Absolute Maximums and Minimums of a Function on a Closed Interval
In the previous activities we discussed the topic of maximums and minimums, and how to find them by way of critical points and the Second Derivative Test.
In this activity we will discuss how to find the absolute 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 including the end points. 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.
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
a, and the absolute minimum occurs at b. 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.
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 f(x) on a closed interval a≤x≤b, follow these steps:
- Calculate all the critical points of
f(x)that are inside the closed interval by solving the equation formed by settingf'(x)equal to 0, and then only keeping the solutions that are greater or equal toaand less than or equal to b. - Calculate
f(a),f(b)and also calculatefat all the critical points you found in the previous step. - The smallest value from step 2 is the absolute minimum of
f(x)on the closed interval. The largest value form step 2 is the absolute maximum of thef(x)on the closed interval.
a, b and all the purple diamonds in the applet above, and the function will update accordingly. Go ahead and try it out!
What you should pay attention to is that no matter how you adjust a, b or the purple points, the absolute maximum and absolute minimum of the function on the closed interval -3≤x≤6 will always occur at either a critical point or an endpoint. You can use the coordinate point to help you explore this.
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.
f(x)=x^2-2x constrained to the closed interval 0≤x≤4, but without looking at it in Geogebra. We'll follow the process described above:
f'(x)=2x-2, so0=2x-2is solved withx=1. So 1 is the only critical point off(x). It is in the closed interval, so we keep it.- Since the critical point
x=1is in the closed interval 0≤x≤4, we will need to check it and the endpoints of the closed interval in the functionf(x). Those calculations aref(0)=0^2-2*0=0andf(1)=1^2-2*1=-1andf(4)=4^2-2*4=8. - Looking at these three outputs of
f, we see that the absolute minimum is -1 and occurs whenxis 1, and the absolute maximum is 8 and occurs at the endpoint whenxis 4.
f'(x) 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 f(x) is complicated.
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 integral.