Extreme Value Theorem & Bolzano's Theorem

Lesson Transcript
Yuanxin (Amy) Yang Alcocer

Amy has a master's degree in secondary education and has taught math at a public charter high school.

Expert Contributor
Robert Ferdinand

Robert Ferdinand has taught university-level mathematics, statistics and computer science from freshmen to senior level. Robert has a PhD in Applied Mathematics.

The extreme value theorem and Bolzano's theorem are two very useful theorems that you can use to help you find solutions as well as maximums and minimums of a function. Learn how to use them in this lesson.

Extreme Value Theorem

The extreme value theorem tells you when a continuous function will have a maximum and a minimum on a closed interval. What does this mean? First, picture a function, any continuous function (i.e. any function that isn't broken up anywhere). Take the graph of the function f(x) = cos x - 2 sin x, for example. Look at the closed interval from x = 0 to x = 6. All that the extreme value theorem is telling you is that, in this interval, your continuous function will reach a highest point and a lowest point. In this case, we have the bottom of the dip and the top of the hill. Now, look at the graph of the function f(x) = x - 5. Applying the extreme value theorem to the closed interval from x = 2 to x = 4, we see that this continuous function reaches a lowest point and a highest point inside this interval. For this continuous function, the lowest point is at x = 2 and the highest point is at x = 4.

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  • 0:00 Extreme Value Theorem
  • 1:00 Finding Maximums and Minimums
  • 2:38 Bolzano's Theorem
  • 3:31 Finding the Solution
  • 4:30 Lesson Summary
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Finding Maximums and Minimums

In math, we call the lowest points minimums and highest points maximums. How can we go about and find these minimums and maximums? There are two steps:

  • Step 1. Find the points where the first derivative of the function equals zero; y' = 0.
  • Step 2. Evaluate the function at the points from above and at the interval end points.

After you have evaluated your function at all your points, then you can find your minimum and maximum points. Your minimum point is the one that evaluates to the lowest answer, while the maximum point is the one that evaluates to the largest answer. For example, to find the maximum and minimum points for the function f(x) = x - 5, on the closed interval from x = 2 to x = 4, we first take the first derivative of the function and solve it for zero. We get f(x) = 1. Since this first derivative simply equals 1, we don't have anything to solve for so we don't have any points to find. Next we evaluate our original function at any points we found from finding the first derivative and at the ends of our closed interval. We don't have points from our first derivative. Our closed interval begins at x = 2 and ends at x = 4, so we can evaluate our function at x = 2 and x = 4. We get f(2) = (2) - 5 = -3 = -3 and f(4) = (4) - 5 = -1 = -1. Looking at all our answers, we see that we have a minimum of -3 at x = 2 and a maximum of -1 at x = 4.

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Additional Activities

Extreme Value Theorem and Bolzano's Theorem: Practice Problems

Key Terms

  • Extreme Value Theorem: The global (absolute) maximum and minimum of a continuous function f(x) on a closed interval exist inside or on the boundary of the interval.
  • Bolazno's Theorem: For f(x) = 0 to have a solution in an interval [a, b], f(a) and f(b) should have opposite signs or f(a) x f(b) < 0.

Materials Needed

  • Paper
  • Pencil
  • Calculator

Example: For the function f(x) = 4 - x^2, find the maximum and minimum values on the interval [-3, 1]. Then, determine whether f(x) = 0 has a solution in [-3, 1].

Solution: Proceed as follows -

Step 1: To find points of maximum and minimum, use the Extreme Value Theorem:

(a) Solve f'(x) = 0 or -2x = 0 which gives us x = 0. Note x = 0 MUST lie inside the interval [-3, 1], which it does, for this x-value to count.

(b) Compare f(x) values at x = 0 and the end-points of the interval, viz. x = -3 and x = 1 to get:

f(0) = 4 - 0^2 = 4.

f(-3) = 4 - (-3)^2 = 4 - 9 = -5.

f(1) = 4 - 1^2 = 3.

From (b) above, the maximum is 4 at x = 0 and the minimum is -5 at x = -3.

Step 2: To determine whether a solution of f(x) = 0 exists in [-3, 1], we use Bolzano's Theorem to get:

f(-3) x f(1) = (-5) x (3) = -15 < 0.

Hence, f(x) = 0 has a solution in [-3, 1].

Practice Problems: Follow the example above and show all your work.

(a) For the function f(x) = (2/3)x - 5, find the maximum and minimum values on the interval [-2, 3]. Then determine whether f(x) = 0 has a solution in [-2, 3].

(b) For the function f(x) = x^2 - 9x + 8, find the maximum and minimum values on the interval [6, 9]. Then determine whether f(x) = 0 has a solution in [6, 9].

Answers (To Check Your Work):

(a) Maximum is -3 at x = 3 while minimum is -19/3 at x = -2. NO, f(x) = 0 does NOT have a solution in [-2, 3].

(b) Maximum is 8 at x = 9 while minimum is -10 at x = 6. YES, f(x) = 0 has a solution in [6, 9].

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