Using the Derivative. Chapter Local Max and Mins. Definition. Let x = c be in the domain of f(x).

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1 Chapter 4 Using the Derivative 4.1 Local Max and Mins Definition. Let x = c be in the domain of f(x). x = c is a local maximum if f(x) apple f(c) for all x near c x = c is a local minimum if f(x) (we allow endpoints) f(c) for all x near c (we allow endpoints) Definition. If x = c is in the domain of f and f 0 (c) = 0, then we call x = c a critical point. We also call the (x, y)-point (c, f(c)) a critical point. We call f(c) the critical value. Example 1. Let f(x) =4x 3 +3x 2 6x. (a) Find the critical points of f(x) algebraically. (b) Use a graph to classify each critical point as a local max/min/neither. Solution. (a) f 0 (x) = 12x 2 +6x 6 12x 2 +6x 6=0 6(2x 2 + x 1) = 0 6(2x 1)(x + 1) = 0 (2x 1) = 0 ) 2x =1) x =1/2 or (x + 1) = 0 ) x = 1 (b) The graph of f(x) isshownbelow 72

2 CHAPTER 4. USING THE DERIVATIVE 73 Before looking at the graph we knew that x = 1 and x =1/2were critical points. Now we can tell that x = 1 is a local max and x =1/2 is a local min. This is where we ended on Wednesday, March 13 Theorem 1 (First derivative test). To find the local max/mins of a function f(x) do the following. 1. First find the critical points. 2. Figure out whether f 0 (x) is + or on each side of each critical point (four cases, lots of pictures): f 0 (x) =+, left of c right of c outcome + ) x = c local max + ) x = c local min + + ) x = c neither ) x = c neither Example 2. Let f(x) =e 2x +5x. (a) This function has one critical point; find it algebraically. (b) Using your calculator (or crude approximations), evaluate f 0 (x) at two x- values, one on the left and one on the right of the critical point. (c) Apply the first derivative test to identify the critical point as a local max/min/neither. Solution. (a) f 0 (x) = 2e 2x +5 2e 2x +5=0 2e 2x =5 e 2x =5/2 ln(e 2x )=ln(5/2) 2x =ln(5/2) x = 1 2 ln(5/2) x 0.46

3 CHAPTER 4. USING THE DERIVATIVE 74 (b) If you use your calculator, you can pick almost any values you want. I m going to pick ones that I can do without a calculator. For instance, I can calculate f 0 (0) f 0 (0) = 2e (reminder: plug into the derivative) =3. I can t probably calculate f 0 (x) for a number to the left of x = 0.46, but I can estimate it enough to see if it s positive or negative. For instance, e is roughly 2 or 3. So, e 20 will be really, really big, like maybe a million (actually even bigger). Therefore 2e 20 will be approximately negative two million. Thus: f 0 ( 10) = 2e = #. million (c) Since f 0 (x) is negative to the left of x = 0.46, and positive to the right, we conclude that x = 0.46 is a local minimum. This is where we ended on Monday, March 18 Definition. A 1D#table (1st Derivative Number Line Table) shows the following: 1. A number line, with each critical point marked on the number line. The critical points divide the line into regions. Any points where the f(x) is discontinuous also divide the line into regions. 2. The following information about the derivitave should be recorded below the line. Each region of the line, and each critical point, should be labeled with f 0 > 0, f 0 < 0 or f 0 = 0 (or something equivalent). 3. The following information about the function should be recorded above the line. Each region of the line should be labeled with f " or f # (or something equivalent). Each critical point should be labeled with l.max, l.min, or neither (or something equivalent). Example 3. Summarize the information from Examples 1 and 2 in 1D#tables. Solution. Examples 1 is summarized below f " max min f # x = 1 x = 1/2 f " f 0 =+ f 0 =0 f 0 = f 0 =0 f =+ Example 2 is summarized below f & local min f % x = 0.46 f 0 < 0 f 0 =0 f 0 > 0

4 CHAPTER 4. USING THE DERIVATIVE 75 Note: I purposely wrote the information a little bit di erently in each table. What is mandatory: that you give the information in some fashion, and that you label it as f or f 0. Example 4. The data below show information about f 0 (x). where the local max and local mins for f(x) are. Use it to estimate x f 0 (x) Solution. We see that f 0 (x) changes sign from + to around 147. So, 147 would be a local max. Similarly, we see that f 0 (x) changes sign from to + around 167, so 167 would be a local min. These are the only two critical points. The endpoints are also local max/mins. The first endpoint, x = 110 is a local min, because f 0 (x) is + to the right. The second endpoint, x = 200 is a local max, because f 0 (x) is+totheleft. Example 5. A restaurant is analyzing their profits to see if they can be improved. Let (d) be their total profit for the current fiscal year on day d. Thus (1) would be how much profit they make on Januray 1, and (365) would be how much profit they ve made over the entire year, up to December 31. In this case, daily profit, i.e. how much profit they made on a single day, is basically the same as marginal profit, i.e. 0 (d) daily profit. The data below shows daily profit. Use it to estimate where the local max and local mins for total annual profit are. day daily profit Solution. We identify daily profit with 0 (d) and apply the first derivative test. We see that 0 (d) changes sign from to + around 127. So, 127 would be a local min. Similarly, we see that 0 (d) changes sign from + to around 152, so 152 would be a local max. These are the only two critical points. The endpoints are also local max/mins. The first endpoint, x = 110 is a local max, because 0 (d) is to the right. The second endpoint, x = 200 is a local min, because f 0 (x) is to the left. Theorem 2 (Second derivative test). To find the local max/mins of a function f(x) try the following. 1. First find the critical points. 2. Figure out whether f 00 (c) is + or (three cases): f 00 (c) outcome + local min local max 0 or DNE test says nothing

5 CHAPTER 4. USING THE DERIVATIVE 76 Example 6. The function f(x) = ln(x) has a critical point at x = e. Use the x second derivative test to identify it as a local max/min. Solution. We start by finding the first derivative, using the quotient rule. Since we state our quotient rule using f and g, maybe it would be nice to use a di erent letter for the original function. Let s use y. y = ln(x) x f g Now let s calculate f 0 and g 0 : Now let s write down y 0 : y 0 = f 0 = 1 x g 0 =1 1 x x ln(x) 1 (x) 2. Now is the time to start simplifying things after taking the derivative. Why? Because we can; that is, this formula can simplify a fair amount. Also, because we need to do something else with it, namely take the derivative again. y 0 = 1 x x ln(x) 1 (x) 2 = 1 ln(x) x 2 To take the derivative again, let s label f and g: y 0 = 1 ln(x) x 2 f g Now we calculate f 0 and g 0 : f 0 = 1 x g 0 =2x Now we write down y 00 : y 00 = 1 x (x2 ) (1 ln(x))(2x) (x 2 ) 2. We don t have a lot of reason to simplify this, unless it would make it easier to plug a number into. Maybe a little simplification would help: y 00 = x 2x(1 ln(x)) x 4.

6 CHAPTER 4. USING THE DERIVATIVE 77 Now, we plug in x = e: y 00 (e) = e 2e(1 ln(e)) e 4. This is complicated enough that I m more worried I ll make a mistake plugging it into my calculator than by simplifying it. So, I ll simplify it: y 00 (e) = e 2e(1 ln(e)) e 4 e 2e(1 1) = e 4 = e 0 e 4 = 1 e 3 = # From this, we see that x = e is a local max.