How do you find the maximum and minimum of a function on an interval?

To find the absolute maximum and minimum of a function on a closed interval, you need to check the function's value at its critical points and at the endpoints of the interval. The highest value you calculate is the absolute maximum, and the lowest is the absolute minimum.

Think of it like hiking a specific section of a mountain trail. The highest and lowest elevations of your hike can only occur at the peaks and valleys along the way (the critical points) or at the exact spots where you started and finished the hike (the endpoints).

The Extreme Value Theorem

Before we calculate anything, we rely on a mathematical guarantee called the Extreme Value Theorem. This theorem states that if a function is continuous on a closed interval [a,b][a, b], it must have both an absolute maximum and an absolute minimum on that interval. A closed interval means the endpoints are included. As long as there are no breaks or holes in the graph, we are guaranteed to find a highest and lowest point.

Finding Critical Points

Critical points are the "peaks and valleys" of our function. To find them, we take the first derivative of the function, f(x)f'(x), and find where it is either equal to zero or undefined. When the derivative is zero, the tangent line to the graph is perfectly flat, which usually indicates a local peak or valley. We only care about the critical points that actually fall inside our specific interval [a,b][a, b].

The Closed Interval Method

The step-by-step process for finding these extremes is called the Closed Interval Method. First, find the derivative of the function. Second, set the derivative to zero or find where it is undefined to get your critical points. Third, plug those critical points into the original function f(x)f(x) to get their yy-values. Finally, plug the endpoints aa and bb into the original function. Compare all the resulting yy-values: the largest is your maximum, and the smallest is your minimum.

Where Students Slip Up

A common mistake is plugging the critical points back into the derivative instead of the original function. Remember, the derivative just tells us the slope, but we are looking for the actual height (yy-value) of the function. Another frequent error is forgetting to check the endpoints. Often, the highest or lowest point of a specific interval isn't a peak or valley at all, but rather the very edge of the graph.

Worked through

Find the absolute maximum and absolute minimum values of the function f(x)=x33x2+1f(x) = x^3 - 3x^2 + 1 on the interval [1,4][-1, 4].

First, we find the derivative of the function: f(x)=3x26xf'(x) = 3x^2 - 6x. Next, we find the critical points by setting the derivative equal to zero: 3x26x=03x^2 - 6x = 0. Factoring out 3x3x, we get 3x(x2)=03x(x - 2) = 0. This gives us critical points at x=0x = 0 and x=2x = 2. Both of these points fall within our interval [1,4][-1, 4], so we will test both of them. Now, we evaluate the original function f(x)f(x) at the critical points and the endpoints. At the left endpoint x=1x = -1: f(1)=(1)33(1)2+1=13+1=3f(-1) = (-1)^3 - 3(-1)^2 + 1 = -1 - 3 + 1 = -3. At the critical point x=0x = 0: f(0)=(0)33(0)2+1=1f(0) = (0)^3 - 3(0)^2 + 1 = 1. At the critical point x=2x = 2: f(2)=(2)33(2)2+1=812+1=3f(2) = (2)^3 - 3(2)^2 + 1 = 8 - 12 + 1 = -3. At the right endpoint x=4x = 4: f(4)=(4)33(4)2+1=6448+1=17f(4) = (4)^3 - 3(4)^2 + 1 = 64 - 48 + 1 = 17. Finally, we compare the results: 3,1,3,17-3, 1, -3, 17. The largest value is 1717, so the absolute maximum is 1717 at x=4x = 4. The smallest value is 3-3, so the absolute minimum is 3-3 at x=1x = -1 and x=2x = 2.

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Where this comes from: OpenStax Calculus Volume 1 · Stewart Calculus: Early Transcendentals · Khan Academy: Absolute maxima and minima

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