3.4 Using the First Derivative to Test Critical Numbers (4.3)

Transcription

1 118 CHAPTER 3. APPLICATIONS OF THE DERIVATIVE 3.4 Using the First Derivative to Test Critical Numbers (4.3) Theory: The rst derivative is a very important tool when studying a function. It is important to know what kind of information it can provide as well as what it can t. Most of the results in this section are based on a very important theorem, called the Mean Value Theorem. Theorem 178 (Mean Value Theorem) If f is a di erentiable function on the interval (a; b) and continuous on [a; b] then there exists a number c in (a; b) such that f 0 (c) = f (b) f (a) b a or equivalently f (b) f (a) = f 0 (c) (b a) This is a very important theorem in mathematics. Not only is it useful here, it is also used in many other parts of mathematics. The quantity f (b) f (a) b a corresponds to the slope of the secant line through the points (a; f (a)) and (b; f (b)). Geometrically, the theorem simply says that there exists a number c between a and b at which the slope of the tangent is the same as the slope of the secant line through the points (a; f (a)) and (b; f (b)). Looking at gure, this seems reasonable. Figure 3.10: Illustration of the Mean Value Theorem De nition 179 (Increasing/decreasing) A function f is called increasing on an interval I if f(x 1 ) < f(x 2 ) whenever x 1 < x 2 in I. Similarly, a function

2 3.4. USING THE FIRST DERIVATIVE TO TEST CRITICAL NUMBERS (4.3)119 f is called decreasing on an interval I if f(x 1 ) > f(x 2 ) whenever x 1 < x 2 in I. A function that is either increasing or decreasing is called monotone. Theorem 180 (Test for increasing/decreasing) on [a; b] and di erentiable on (a; b) : Suppose f is continuous 1. If f 0 (x) > 0 for every x in an interval, then f is increasing on that interval. 2. If f 0 (x) < 0 for every x in an interval, then f is decreasing on that interval. Though this theorem is intuitively easy to understand, proving it is a little bit more challenging. It requires the use of the Mean Value Theorem. The next theorem summarizes the discussion in section on page 107. Theorem 181 (First derivative test) Suppose c is a critical number of a function f and that f is continuous at c. We have: 1. If f 0 changes sign from + to at c, then f has a local maximum at c. 2. If f 0 changes sign from to + at c, then f has a local minimum at c. 3. If f 0 does not change sign, then f has neither a local maximum nor a local minimum. Figure 3.11 illustrates this theorem. Remark 182 The rst derivative test has several advantages over other tests we will see later: It can be used to test every critical number. It always gives an answer i.e. we know for sure that a critical number is either an extreme value or is not. Remark 183 To nd where a function is increasing/decreasing, or to test critical numbers involves roughly the same type of work since both are determined by knowing the sign of the rst derivative. You should follow the following steps Compute the rst derivative Find the critical numbers The critical numbers determine intervals on the real line. The sign of the rst derivative is constant in each interval. Find the sign. Conclude depending on the sign of the rst derivative. We illustrate the above procedure with several examples.

3 120 CHAPTER 3. APPLICATIONS OF THE DERIVATIVE Figure 3.11: Illustration of the First Derivative Test

4 3.4. USING THE FIRST DERIVATIVE TO TEST CRITICAL NUMBERS (4.3)121 Example 184 Find where f (x) = x 3 3x is increasing, decreasing. For this rst example, we explain every detail of the procedure involved. We know that nding where a function is increasing or decreasing amounts to studying the sign of its derivative. This is done by nding the critical numbers of the function. These critical numbers will determine intervals in which the sign of f 0 will be constant. Therefore, to nd the sign of f 0 in each interval, it is enough to nd the sign of f 0 at a point of each interval. In other words, it is enough to evaluate f 0 (d) where d is a point inside each interval determined by the critical numbers. The sign of the answer will be the sign of f 0 in that interval. This sounds complicated, but it is easy. We will set it up as a table. Finding the critical numbers. For this, we nd f 0 (x) and look for points where f 0 (x) = 0 or f 0 (x) unde ned. f 0 (x) = x 3 3x = 3x 2 6x = 3x (x 2) Since f 0 (x) is a polynomial, it is always de ned. Also, f 0 (x) = 0 when x = 0 or x = 2. Therefore, 0 and 2 are the two critical numbers. Studying the sign of f 0. We do it using a table like the one below: Interval 1 < x < 0 0 < x < 2 2 < x < 1 Test point d f 0 (d) Sign of f Behavior of f % & % Thus, we can see that f is increasing on ( 1; 0] [ [2; 1). It is decreasing on [0; 2]. We can verify this looking at the graph of f below.

