval -3≤x≤6. It doesn't matter what the function is.

As you can see, the absolute maximum occurs at [code]a[/code], and the absolute minimum occurs at [code]b[/code]. Even though the function has a critical point at about 2.4, this is NOT the absolute maximum. Slide the point with coordinates to see that the tangent line has a slope of 0 somewhere around 2.4. The exact value doesn't matter.[br][br]It turns out that the absolute maximum and the absolute minimum of a function constrained to a closed interval like above must occur at either the endpoints of the interval, or at critical points inside the interval. In light of this fact, there's a very straightforward algorithm for finding absolute minimums and maximums of such functions. To find the absolute maximum and absolute minimum of a function [code]f(x)[/code] on a closed interval [code]a≤x≤b[/code], follow these steps: [br][br][list=1][*]Calculate all the critical points of [code]f(x)[/code] that are inside the closed interval by solving the equation formed by setting [code]f'(x)[/code] equal to 0, and then only keeping the solutions that are greater or equal to [code]a[/code] and less than or equal to b.[/*][*]Calculate [code]f(a)[/code], [code]f(b)[/code] and also calculate [code]f[/code] at all the critical points you found in the previous step.[/*][*]The smallest value from step 2 is the absolute minimum of [code]f(x)[/code] on the closed interval. The largest value form step 2 is the absolute maximum of the [code]f(x)[/code] on the closed interval.[/*][/list][br]To see why this is true, take another look at the applet above. To help you see that the process described above will [i]always[/i] work (not just for a specially constructed function of [i]my[/i] choosing), I've made it so that [i]you[/i] can adjust [code]a[/code], [code]b[/code] and all the purple diamonds in the applet above, and the function will update accordingly. Go ahead and try it out! [br][br]What you should pay attention to is that no matter how you adjust [code]a[/code], [code]b[/code] or the purple points, the absolute maximum and absolute minimum of the function on the closed interval [code]-3≤x≤6[/code] will always occur at either a critical point or an endpoint. You can use the coordinate point to help you explore this.[br][br]Thus, the conclusion you can draw is that the process above will always work to pick up the absolute maximums and minimums of a function constrained to a closed interval. [br][code][/code][br]Let's try it out by trying to find the absolute maximum and absolute minimum of [code]f(x)=x^2-2x[/code] constrained to the closed interval [code]0≤x≤4[/code], but without looking at it in Geogebra. We'll follow the process described above:[br][br][list=1][*][code]f'(x)=2x-2[/code], so [code]0=2x-2[/code] is solved with [code]x=1[/code]. So 1 is the only critical point of [code]f(x)[/code]. It is in the closed interval, so we keep it.[/*][*]Since the critical point [code]x=1[/code] is in the closed interval 0≤x≤4, we will need to check it and the endpoints of the closed interval in the function [code]f(x)[/code]. Those calculations are [code]f(0)=0^2-2*0=0[/code] and [code]f(1)=1^2-2*1=-1[/code] and [code]f(4)=4^2-2*4=8[/code].[/*][*]Looking at these three outputs of [code]f[/code], we see that the absolute minimum is -1 and occurs when [code]x[/code] is 1, and the absolute maximum is 8 and occurs at the endpoint when [code]x[/code] is 4.[/*][/list][br]The only potential trouble spot in general with this process is finding the critical points in step 1 because it requires you to solve the equation formed by setting [code]f'(x)[/code] equal to 0, and this can be a really challenging algebra puzzle. This time it wasn't, but it easily could be quite hard if [code]f(x)[/code] is complicated.[br][br]There's plenty more we could study about derivatives, but this is a good overview of the core ideas. It's time to move forward and study the [b]integral[/b].