Chapter 12 Overview: Review of All Derivative Rules

Transcription

1 Chapter 12 Overview: Review of All Derivative Rules The emphasis of the previous chapters was graphing the families of functions as they are viewed (mostly) in Analytic Geometry, that is, with traits. The focus of this chapter is sketching graphs of the various functions, but with a Calculus outlook. First, it would be best to review the traits of each kind. REMEMBER TRAITS: 1. Domain 2. Range 3. y-intercept 4. Zeros 5. Vertical Asymptotes 6. Points of Exclusion 7. End Behavior 8. Extreme Points In this chapter, graphs are determined by the sign patterns of the function, its derivative, and the derivative of the derivative. Rather than the traits above, the sign patterns emphasize what is referred to as KEY TRAITS: Zeros and/or VAs Extreme Points Points of Inflection The interpretations of the sign patterns of these key traits are connected by way of the First Derivative Test and the Concavity Test. That is, the zeros of the 1st Derivative are the critical values of the original function; the increasing and decreasing intervals of the derivative are the intervals of concavity of the original function, etc. 673

2 Interpretation of Sign Patterns: Sign + 0 f ( x) Curve above x-axis Zero Curve below x-axis f!( x) Increasing Critical Value Decreasing f!( x) Concave Up POI Concave Down 674

3 12-1: Second Derivatives This section explores some of the meanings of the 2nd derivative (that is, the derivative of the derivative). But before that, it is best recap all derivative rules: The Power Rule: The Product Rule: The Quotient Rule: The Chain Rule: d dx! " sinu # = ( cosu) du dx d dx! " cosu # = (%sinu) du dx d dx! " tanu # = ( sec 2 u) du dx d! dx " eu d! dx " au # = eu du dx du # = au i lna dx d " dx # sin!1 u % = 1 1! u i D 2 u d " dx # cos!1 u % =!1 1! u i D 2 u d " dx # tan!1 u % = 1 u 2 +1 i D u d dx un = nu n!1 i du dx d ( dx u i v ) = u i dv dx + v i du dx d dx d dx! # "! " u x v x & = % f g ( x ) du vi dx ' u i dv dx v 2 # = f % ( g ( x ))i g% x d dx! " cscu # = (%cscucotu) du dx d dx! " secu # = ( secu tanu) du dx d dx! " cotu # = (%csc 2 u) du dx d dx!" lnu# = % 1 ( du & ' u ) * dx d dx!" log a u# = % 1 ( du & ' u i lna) * dx d " dx # csc!1 u % =!1 u u 2!1 i D u d " dx # sec!1 u % = 1 u u 2!1 i D u d " dx # cot!1 u % =!1 u 2 +1 i D u 675

4 What had previously been called the derivative is actually the 1st derivative. There can be successive uses of the derivative rules, and they have meanings other than the slope of the tangent line. This section will explore the process of finding the 2nd derivative and will explore the meanings in subsequent sections. Vocabulary: 1. Second Derivative the derivative of the derivative Just as with the 1st derivative, there are several symbols for the 2nd derivative: Second Derivative Symbols: d 2 y dx 2 d 2 dx 2 f!( x) = f double prime of x y! = y double prime LEARNING OUTCOMES Find the second derivative. Apply the Second Derivative Test. EX 1 d2 " x 4! 7x 3! 3x 2 + 2x! 5 dx 2 # % d 2 " x 4! 7x 3! 3x 2 + 2x! 5 dx 2 # % = d " d dx & " dx # x4! 7x 3! 3x 2 + 2x! 5 # = d " dx # 4x3! 21x 2! 6x + 2 % =12x 2! 42x! 6 %' % 676

5 More complicated functions, in particular composite functions, have a complicated process. When the Chain Rule is applied, the answer becomes a product or quotient. Therefore, the 2nd derivative will require the Product or Quotient Rules as well as, possibly, the Chain Rule again. EX 2 y = e 3x2, find d2 y dx 2. dy dx = e3x2 i 6x = 6xe 3x2 d 2 y dx = 6x ( 2 e3x2 i 6x)+ e 3x2 i 6 = 36x 2 e 3x2 + 6e 3x2 = 6e ( 3x2 6x 2 +1) EX 3 y = sin 3 x, find d2 y dx 2 y! = 3sin 2 x i cosx y" = 3sin 2 x (#sin x)+ cosx 6sin x i cosx = 3sin x( 2cos 2 x! sin 2 x) 677

6 EX 4 f ( x) = ln( x 2 + 3x!1), find f!( x). f!( x) = f # x 1 ( x 2 + 3x "1 2x + 3 ) = 2x + 3 x 2 + 3x "1 " ( 2x + 3) ( 2x + 3) 2 ( x 2 + 3x "1) 2 " ( 4x 2 +12x + 9) ( x 2 + 3x "1) 2 = x2 + 3x "1 = 2x2 + 6x " 2 =!2x2! 6x!11 x 2 + 3x!1 2 EX 5 g( x) = 4x 2 +1, find g! ( x). g! ( x) = 1 2 ( 4x2 +1) " 1 2 ( 8x) = g# ( x) = = ( 4x 2 +1) 12 ( 4)" 4x 4x ( 4x 2 +1) 1 2 & % " 1 2 8x 1 2 4x & % ( 4x 2 +1) 12 ( 4)" 4 ( 4x 2 +1) 3 2 4x 2 +1 ' ) ( 16x 2 ( 4x 2 +1) 1 2 4x 2 +1 = 4x2 +1 "16x = ( 4x 2 +1) 3 2 ' ) ( 678

