Lecture 9 - Increasing and Decreasing Functions, Extrema, and the First Derivative Test

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1 Lecture 9 - Increasing and Decreasing Functions, Extrema, and the First Derivative Test 9.1 Increasing and Decreasing Functions One of our goals is to be able to solve max/min problems, especially economics related examples. We start with the following definitions: Definition 9.1 A function f is called increasing on an interval (a, b) if for any x 1, x 2 (a, b), we have that x 1 < x 2 f(x 1 ) < f(x 2 ) A function f is called decreasing on an interval (a, b) if for any x 1, x 2 (a, b), we have that x 1 < x 2 f(x 1 ) > f(x 2 ) Note that some books call these strictly increasing and strictly decreasing respectively. Examples: 1. f(x) = x 2. f(x) is increasing on (0, ) and decreasing (, 0). 2. f(x) = x 3. f(x) is increasing on (, ). As you may have guessed, we can use the derivative to test for increasing/decreasing. Let f be differentiable on the interval (a, b). 1. If f (x) > 0 for all x in (a, b), then f is increasing on (a, b). 2. If f (x) < 0 for all x in (a, b), then f is decreasing on (a, b). 3. If f (x) = 0 for all x in (a, b), then f must be constant on (a, b). FACT: If f is a continuous function, then f (x) can only change signs at values of x where f (x) = 0 or f (x) doesn t exist. Examples: 1. Let f(x) = x3 3x. Find the intervals on which f(x) is increasing or decreasing. 4 solution: We will use the test and fact above. The derivative is f (x) = 3x2 4 3 = 3 4 (x2 4) = 3 4 (x 2)(x + 2) Because of the above, we split our real line up where f (x) = 0:

2 2 2 f(x) On (, 2), (x 2) and (x + 2) are both negative, so f (x) = 3 (x 2)(x + 2) > 0, 4 so f is increasing. On ( 2, 2), (x 2) is negative while (x + 2) is positive, so f (x) = 3 (x 2)(x + 2) < 0, 4 so f is decreasing. On (2, ), (x 2) and (x + 2) are both positive, so f (x) = 3 (x 2)(x + 2) > 0, so f 4 is increasing. To summarize, f(x) is increasing on (, 2) and (2, ) while f(x) is decreasing on ( 2, 2). 2. Let g(x) = x + 32 x 2. Find the intervals on which g(x) is increasing and decreasing. solution: We first need to find the derivative. g (x) = 1 64 x 3 Note that g (x) = 0 when x 3 = 64 x = 4. The derivative is also not defined when x = 0, so these are the places we divide our real line. 0 4 g(x) To find out what sign the derivative is on (, 0), we can substitute in a test point in that interval and see if the result is positive or negative. The point x = 1 is a convenient one to use: g ( 1) = 1 64 ( 1) 3 = 65 > 0 Because of our fact, g (x) will be positive everywhere in (, 0), and thus g is increasing on (, 0). Similarly, g (1) = 1 64 (1) 3 = 63 < 0 So g (x) will be negative everywhere in (0, 4), and thus g is decreasing on (0, 4).

3 Lastly, g (5) = 1 64 (5) 3 = > 0 So g (x) will be positive everywhere in (4, ), and thus g is increasing on (4, ). 9.2 Extrema and the First Derivative Test We now have enough information to sketch these graphs. 1. First let s graph f(x) = x3 4 3x. First we find the intercepts. To get the x-intercept we solve f(x) = 0: ( ) x 3 x 2 4 3x = 0 x 4 3 So the x-intercepts are x = 0, ± 12 ±3.5. The y-intercept is at (0, 0). We know that the function changes from increasing to decreasing and vice-versa when x = ±2. Plugging those values into our function we see that the points (2, 4) and ( 2, 4) are on our graph. = 0 2. Similarly, g(x) = x + 32 x 2. This function has no y-intercept, because 0 is not in its domain. To find the x-intercept we solve g(x) = 0: x + 32 x 2 = 0 x3 = 32 x =

4 We have a vertical asymptote at x = 0 because lim g(x) = + = lim g(x). x 0 + x 0 Finally, we calculate the value of g at x = 4 and find that the point (4, 6) is on the graph. The point (4, 6) in the above graph is called a relative minimum. Definition 9.2 Let f be a function defined at a point c. Then 1. We say that f has a relative minimum (or local minimum) at x = c if there is an interval (a, b) containing c such that f(c) < f(x) for all x in (a, b). 2. We say that f has a relative maximum (or local maximum) at x = c if there is an interval (a, b) containing c such that f(c) > f(x) for all x in (a, b). The word relative is significant here. We are not saying that a relative max is largest value f(x) ever takes. We will get to such points soon. Together, relative maxes and mins are called relative extrema. Theorem 9.1 If f(x) has a relative max or min at x = c, then either f (c) = 0 or f (c) doesn t exist. Definition 9.3 If a point c is in the domain of f and either f (c) = 0 or f (c) doesn t exist, then we call x = c a critical point or critical value of f.

5 Examples: 1. f(x) = x 2. x = 0 is a critical point and a local min. f (0) = 0 2. f(x) = x. x = 0 is a critical point and a local min. f (0) is undefined. We notice that in the examples above that we have local mins with intervals of negative slope to the left and positive slope to the right. Under some conditions, this is what happens in general.

