First Derivative Test

Transcription

1 MA 2231 Lecture 22 - Concavity and Relative Extrema Wednesday, November 1, 2017 Objectives: Introduce the Second Derivative Test and its limitations. First Derivative Test When looking for relative extrema among a function s critical points, we can tell if a critical point is a relative maximum, relative minimum, or a saddle, by looking at the signs of the derivative. In particular, if the derivative changes from + to, the critical point is a relative maximum. If it changes from to +, the critical point is a relative minimum. Otherwise, we have a saddle. This way of checking is often called the (First) Derivative Test. Theorem 1. (First Derivative Test) Given a function f(x) and f (a) = 0 (i.e., we have a critical point at x = a), then there is a relative maximum at x = a, if the signs of f change from positive to negative, a relative minimum at x = a, if the signs of f change from negative to positive, and a saddle at x = a, if the signs are positive on both sides or negative on both sides. Second Derivative Test There is another way to check called the second derivative test. It s generally a bit easier to use, but it doesn t always give us an answer, and the reason for that is an important concept in itself. It never tells us that we have a saddle, for example. It does tell us if we have a max or min most of the time. Here s the basic idea. At a relative maximum, the derivative changes from + to. In particular, f (x) is decreasing. The opposite is true at a relative minimum, where the derivative changes from to +, and so f (x) is increasing. Remember that the sign of f tells us if f is going up or down. Now, we want to know if f is decreasing and increasing, and so the derivative of f (x), the second derivative f (x), indicates this by being negative or positive. Visually, if f (x) < 0 around a critical point, the graph is concave down, and so we must be looking at a relative maximum. Similarly, if f (x) > 0 around a critical point, the graph is concave up, and so we must have a relative minimum. Remember that concave up looks like this. And concave down looks like this. Theorem 2. (Second Derivative Test) Suppose that we have a function f, and f (a) = 0 (we have a critical point). If f (a) > 0, then there is a relative minimum at x = a. If f (a) < 0, then there is a relative maximum at x = a. If f (a) = 0, then we may have a max, min, or saddle, but the second derivative doesn t tell us which. The second derivative will be zero at a saddle, but if a relative max or min is particularly flat, the second derivative may be zero there also. 1

2 MA 2231 Lecture 22 - Concavity and Relative Extrema 2 Example 1. Consider the function f(x) = x 4. The derivative of this function is f (x) = 4x 3. Clearly, the only critical point is at x = 0. The second derivative is f (x) = 12x 2, and f (0) = 12(0) 2 = 0. This is a case, where we have a relative minimum, but the graph is so flat, the second derivative cannot detect any concavity. You can kind of see this in the graph of f(x) = x 4 shown below. Basic Principle 1. The first derivative gives us information about the tangent line, so it only tells us stuff that a straight line can see. The second derivative gives us information about how well a quadratic curve can fit the graph of the function, and so it can only tell us stuff that a quadratic can see. Example 2. Find the critical numbers for the following function, and determine if there are relative maximums or minimums at each. (1) f(x) = x 3 6x 2 + 9x + 4. Taking the derivative, we have (2) f (x) = 3x 2 12x + 9 = 3(x 2 4x + 3) = 3(x 1)(x 3), and so there critical points at x = 1, 3. The second derivative for this function (It s easiest to find the second derivative from the expanded form of f ) is (3) f (x) = 6x 12. Checking each critical number with the second derivative gives us (4) (5) f (1) = 6(1) 12 = 6 < 0 and f (3) = 6(3) 12 = 6 > 0 We must, therefore, have a relative max at x = 1 and a relative min at x = 3. Example 3. Consider the function (6) f(x) = x 3 3x 2 + 3x + 7.

3 MA 2231 Lecture 22 - Concavity and Relative Extrema 3 The (first) derivative is (7) f (x) = 3x 2 6x + 3 = 3(x 2 2x + 1) = 3(x 1) 2, we only have one critical point, which is at x = 1. The second derivative is (8) f (x) = 6x 6, and checking x = 1 in the second derivative, we get (9) f (1) = 6(1) 6 = 0. The second derivative tells us nothing. Since the derivative is clearly always positive, we know there is a saddle here. So we ve seen a relative min and a saddle give us a zero in the second derivative. The graph of f is shown below. Quiz 22 Consider the function (10) f(x) = 3x x x Find the critical numbers. 2. Use the 2nd Derivative Test. 3. Use 1st Derivative Test, if necessary. Homework 22 For each of the given functions, find the critical points, and use the second derivative test. Your answers should be relative max, relative min, or don t know. 1. f(x) = x f(x) = x 2 6x f(x) = x 2 + 2x f(x) = x f(x) = 2x 3 9x x f(x) = x 3 + 3x 2 9x f(x) = 3x 4 4x

4 MA 2231 Lecture 22 - Concavity and Relative Extrema 4 Odd Answers. 1. f(x) = x 2 3. f (x) = 2x, so x = 0 is the only critical point. The second derivative is f (x) = 2, and f (0) = 2 > 0. Therefore, we have a relative minimum at x = 0. Compare this to the graph of f shown below. 3. f(x) = x 2 + 2x 7. f (x) = 2x + 2 = 2(x + 1), so the only critical point is at x = 1. The second derivative is f (x) = 2, and f ( 1) = 2 > 0. Therefore, we have a relative minimum at x = 1. Compare this to the graph of f shown below.

5 MA 2231 Lecture 22 - Concavity and Relative Extrema 5 5. f(x) = 2x 3 9x x + 4. f (x) = 6x 2 18x + 12 = 6(x 2 3x + 2) = 6(x 1)(x 2), so we have critical points at x = 1, 2. The second derivative is f (x) = 12x 18, and f (1) = 6 < 0 and f (2) = 6 > 0. Therefore, we have a relative maximum at x = 1 and a relative minimum at x = 2. Compare this to the graph of f shown below. 7. f(x) = 3x 4 4x f (x) = 12x 3 12x 2 = 12x 2 (x 1), so we have critical points at x = 0, 1. The second derivative is f (x) = 36x 2 24x, and f (0) = 0, and f (1) = 12 > 0. Therefore, the second derivative tells us nothing about the critical point at x = 0, but there is a relative minimum at x = 1. Compare this to the graph of f shown below.