7 The Second Derivative Test: The 1st Derivative Test (which has been used through previous chapters) is not the only way to test whether critical values are associated with maximum or minimum points. There is a 2nd Derivative Test, which determines the same thing, but it does not use a sign pattern. The sign pattern of the 2nd derivative implies something other than whether the critical value is at a maximum point vs. at a minimum point. What it implies (and why) will be discussed in the next section; for now, here is the Second Derivative Test: For a function f ( x), 1. If f c value at x = c. 2. If f c value at x = c. The Second Derivative Test: = 0 and f!( c) > 0, then f ( x) has a relative minimum = 0 and f!( c) < 0, then f ( x) has a relative maximum Note that the 2 nd Derivative Test does NOT work for critical values determined by f c not existing or for endpoints. 679

8 EX 6 Use the 2nd Derivative Test to determine if the critical values of g t = 27t! t 3 are at a maximum or minimum value. g ( t) = 27 3t 2 = 0 t = ±3 So t =!3 and t = 3 are critical values. Next, find the g! ( t) and substitute the critical values: g! t = "6t = "6("3) =18 = "6( 3) = "18 g! "3 g! 3 Therefore, g has a minimum value at t =!3 and has a maximum value at t = 3. (Note that the numerical value 18 is irrelevant. Only the sign matters.) 680

9 EX 7 Determine if the critical values of y = ( 3! x 2 )e x are at a maximum or a minimum value. First, find the critical values: The derivative is never DNE. dy = ( 3! x2 )e x + e x!2x dx =!e x x 2 + 2x! 3 = 0 Critical values: x =!3,!1 Find the 2nd derivative and substitute the critical values: dy dx =!ex x 2 + 2x! 3 d 2 y dx = 2!ex ( 2x + 2) + x 2 + 2x! 3 (!e x ) =!e x x 2 + 4x!1 d 2 y =!4e dx 2 x=1 d 2 y = 4e!3 dx 2 x=!3 Therefore, x = 1 is at a maximum value and x =!3 is at a minimum value. 681

10 12-1 Multiple Choice Homework 1. The graph of f is shown in the figure to the right. Which of the following could be the graph of the derivative of f? 2. The maximum acceleration attained on the interval 0! t! 3 by the particle whose velocity is given by v t = t 3! 3t is (a) 9 (b) 12 (c) 14 (d) 21 (e)

11 3. If d dx! " f ( x) # = g( x) and d dx g( x) + 2g( x 2 ) + 2g( x 2 ) (a) 4x 2 f 3x 2 (b) f 3x 2 (c) f x 4 (d) 2xf 3x 2 (e) 2xf 3x 2! " # = f 3x, then d 2! f x 2 dx 2 " # is 4. If y = e kx, then d 5 y dx 5 = (a) k 5 e x (b) k 5 e kx (c) 5!e kx (d) 5!e x (e) 5e kx = 1, f!( 2) = 4, and = 3. What is the value of the approximation of f ( 1.9) using the line tangent 5. The function is twice differentiable with f 2 f!! 2 to the graph of at x = 2? (a) 0.4 (b) 0.6 (c) 0.7 (d) 1.3 (e)

12 12-1 Free Response Homework Evaluate the 2nd derivative. 1. d 2 " dx 2 # 5x4 + 9x 3! 4x 2 + x! 8 % 2. d 2 " 4x 7! 3x 5 + 3x 3 + 6x!1 dx 2 # % 3. y = cos x 2, find y!! 4. y = tan 2 x, find y!! 5. y = sec3x, find d2 y dx 6. y = 2 xe2x, find d2 y dx 2 7. f ( x) = ln x h x, find f!! ( x) 8. g x = x 2 + 5, find h!! ( x) 10. F x, find g!! ( x) = ln x 2! 4x + 4 = 3x 2! 2x +1, find F!! 11. y = x2! 3 x 2!10, find d2 y 3x y = dx x 3 +1, find d2 y dx 2 ( x) Find the critical values of the following functions and use the 2nd Derivative Test to determine which are at minimum points and which are at maximum points. 13. y = 2x 3 + 9x 2!168x 14. y = x x y = 9x 3! 4x 2! 27x y = 3x 9! x y = x 2 i 3 9! x y = ( x! x 2 )e x 19. y = 3 2 x + cos x on x! "2#, 2# % & ' 684

13 Find a( t) if the given equation describes the position of a particle at time t. 20. x( t) = t 3! 6t 2! 63t x( t) = 3t 5! 38t t 3!180t x t 23. x( t) = # = 4 + 9sin! ( 4 t " 5 ) t t x( t) = 1+ e t % & ( ' Find a( t). 25. x t # % =!5 + 8cos " ( 4 t! 3 ) 26. x( t) = 1+ ln( t 2 ) ; c =!1 & ( ; c = 6 ' 685