6 Theorem 9.2 (The First Derivative Test): Let f be continuous on the interval (a, b) and suppose that c is the only critical point of (a, b). Suppose f is differentiable on (a, b) except possibly at c. Then: 1. If f (x) is negative to the left of c and positive to the right of c, then f has a local min at c. 2. If f (x) is positive to the left of c and negative to the right of c, then f has a local max at c. 3. If f (x) is the same sign to the right and left of c then c is neither a max nor a min. In cases like 3 above we call the point c a saddle point if f (c) = 0. Examples: Find the critical points and identify their type for the following functions. 1. f(x) = 2x 3 + 9x 2 108x solution: We first have to find the critical points f (x) = 6x x 108 = 6(x 2 + 3x 18) = 6(x + 6)(x 3) So the critical points are x = 6, x = 3. To fill in our real line we plug in some test points from (, 6), ( 6, 3) and (3, ): Thus we have the following f ( 7) = 6( 7 + 6)( 7 3) = 60 > 0 f (0) = 6(6)( 3) = 108 < 0 f (4) = 6(4 + 6)(4 3) = 60 > f (x) f(x) So by the first derivative test we can conclude that f(x) has a local max at x = 6 and a local min at x = 3.

7 2. h(x) = x 2 e 3x. solution: We first have to find the critical points: h (x) = 3x 2 e 3x + 2xe 3x = x(3x + 2)e 3x So the critical points are x = 0, x = 2. Notice that the 3 e3x doesn t give us any critical points because it is defined everywhere and is never 0. To fill in our real line we plug in some test points from (, 2), ( 2, 0) and (0, ): 3 3 h ( 1 2 Thus we have the following h ( 1) = ( 1)( 3 + 2)e 3 = e 3 > 0 ) ( = 1 2 ) ( ) e 3/2 = 1 4 e 3/2 < 0 h (1) = (1)(3 + 2)e 3 = 5e 3 > h (x) h(x) So by the first derivative test we can conclude that h(x) has a local max at x = 2 3 and a local min at x = f(x) = x 1/3 e x/6. solution: Notice that this function is defined everywhere, i.e. its domain is R. We once again find the critical points: ( ) 1 f (x) = x 1/3 6 ex/ x 2/3 e x/6 ( x = e x/6 1/ ) 3x 2/3 = e x/6 ( x 6x 2/3 + 2 = e x/6 ( x + 2 6x 2/3 ) 6x 2/3 So f (x) = 0 at x = 2 and is undefined at x = 0, and thus we have two critical points. To fill in our real line we plug in some test points from (, 0), (0, 2) and (2, ): ( ) 1 f ( 3) = e 3/6 < 0 6( 3) 2/3 )

8 Notice that the denominator is always positive (why?). ( ) f ( 1) = e 1/6 1 > 0 6( 1) 2/3 ( ) 3 f (1) = e 1/6 > 0 6(1) 2/3 Thus we have the following 2 0 f (x) 0 + UN + f(x) So there is a local min at x = 2 and at x = 0 there is neither a max nor a min (notice that there is not a saddle point there because the derivative doesn t exist there). It is important to note that x = 0 is still a critical point. THIS IS A CRUCIAL ISSUE In the previous example, there was a critical point at x = 0 since f was undefined at x = 0, but 0 was not in the domain of f. If x = c is not in the domain, you cannot have a critical point at x = c. Example: Let f(x) = 1 x. Then f (x) = 1 x 2. f is undefined at x = 0, but there is no critical point there since f is also undefined at x = 0. Absolute Extrema In ecomonics applications, we are generally uninterested in local maxes and mins. Instead we want to know when our function is maximal or minimal on its entire domain. Definition 9.4 Let f be defined on an interval I (possibly all of R) containing c. Then f is said to have an absolute maximum or global maximum on I if f(c) f(x) for every x in I. There is an analogous statement for absolute/global minimums. The next theorem is both deep and extremely important.

9 Theorem 9.3 (Extreme Value Theorem): If f is continuous on a closed interval [a, b], then f has both an absolute max and an absolute min on [a, b]. Note that the word extreme in the name of the theorem above is modifying value and not theorem. I am not making any claims as to how extreme this theorem is. The theorem is certainly false for other intervals like R. f(x) = e x has no maxes or mins on R. To find extrema on a closed interval [a, b] proceed as follows: 1. Evaluate f at each critical point in [a, b]. 2. Evaluate f at the endpoints a and b. 3. The least of these values is the absolute min and the greatest is the absolute max. Examples: 1. Find the absolute max and min of f(x) = 2x 3 3x 2 36x + 2 on the interval [0, 5]. solution: We first find the critical points f (x) = 6x 2 6x 36 = 6(x 2 x 6) = 6(x 3)(x + 2) So the critical points are x = 3 and x = 2. We can throw away x = 2 because it isn t in [0, 5]. Now we calculate f at all relevant points. f(0) = 2 f(3) = 2(27) 3(9) 36(3) + 2 = 79 f(5) = 2(125) 3(25) = 3 So the absolute max occurs at x = 0 and the absolute min is at x = 3. Note that the absolute maxes and mins can occur at endpoints. 2. A company finds that the profit P (x) where x represents thousands of units, is given by P (x) = x 3 + 9x 2 15x 9 If the company can only make a maximum of 6000 units, what is the absolute maximum profit? solution: Since x is in thousands of units, we must find the absolute max of P (x) on the interval [0, 6]. P (x) = 3x x 15 = 3(x 2 6x + 5) = 3(x 5)(x 1)

10 So there are two critical points at x = 1, 5. They are both in [0, 6], so we have four points to calculate: P (0) = 9 P (1) = (1) 3 + 9(1) 2 15(1) 9 = 16 P (5) = = 16 P (6) = = 9 The absolute max occurs at x = 5, so the company should make 5000 units to maximize profit. We close out this section with a helpful theorem. Theorem 9.4 If f is a continuous function on an interval and has only one critical value in that interval, then a relative max or min is also an absolute max or min.