5 122 CHAPTER 3. APPLICATIONS OF THE DERIVATIVE y x Remark 185 Let us make several remarks regarding this table. 1. To come up with the correct intervals, simply order the critical numbers found. The intervals will then be the regions between these numbers. If the intervals you come up with overlap, then you made a mistake. 2. The ultimate goal is to nd the sign of f 0 in a given interval. If you can nd it without plugging in a point, then there is no need to have a test point. In the case above, to nd the sign of f 0 (x) = 3x (x 2) in the interval ( 1; 0], it is enough to notice that in this interval, both 3x and x 2 are negative and that the product of two negative numbers is positive. Example 186 Find the local extrema of f (x) = x 3 6x 2 + 9x + 1. From Fermat s theorem, we know that these can only happen at a critical number. From the rst derivative test, we know that to test the critical numbers, we need to know the sign of f 0 (x) around the critical points. Therefore, we need to nd the critical numbers of f and study the sign of f 0. This is exactly what we did in the previous example. First, we nd f 0 (x). f 0 (x) = x 3 6x 2 + 9x = 3x 2 12x + 9 = 3 x 2 4x + 3 = 3 (x 1) (x 3) f 0 (x) is always de ned. f 0 (x) = 0 when x = 1 or x = 3. We now study the sign of f 0 using a table as in the previous example.

6 3.4. USING THE FIRST DERIVATIVE TO TEST CRITICAL NUMBERS (4.3)123 Interval 1 < x < 1 1 < x < 3 3 < x < 1 Test point d f 0 (d) Sign of f Behavior of f % & % Thus, we see that f has a local maximum at x = 1. The local maximum is f (1) = 5. f has a local minimum at x = 3. The local minimum is f (3) = 1. We can check this by looking at the graph of f below.x 3 6x 2 + 9x + 1 y x 10 Remark 187 It should be clear to the reader by looking at the two previous examples that the work to nd extreme values or to nd where a function is increasing or decreasing is exactly the same. Example 188 Given f (x) = x 4 4x 3, answer the questions below: 1. Find where f is increasing, decreasing. 2. Find the local extreme values of f. 3. Does f have a global minimum? Does it have a global maximum? 4. Finally, nd the global extreme values of f on the interval [ 1; 5]. From the two examples above, we should now have a good idea how to proceed, at least for the rst two questions. In fact, the work needed to answer the rst question will also provide the answer for the second question.

7 124 CHAPTER 3. APPLICATIONS OF THE DERIVATIVE Answer to questions 1 & 2 We begin by nding the critical numbers. f 0 (x) = x 4 4x 3 0 = 4x 3 12x 2 = 4x 2 (x 3) f 0 (x) is always de ned. f 0 (x) = 0 when x = 0 or x = 3. Next, we study the sign of f 0 using a table. Interval 1 < x < 0 0 < x < 3 3 < x < 1 Test point d f 0 (d) Sign of f Behavior of f & & % Thus we see that f in decreasing on ( 1; 3) and increasing on (3; 1). The critical number 0 does not correspond to an extreme value because f 0 does not change sign. f has a local minimum at x = 3. The local minimum is f (3) = 27. Answer to question 3 If we had the graph of f, we would know the answer. There is another way we can derive it. We can look at the end behavior of f. lim f (x) = lim x! 1 x! 1 x4 4x 3 = 1. Similarly, lim x!1 x4 4x 3 = 1. Therefore, f cannot have a global maximum. It will have a global minimum, which will be the smallest of its local minima. Since there is only one local minimum, it is also the global minimum. Answer to question 4 In this case, it is easy. We compare the value of f at the endpoints of the given interval with the value of f at the critical numbers. This is summarized in the table below: Point Value of f at the point Left endpoint: 1 f ( 1) = 5 Right endpoint: 5 f (5) = 125 First critical number: 0 f (0) = 0 Second critical number: 1 f (3) = 27 We see that on [ 1; 5], f has a global maximum at x = 5. The global maximum is f (5) = 125. f has a global minimum at x = 3. The global minimum is f (3) = 27. Remark 189 The rst examples might suggest that the sign of f 0 is always alternating. If this were true, it would mean that it is enough to nd it in the rst interval, thus saving quite a bit of time. However, this is not true, as the third example illustrates. So, never assume that the sign of f 0 alternates. You must nd it in each interval. Summary: what the rst derivative does for us

8 3.4. USING THE FIRST DERIVATIVE TO TEST CRITICAL NUMBERS (4.3)125 The rst derivative tells us where a function is increasing/decreasing. The rst derivative allows us to test the critical numbers of a function f to determine if f has extreme values Things to know: Know how to nd the number c in the mean value theorem. Given a function, be able to nd where it is increasing, decreasing. Given a function, be able to nd its extreme values. Be able to sketch the graph of a function using the information provided by the rst derivative. Be able to do problems such as # 3, 7,11, 17, 19, 25, 27, 29, 37 on pages In addition, be able to do problems such as: 1. True or False: If c is a critical number of a function f, then f must have an extremum at c. 2. True or False: If a function f has a maximum at x = c, then c is a critical number. 3. True or False: If a function f has a maximum at x = c, then f 0 (c) = True or False: If a function f has a minimum at x = c, then the minimum is f(c). 5. Draw the graph of a function y = f(x) such that f 0 (c) = 0 but the function does not have an extremum at x = c. 6. Draw the graph of a function y = f(x) such that f 0 (c) = 0 and the function has a maximum at x = c. 7. Draw the graph of a function y = f(x) such that f 0 (c) = 0 and the function has a minimum at x = c. 8. Draw the graph of a function y = f(x) such that f 0 (c) does not exist and the function does not have an extremum at x = c. 9. Draw the graph of a function y = f(x) such that f 0 (c) does not exist and the function has a maximum at x = c. 10. Draw the graph of a function y = f(x) such that f 0 (c) does not exist and the function has a minimum at x = c.