14 12-2: Concavity and Points of Inflection One trait of algebraic and transcendental curves, which was not previously explored, is concavity. Vocabulary: 1. Concavity the way a curve bends 2. Concave Up the curve bends counter-clockwise 3. Concave Down he curve bends clockwise 4. Inflection Value the x-value at which the concavity of the curve switches. This can be a vertical asymptote. 5. Point of Inflection the point at which the concavity of the curve switches from up to down or down to up The easiest way to see this is with a sine wave. On x!( 0, " ), the curve is concave down. On x!(", 2" ), it is concave up. The 2nd derivative is what is used to analyze this trait. Points of inflection are to the 2nd derivative as extreme points are to the 1st derivative, for the most part. 686

15 If x = c is at a Point of Inflection (POI), then i) f! x or ii) f! x = 0 does not exist. NB. Endpoints on a domain CANNOT be Points of Inflection. Also, intervals of concave up or down are comparable to intervals of increasing or decreasing, respectively. LEARNING OUTCOME Find points of inflection and intervals of concavity. EX 1 Given this sign pattern for the second derivative of F x determined about F x F x? F! ( x) "!!!! 0!+!0! " 0!!!!!+ %%%%%%%%%%% x "1!! 0! 2 #, what can be is concave up where the second derivative is positive namely, x!("1, 0)# ( 2, ) is concave down where the second derivative is negative namely, x!("#, "1) ( 0, 2) F x F( x) has inflection values where the signs change namely, x =!1,!0,!and!2 Note that this use of the sign pattern of the 2nd derivative is known as the Second Derivative Test for Concavity, not the Second Derivative Test. 687

16 EX 2 Given this sign pattern concave down? y! + DNE " 0 + DNE " #%%%%%%%%%%%%%, where is y x " 2! 0!!! 2 It can be concluded that y is concave down on x!("2, 0) and ( 2, #). EX 3 Find the points of inflection of y = x 3 + 4x 2! 3x +1. y! = 3x 2 + 8x " 3 y# = 6x + 8 = 0 x =! 4 3 y! is never DNE. y! " 0! + x " 4 3!!!! #%%%%%%%% x =! 4 3 " y = " # % &! 4 3, ' is the POI. 688

17 EX 4 Find the points of inflection of y = sin x on x!( 0, 2" ). dy dx = cos x d 2 y dx 2 =!sin x = 0 x = 0 ± "n d 2 y 2 always exists (result is never DNE). dx On the given interval, the only inflection value is x =!. y! END # 0 + END x 0 " 2" Because of the sign change in the sign pattern x =! is the x-coordinate of the POI. x =! " y = 0, so (!, 0) is the POI. EX 5 Find the points of inflection of y = y! = x2 + 4 " 5x x x x ( 2x) = "5x ( x 2 + 4) 2 " ("5x ) 2( x 2 + 4) 2x ( x 2 + 4) 4 ("10x)#( x 2 + 4)" 2( x 2 " 4 & % )'( ( x 2 + 4) 4 = 0 ) x = 0, ± 2 3 ( x 2 " 4) 3 ( y!! = x2 + 4) 2 "10x = x2 + 4 = 2x "x2 +12 ( ) 689

18 y!! +!0! " 0 +!!!0! " #%%%%%%%%%%%%%%% x " 3! 0!!! 3 Because of the change in the sign pattern, x = 0, ± 2 3 are the x-coordinates of the POIs. " ( 0, 0),! 3,!5 3 7 # % ' &! 5 3, and # 3, 7 " & % are the POIs y = 5x x The graph shows that ( 0, 0) is a point of inflection. But it shows something else. Namely, that the concavity switches (as it often does) on either side of the vertical asymptotes. The vertical asymptotes cannot be points of inflection, because they are not points, but one must be aware of this in cases where the question is about concavity rather than about POI. The process is similar to the 1st Derivative Test, including the need to check where the derivative does not exist. It will be referred to as the 2nd Derivative Test for Concavity. 690

19 EX 6 Find the intervals of concavity for y = xe 2x. ( ) + e 2x ( 1) = e 2x ( 2x +1) dy dx = x e2x 2 d 2 y dx = 2 e2x ( 2) + 2x +1 = e 2x!( 2) + 2x +1 e 2x ( 2) " ( 2) = 0 = e 2x 4x + 4 # The inflection value for the POI is x =!1 and the 2nd derivative is never DNE. So, y! " 0 + #%%%%%%% x "1! and and y is concave down on x! "#, "1 y is concave up on x!("1, #). y = xe 2x 691

20 EX 7 Find the points of inflection of y =!x x 2! 4. ( y! = x2 " 4) ("1) " ("x)( 2x) = x2 + 4 ( x 2 " 4) 2 ( x 2 " 4) 2 ( y!! = x2 " 4) 2 ( 2x) " ( x 2 + 4) 2( x 2 " 4) 2x ( x 2 " 4) 4 ( = x2 " 4) ( 2x) #( x 2 " 4) " 2( x & % )'( ( x 2 " 4) 4 = 2x ( "x2 "12) = 0 ) x = 0 ( x 2 " 4) 3 y!! DNE " x = ±2 ( ) But from a previous chapter, x = ±2 are vertical asymptotes; hence they cannot be x-coordinates for POIs. y! + DNE " 0 + DNE " #%%%%%%%%%%%%% x " 2! 0!!! 2 Because of the sign change in the sign pattern we know x = 0 is the x-coordinate of a POI. x = 0! y = 0, so ( 0,!0) is the POI. 692

