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Using the Critical Number Theorem

Instructor: Carmen Andert

Carmen has two master's degrees in mathematics has has taught mathematics classes at the college level for the past 9 years.

In this lesson, we will review the statement of the critical number theorem and give several typical examples of its application in locating maximum and minimum values of a function.

Maximizing Area

A farmer needs to build a new pen for his goats, chickens, and pigs. To save money, he decides to fence in a rectangular area and divide the region with two additional pieces of fence as shown.


Rectangular Animal Pen
rectangular fenced region


If the farmer has 3600 yards of fence available, what dimensions of the pen enclose the largest area?

To set up this problem, let's label one side of the large rectangle x as shown. The farmer will be using that same length of x on the opposite side of the rectangle and for the two additional pieces that divide it into regions, so he's left with 3600 - 4x yards of fence for the two horizontal sides in the diagram. Thus, each horizontal side has length (3600 - 4x)/2.

So, the area is given by x(3600 - 4x)/2, or 1800x - 2x2. At this point, we just need to find the ''best'' x-value to get as large an area as possible. But where do we start?

Critical Number Theorem

The critical number theorem gives us a way to locate the maximum and minimum values of a function. More precisely, it states that if a function f(x) has a relative extremum (maximum or minimum) at x = c, then c is a critical number of f(x) or c is an endpoint of the function's domain.

For this to work, the critical number c needs to meet the following requirements:

  • The function's first derivative f '(c) is either 0 or is undefined.
  • c is in the domain of f(x): this means that c is an x-value for which the function f(c) is defined.

How does this help us? This theorem is basically saying that we only need to look at the critical numbers of a function and the endpoints of its domain to find its extrema. Many functions do not have domains with endpoints, so the real focus is on the critical numbers. To find the values of c, we just need to focus on the spots where the derivative f '(c) is 0 or undefined.

We're going to get back to our farmer and his rectangular pen problem, but first let's hone our new skills with a couple examples.

Example 1

Find all local extrema for the function f(x) = x4 - 18x2 - 4.

Notice that the domain of f(x) is all real numbers, so it has no endpoints. Let's find the critical numbers of f(x). First, find the derivative: f '(x) = 4x3 - 36x. Setting this equal to 0, we solve:

  • 4x3 - 36x = 0
  • 4x(x2 - 9) = 0
  • 4x(x + 3)(x - 3) = 0
  • x = -3, 0, 3

Now we use the first derivative test to classify the critical numbers as local maxima or minima:

  • Since f '(x) changes from negative to positive at x = -3, f has a local minimum at -3.
  • Since f '(x) changes from positive to negative at x = 0, f has a local maximum at 0.
  • Since f '(x) changes from negative to positive at x = 3, f has a local minimum at 3.

Last but not least, we must plug each critical number (-3, 0, and 3) into f to find the corresponding maximum and minima:

  • f(-3) = (-3)4 - 18(-3)2 - 4 = 81 - 162 - 4 = -85
  • f(0) = (0)4 - 18(0)2 - 4 = -4
  • f(3) = (3)4 - 18(3)2 - 4 = 81 - 162 - 4 = -85

Therefore, this function has a local maximum of -4 (at 0) and a local minimum of -85 (at -3 and 3)

Example 2

Find all critical numbers for the function f(x) = 3x/ln(x).

The domain of f(x) is (0, 1) ∪ (1, ∞), which means that f(x) is not defined at x = 1 (since the denominator would be 0). To find this function's critical numbers, we find f '(x) using the quotient rule:

f '(x) = (ln(x)*3 - 3x*1/x)/(ln(x))2 = (3ln(x) - 3)/(ln(x))2.

f '(x) is undefined where the bottom of the fraction is 0:

  • (ln(x))2 = 0
  • ln(x) = 0
  • x = 1

But since 1 is not in the domain of f(x), 1 cannot be a critical number of f(x).

And f '(x) is 0 where the top is 0:

  • 3ln(x) - 3 = 0
  • 3ln(x) = 3
  • ln(x) = 1
  • x = e

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