# Extreme Value Theorem & Bolzano's Theorem

## 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.

## 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.

## Bolzano's Theorem

Where the extreme value theory helps you find your maximums and minimums, **Bolzano's theorem** helps you find your solutions. The theorem tells you that if a continuous function on a given defined interval changes sign, then it must equal zero at some point in the interval. For example, because the function f(x) = cos x - 2 sin x, changes sign on the interval beginning at x = 0 and ending at x = 1, then according to Balzano's theorem, this function must equal zero at some point in this interval. As you can see from the graph, this function is first positive and then changes to negative in this interval. And you can see that it does cross the x-axis and, therefore, equals zero at what looks like x = 0.5. This then is one of the solutions of the function. Remember that our solutions are where the function equals zero or crosses the x-axis.

## Finding the Solution

Bolzano's theorem doesn't tell you what the solution is, but it does tell you whether a solution is present within a certain interval. To find the solution, you can either look at the graph and estimate or you can set the function equal to zero and then solve it.

For functions that are cyclical, like the sine and cosine function, you look for the answer that is included in the interval. For our function f(x) = cos x - 2 sin x, the solution inside our interval is 0.464. This particular function also has solutions at 3.605 and 6.747. Because this function is cyclical, it has answers that repeat every so often. But since we're only interested in answers inside the closed interval from x = 0 to x = 1, the only solution that fits this criteria is the 0.464 solution. As you can see, this 0.464 solution is very close to 0.5, which is what we see from the graph.

## Lesson Summary

Let's review what we've learned. The **extreme value theorem** tells you when a continuous function will have a maximum and a minimum on a closed interval. In math, we call the lowest points **minimums** and the highest points **maximums**. To find these minimums and maximums, we follow these 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.

**Bolzano's theorem** tells you that if your continuous function on a given defined interval changes sign, then it must equal zero at some point in the interval. To find the solution, you can either look at the graph and estimate or you can set the function equal to zero and then solve it.

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## 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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