21 y =!x x 2! 4 693

22 12-2 Multiple Choice Homework 1. An equation of the line tangent to the graph of y = x 3 + 3x at its point of inflection is (a) y =!3x + 1 (b) y =!3x + 7 (c) y = x + 5 (d) y = 3x + 1 (e) y = 3x The number of inflection points for the graph of y = 2x + cos( x 2 ) in the interval 0! x! 5 is (a) 6 (b) 7 (c) 8 (d) 9 (e) On which of the following intervals is the graph of the curve y = x 5! 5x x + 15 concave up? I. x < 0 II. 0 < x < 3 III. x > 3 (a) I only (b) II only (c) III only (d) I and II only (e) II and III only 694

23 = x 2 ( x " 3) x " 6 4. Let f be a function with a second derivative given f!! x What are the x-coordinates of the points of inflection of the graph of f? (a) 0 only (b) 3 only (c) 0 and 6 only (d) 3 and 6 only (e) 0, 3, and The derivative of the function is given by f!( x) = x 2 cos( x 2 ). How many points of inflection does the graph of have on the open interval!2, 2 (a) One (b) Two (c) Three (d) Four (e) Five? 12-2 Free Response Homework 1. Given this sign pattern for the second derivative of F x interval(s) is F x concave down? F! ( x) "!!! 0!+!!0! " 0!!!!!+ %%%%%%%%%%% x " # 2. Given this sign pattern for the second derivative of G x interval(s) is G x concave down? G!! ( x) " 0 " x 2 " # %%%%%%%%%%%%, on what, on what 695

24 3. The sign pattern for the second derivative of H ( x) is given. a) Is x =!4 a point of inflection? b) Is x =!1 a point of inflection? d 2 H dx 2 x + 0! 0! 0 + " #! 4! Given these two sign patterns: dy dx x d 2 y dx 2 x!!!! 0 +!!0!!!!!0! 0 + " #!!! 9! 3!!!!!1!! 7! + 0! 0 + 0! 0 + " #! 7! a) On what interval(s) is y decreasing? b) On what interval(s) is y concave up? c) On what interval(s) is y both increasing and concave down? Find the sign pattern of the second derivative and identify the intervals of concavity. 5. y = x 3 + x 2! 7x!15 6. y = 3x 4! 20x x 2! 36x y =!4x x y = x2!1 x 2! 4 9. y = x 8! x y = 1 2 x + sin x on x!( 0, 2" ) 11. y = xe!x 12. y = e!x2 13. y = x x 2! y = 2x! x

25 12-3: Advanced Curve Sketching One now has just about everything to do an accurate, complete curve sketch without a calculator. The zeros, POEs, extreme points, POIs, and vertical asymptotes are specific values that are used to plot the graph. These are referred to as the key traits. What was not explored previously is how to connect the dots more accurately. The intervals and how they interrelate dictate how the specific values are connected. Below are the ways that the intervals of increasing and decreasing can mix with the intervals of concave up and concave down as well as how they affect the graph. Increasing and Concave Up Increasing and Concave Down Decreasing and Concave Up Decreasing and Concave Down 697

26 As seen in the homework of the last section, the sign patterns are not necessarily the best way to organize all the information. A table, with the 1st and 2nd derivative signs and conclusions as the column (or row) headings and the intervals and key traits in the first row (or column), often helps. The columns with the 1st and 2nd derivative signs can be read the same as the sign patterns from earlier. Whether the table is oriented horizontally (see EX 2), like the sign patterns, or vertically (see EX 3) is a matter of personal choice, but, a table with many traits probably should be oriented vertically for all the information to fit. LEARNING OUTCOME Find the key traits and a more complete sketch of any function. EX 1 Sketch a graph that matches the traits given in this table: x x =!3!3,!! 3 f x x =! 3 (! 3,!0) x = 0 ( 0,! 3) x =! 3 ( 3,!3) x = f! x f! x

27 EX 2 Sketch the graph of the function whose traits are given below. x f x DNE 0 DNE DNE DNE + 0 DNE 0 + DNE f! x f! x 6 y x!3!2! !1!2!3!4!5!6 Note that the information given in the table was not from a given equation; therefore, it is not a problem that this graph is not smooth. 699

28 EX 3 Make a table of key traits and a complete sketch of y = xe 2x. Zeros: xe 2x = 0! x = 0! ( 0, 0) y-int: x = 0! y = 0! ( 0, 0) Extreme Points: dy dx = xe2x 2 xe 2x ( 2)+ e 2x 1 + e 2x ( 1) = 0 = 0 e 2x 2x +1 x =! 1 2 " y =!0.184 "! 1 # 2,! % & ' POI: ( ) d 2 y dx = 2 e2x ( 2)+ ( 2x +1) e 2x 2 = e 2x ( 4x + 4) = 0 x =!1" y =!0.135 (!1,! 0.135) " % x (!",!!1 # 2 & ' x =! 1! 1 2 # 2,!0 & ' x = 0 ( 0,!! ) f x f! x f! x Conclusion POI min pt. zero " % 700

29 y = xe 2x Many prefer to make the table horizontal, since this way it matches the sign patterns more directly. The only disadvantage to the horizontal table is whether it fits on the page or not. (The homework answers have vertical tables for that reason.) EX 4 Make a table of key traits and a complete sketch of y = x 4! 2x 3. Zeros: x 4! 2x 3 = 0 " x = 0 or 2 " ( 0, 0), ( 2, 0) Extreme Points: dy dx = 4x3! 6x 2 = 0 " x = 0 or 3 2 " ( 0, 0 ), # % 3 2,! 27 & 16 ' ( POI: d 2 y dx 2 =12x2!12x = 0 " x = 0 or 1" 0, 0, 1,!1 There are several key trait values, so a vertical table might be better. 701

30 Trait y y! y! Conclusion x < Above the x-axis, decreasing and concave up x = Zero and POI 0 < x <1 Below the x-axis, decreasing and concave down x = POI 1< x < Below the x-axis, decreasing and concave up x = 3! Minimum Point 3 < x < 2 2 Below the x-axis, decreasing and concave down x = 2 0 Zero x > Above the x-axis, increasing and concave up y = x 4! 2x 3 702

31 EX 5 Make a table of key traits and a complete sketch of y = Zeros: y = 0! x = "1! ("1, 0) VAs: y =! " x = 3 or # 2 x +1 x 2! x! 6. Extreme Points: ( 1)! ( x +1) ( 2x!1) = 0 ( x 2! x! 6) 2 dy dx = x2! x! 6!x 2! 2x! 5 = 0 x = 2 ± 4! 20 2 = no real solutions POI: 2!2x! 2 d 2 y dx = x2! x! 6 2 2( x 2! x! 6) 2x!1 ( x 2! x! 6) 4!!x 2! 2x! 5 d 2 y dx = 2x3 + 6x x x 2! x! 6 x =! = 0 ( ) ( Must be solved with graphing calculator. ) Trait f ( x) f!( x) f!( x) Conclusion (!",! 2) Below axis, decreasing, concave down x =!2!!! VA (!2,!1) + + Above axis, decreasing, concave up x =!1 0 + Zero (!1,! 0.068) + Below axis, decreasing, concave up x =! POI (!0.068, 3) Below axis, decreasing, concave down x = 3!!! VA ( 3,!) + + Above axis, decreasing, concave up 703

32 y = x +1 x 2! x! 6 Of course, the function s equation need not be given in order to find the graph. The table or sign patterns would suffice, especially if the function is piece-wise defined. 704

33 12-3 Multiple Choice Homework 1. Consider the function f x 3 for all real numbers x. The number = x 2! 5 of inflection points for the graph of f is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 2. The graph of y = 3x 5! 10x 4 has an inflection point at and ( 2,! 64 ) and ( 3,! 81 ) only only only (a) 0, 0 (b) 0, 0 (c) 0, 0 (d) 3,! 81 (e) 2,! How many points of inflection does the graph of y = cos x cos 3x! 1 cos5x have on the interval 0! x! "? 5 (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 705

34 4. The graph of the function f is shown in the figure to the right. Which of the following could be the graph of f!, the derivative of f? 5. The graph of the derivative of a function f is shown in the figure at right. The graph has horizontal tangent lines at x =!1, x = 1 and x = 3. At which of the following values of x does f have a relative maximum? (a) 2 only (b) 1 only (c) 4 only (d) 1 and 3 only (e) 2, 1, and 4 706

35 12-3 Free Response Homework Sketch the possible graph of a function that has the traits shown below x f ( x) f!( x) f!( x) x f ( x) f!( x) f!( x) x < x <! x = x =!1 3 DNE DNE 2 < x < 3 +!1 < x < x = x = < x < < x < x = x = 9 3 DNE DNE 5 < x < x > x = < x < 9 + x = < x < x = x > 10 + Find the key traits, make a table, and sketch. 3. y = 3x 4!15x y = x 3 + 5x 2 + 3x! 4 5. y = x +1 x 2! 2x! 3 6. y = x 2!1 x 2 + x! 6 7. y = 5x 2 3! x y = x i tan x on x!# 0, " % & {Hint: Solve using a graphing calculator.} 9. y = ( x 2 ) 4! x 10. y = x 2 e!x 707

36 12-4: Traits of a Function from Its Derivative Until now, the derivative has been treated as a by-product of some original function. But the derivative can be considered a function itself. As such, it has all the traits that any function of its type would have. For example, there is no graphical distinction between y = cos x as a function and y = cos x as the derivative of y = sin x. In fact, viewing the graphs shows that some of the traits of y = cos x correspond directly to traits of the y = sin x Most obvious is the fact that the zeros of one curve are the critical values of the other. This should make sense since the way to find a critical value is to set the derivative equal to zero. There are several other comparisons between the curves that can be made: 708

37 EX 1 Find the sign patterns of f x and compare them. = 9x! x 3, f!( x) = 9 " 3x 2, and f!( x) = "6x y + 0! 0 + 0! " # x! y! " " #%%%%%%%%%%%% x " 3 3 y! + 0 % x 0 " # EX 2 Compare the graphs below of f x f! x = "6x. = 9x! x 3, f!( x) = 9 " 3x 2, and It can be seen that the extreme points of f x the extreme point of f! x Similarly, the positive parts of f! x concave up part of f ( x). line up with the zeros of f! x lines up with the zero of f!( x) and the POI of f ( x). match the increasing part of f!( x) and the, and 709

38 Almost any trait can be found from a known first derivative curve, but, unfortunately, the derivative is not enough to tell the exact original curve, because, though the critical values and x-values of the POI are known, their y-values cannot be found. Similarly, the zeros of the original curve cannot be found from its derivative. The Chart: This can help organize what each piece of information can reveal f ( x y value ) positive f!( x) f!( x) y value zero y value negative interval of increasing y! value positive max. or min. y! value zero interval of decreasing y! value negative concave up interval of increasing y! value positive POI max. or min. y! value zero concave down interval of decreasing y! value negative LEARNING OUTCOME Find traits of a function from its derivative. 710

39 EX 3 If the curve below is y = f!( x), what are a) the critical values for the maximum and minimum points, b) intervals of increasing and decreasing, c) the x- coordinate(s) of the POIs, and d) the intervals of concavity of y = f ( x)? y y=f'(x) x What is pictured here is a two dimensional representation of the sign patterns of the first AND second derivatives. dy dx x! 0 + 0! 0! 0 + " #! 9! d 2 y dx 2 x + 0! 0 + 0! 0 + " #! 7! a) 9 and 10 must be the critical values of the minimum points, because they are zeros of the derivative and the sign of the first derivative changes from to + (the 1st Derivative Test). 3 is the critical value of the maximum point. 0 is not at an extreme point because the sign does not change on either side of it. 711

40 and x! 10, " and ("3, 10). b) f ( x) is increasing on x! "9, " 3 decreasing on x! "#, " 9, and it is The intervals of increasing and decreasing on f! x concavity on f ( x). So signs of the slopes of f! x derivative sign pattern: are the intervals of make up the second c) 7, 2, 0 and 7 are points of inflection, because they are where the second derivative equals 0 and the second derivative signs change. d) f ( x) is concave up on x!("#, " 7), "2, 0 concave down on x! "7, " 2 and ( 0, 7). and 7, #, and it is 712

41 EX 4 Given the same graph of y = f! x for f ( x) on x!# "2, 2% &. On x!# "2, 2% &, f! x Since f ( x) is decreasing and f 0 in EX 1, if f ( 0) = 0, sketch a likely curve is negative so f ( x) in decreasing on that interval. is the zero, the curve must be above the x-axis on x!# "2, 0) and below the x-axis on x!( 0, 2" #. On x!("2, 0), the slope of f!( x) ( which is f!( x) ) is positive, so f ( x) is concave up. Similarly, on x!( 0,!2), f!( x) is negative since f!( x) is decreasing, so f ( x) is concave down. So ( 0, 0) is not only a zero, it is also a point of inflection. Putting it all together, this is a likely sketch: Note that there are not markings for scale on the y-axis. This is because the y-values of the endpoints of the given domain cannot be known, from the given information. 713

42 EX 5 If the following graph is the velocity of a particle in rectilinear motion, what can be deduced about the acceleration and the distance? y=v(t) The particle is accelerating until t =!6, decelerating from t =!6 to t = 6, and then accelerating again. The distance is at a relative maximum value at t =!3 and at a relative minimum value at t =!9 and t = 10, but what the distances from the origin are cannot be known. 714

43 12-4 Multiple Choice Homework 1. The figure shows the graph of f!, the derivative of a function f. The domain of is the interval!4 " x " 4. Which of the following are true about the graph of f? I. At the points where x =!3 and x = 2 there are horizontal tangents. II. At the point where x = 1 there is a relative minimum point. III. At the point where x =!3 there is an inflection point. (a) None (b) II only (c) III only (d) II and III only (e) I, II, and III 2. The graph of f!, the derivative of a function f, is shown below. Which of the following statements are true about the function f? I. f is increasing on the interval (!2,!1). II. f has an inflection point at x = 0. III. f is concave up on the interval (!1, 0). (a) I only (b) II only (c) III only (d) I and II only (e) II and III only 3. The derivative of the function g is g! ( x) = cos( sin x). At the point where x = 0 the graph of g I. is increasing II. is concave down III. attains a relative maximum point (a) I only (b) II only (c) III only (d) I and III only (e) I, II, and III 715

44 4. The graph of the derivative of a function is f shown below. Which of the following is true about the function f? I. f is increasing on the interval (!2,1). II. f is continuous at x = 0. III. f has an inflection point at x =!2. (a) I only (b) II only (c) III only (d) II and III only (e) I, II, and III 5. The graph of the second derivative of a function f is shown at right. Which of the following is true? I. The graph of f has an inflection point at x =!1. II. The graph of f is concave down on the interval (!1, 3). III. The graph of the derivative function f! is increasing at x = 1. (a) I only (b) II only (c) III only (d) I and II only (e) I, II, and III 716

45 12-4 Free Response Homework If the curve below is y = f!( x), show the sign patterns of the first and second derivatives. Then find a) the critical values for the maximum and minimum points, b) x-coordinates of the POIs, c) intervals of increasing and decreasing, d) the intervals of concavity of y = f ( x), and e) sketch a possible curve for f x. with y-intercept 0,

46

47 The Second Derivative Practice Test Part 1: CALCULATOR REQUIRED Round to 3 decimals places. Show all work. Multiple Choice (3 pts. each) 1. On which interval is the graph of f x increasing? (a) ( 0, 1)! (b) 0, 1 " # 2% &! (c) 0, 1 " # 4% & (d) (e)! 1 4, 1 " # 2% &! 1 " # 4, 1 % & 2. Consider the function f ( x) = x4 maximum values at (a) 3 (b) 4 (c) 5 (d) 0 (e) There is no maximum value 3. At which point on the graph of y = g( x), shown to the right, is g! ( x) = 0 and g!! ( x) = 0? = 4x 3 2! 3x 2 both concave down and 2! x5. The derivative of f attains its 10 (a) A (b) B (c) C (d) D (e) E 719

48 Free Response (10 pts. each) "!3x % 1. Find zeros and intervals above/below the x-axis for y = ln # x 2 + 9& '. 2. Find the extreme points and intervals of increasing/decreasing of "!3x % y = ln # x 2 + 9& '. "!3x % 3. Find the points of inflection and intervals of concavity for y = ln # x 2 + 9& '. 720

49 The Second Derivative Practice Test Part 2: NO CALCULATOR ALLOWED Round to 3 decimals places. Show all work. Multiple Choice (3 pts. each) 4. What is the x-coordinate of the point of inflection on the graph of y = 1 3 x3 + 5x ? (a) 5 (b) 0 (c)! 10 3 (d)!5 (e)!10 5. Let f be a function that has domain "#!2, 5 %. The graph of f! is shown at the right. Which of the following statements are TRUE? I. f has a relative maximum at x =!1. II. f has an absolute minimum at x = 0. III. f is concave down for!2 < x < 0. IV. f has inflection points at x = 0, x = 2, and x = 3. (a) I, II, IV (b) I, III, IV (c) II, III, IV (d) I, II, III (e) I, II, III, IV 721

50 6. The graph of the twice-differentiable function f is shown in the figure below. Which of the following is true? (a) f 1 (b) f 1 (c) f! 1 (d) f!! 1 (e) f!! 1 < f!( 1) < f!! ( 1) < f!! ( 1) < f!( 1) < f ( 1) < f!! ( 1) < f ( 1) < f!( 1) < f!( 1) < f ( 1) Free Response (10 pts. each) "!3x % 4. Using the traits found in FR #1-3, sketch y = ln # x 2 + 9& '. 722

51 The Second Derivative Practice Group Test Page 1 Round at 3 decimal places. Show all work. 1. Find the extreme points of y = 2x 2 e 1 2 x. 2. Find the points of inflection of y = ln( x 3! 4x). 723

52 3. Make a key trait table for y = 4x x Sketch y = x x! x 2, labeling all the key traits. 724

53 The Second Derivative Practice Group Test Page 2 Round at 3 decimal places. Show all work. 1. Find the extreme points of y = x x! x Find the points of inflection of y = 2x 2 e 1 2 x. 725

54 3. Make a key trait table for y = ln( x 3! 4x). 4. Sketch y = 4x x 2, labeling all the key traits

55 The Second Derivative Practice Group Test Page 3 Round at 3 decimal places. Show all work. 1. Find the extreme points of y = 4x x Find the points of inflection of y = x x! x

56 3. Make a key trait table for y = 2x 2 e 1 2 x. 4. Sketch y = ln( x 3! 4x), labeling all the key traits. 728

57 The Second Derivative Practice Group Test Page 4 Round at 3 decimal places. Show all work. 1. Find the extreme points of y = ln( x 3! 4x). 2. Find the points of inflection of y = 4x x

58 3. Make a key trait table for y = x x! x Sketch y = 2x 2 e 1 2 x, labeling all the key traits. 730

59 The Second Derivative Homework Answer Key 12-1 Multiple Choice Homework 1. A 2. D 3. A 4. B 5. B 12-1 Free Response Homework 1. 60x x! x 5! 60x 3 +18x 3. y! = "2 ( sin x 2 + 2x 2 cosx 2 ) 4. y! = 2sec 2 x ( 2tan 2 x + sec 2 x) 10. F!! ( x) = 2 ( 3x 2 " 2x +1) d 2 y dx = 14 3x x 2!10 d 2 y dx = 18x2!18x 2 x 2! x x =!7 a maximum point x = 4 a minimum point 14. x = 3 a maximum point x =!3 a minimum point 5. d 2 y dx = 9sec3x ( 2 tan2 3x + sec 2 3x) 15. x =!0.863 a maximum point x = ± 3, 9 4 minimum points f!! 8. g!! 9. h!! d 2 y dx 2 = 4e2x x +1 2 ( x) = "2 x2 " 3 x ( x) = ( x) = "2 ( x " 2) 2 5 ( x 2 + 5) x = 0 a minimum point 17. x = ± 27 maximum points 2 x = 0 a minimum point 18. x = a maximum point x =!2.999 a minimum point 19. x =! 5" 3, ", 2" max points 3 x =!2",! 4" 3, 2" 3 min pts 20. a( t) = 6t!12 731

60 21. a( t) = 60t 3! 456t t! a( t) =!9" a t 24. a t # 16 sin " % 4 t! 5 = 2t t 2! 3 ( t 2 +1) 2 = et " 2! e t # 2 1+ e t 25. a( 6) =! a (!1) =!2 % & ' Multiple Choice Homework 1. A 2. C 3. C 4. D 5. E 12-2 Free Response Homework 1. x!("#, " 3) ( 0, 3) " 2. x! 2 # 3, 2 % & ' ( 2, ) 3a. Yes b. No & ( ' 4a. x!("#, " 9) ("3, "1) "1, 7 b. x!("#, " 7) ("2, 0) ( 5, #) c. x! "7, " 3 " 0 + y!! #%%%%%% x " 1 3 CD: x! "#, " 1 ' % & 3( ) CU: x! " 1 3, # ' % & ( ) + 0 % 0 + y!! " # x " CD: x! 1, 7 % # 3& ' " CU: x!((), 1)* 7 3, ) % # & ' 7. y + 0 " "!! #%%%%%%%%%%%%%%% x " CD: x!("3.464, 0)# ( 3.464, ) CU: x!(", 3.464)# ( 0, 3.464) " 0 + y!! x " 2 2 #%%%%%%%% CD: x! "2, 2 CU: x!("#, " 2) 2, # 9. Concave down for all domain 732

61 y!! x End " 0 + End CD: x! 0, " 0 # 2# CU: x! ", 2" + 0 % y!! " # x 2 CD: x! 2, " CU: x! #", " 0 + y!! x " # CD: x! " 1 2, 1 & % ( 2 ' #%%%%%%%%% # CU: x! "), " 1 % 2 ' ( * 1 % 2, ) & # & ( ' 12-3 Multiple Choice Homework 1. D 2. E 3. E 4. B 5. C 12-3 Free Response Homework " DNE + 0 " DNE + y!! #%%%%%%%%%%%%%%% x " CD: x!("#,"3) ( 0, 3) CU: x!("3, 0) ( 3, #) 14. Concave up for all domain 733

62 3. y = 3x 4!15x x y y! y!! Conclusion!",! above, decreasing, CU x =! zero!2.116,! below, decreasing, CU x =! minimum pt!1.581,! below, increasing, CU x =! POI!0.913,! below, increasing, CD x =! zero!0.722, above, increasing, CD x = maximum pt. 0, above, decreasing, CD x = zero 0.722, below, decreasing, CD x = POI 0.913, below, decreasing, CU x = minimum pt , below, increasing, CU x = zero 2.116,! above, increasing, CU 734

63 4. y = x 3 + 5x 2 + 3x! 4 x y y! y!! Conclusion (!",! 4) + below, increasing, CD x =!4 0 + zero (!4,! 3) + + above, increasing, CD x =!3 5 0 maximum pt. "!3,! 5 % # 3& ' + above, decreasing, CD x =! POI "! 5 # 3,!1.618 % & ' + + above, decreasing, CU x =! zero "!1.618,! 1 % # 3& ' + below, decreasing, CU x =! minimum pt. " #! 1 3, % & ' + + below, increasing, CU x = zero 0.618,! above, increasing, CU 735

64 5. y = x +1 x 2! 2x! 3 x y y! y!! Conclusion (!",!1) below, decreasing, CD " x =!1 DNE 0 POE!1,! 1 % # 4 & ' (!1, 3) below, decreasing, CD x =!3 DNE DNE DNE VA 3,! above, increasing, CU 736

65 6. y = x2!1 x 2 + x! 6 x y y! y!! Conclusion!",! above, increasing, CU x =!3!!! VA!3,!1 + below, increasing, CD x =!1 0 + zero!1, above, increasing, CD x = maximum pt , 1 + above, decreasing, CD x =1 0 zero 1, 2 below, decreasing, CD x = 2!!! VA 2, above, decreasing, CU x = minimum pt , above, increasing, CU x = POI ,! + + above, increasing, CD 737

66 7. y = 5x 2 3! x 5 3 x y y! y!! Conclusion (!",!1) + + above, decreasing, CU x =!1 6 0 POI (!1, 0) + above, decreasing, CD x = 0 0!! zero, minimum pt. ( 0, 2) + + above, increasing, CD x = maximum pt. ( 2, 5) + above, decreasing, CD x = zero ( 5,!) below, decreasing, CD 738

67 8. y = x i tan x on x! 0, " # % & x y y! y!! Conclusion x = zero, endpoint, minimum pt. " 0,! % # 2 & ' above, increasing, CU x =! 2 "! # 2, % & ' DNE DNE DNE VA + below, increasing, CD x = POI 2.798, below, increasing, CU x =! zero, endpoint, maximum pt. 739

68 9. y = ( x 2 ) 4! x x y y! y!! Conclusion (!", 0) + + above, decreasing, CU x = zero, minimum pt. ( 0, 2) above, increasing, CU x = POI! 2, 16 " # 3 % & + + above, increasing, CD x = maximum pt.! 16 " # 3, 4 % & above, decreasing, CD x = 4 0 zero, endpoint, minimum pt. 740

69 10. y = x 2 e!x x y y! y!! Conclusion (!", 0) + + above, decreasing, CU x = zero, minimum pt. (!1, 0) above, increasing, CU x = POI ( 0.586, 2) + + above, increasing, CD x = maximum pt. ( 2, 3.41) + above, decreasing, CD x = POI ( 3.41,!) + + above, decreasing, CU 741

70 12-4 Multiple Choice Homework 1. D 2. D 3. A 4. D 5. D 12-4 Free Response Homework 1a. min x = ±2 max x = 0 b. x = ±1 c. inc: x! "2, 0 # ( 2, 3) # ( 0, 2) # ( 1, 3) dec: x! "3, " 2 d. CU: x! "3, "1 CD: x! "1, 1 3a. min x = 2 max x =!2 b. x = ±1.4, 0 c. inc: x! "3, " 2 dec: x! "2, 0 # ( 2, 3) # ( 0, 2) d. CU: x! "1.4, 0 # ( 1.4, 3) # ( 0, 1.4) CD: x! "3, "1.4 2a. min x = 2 b. x = 0, 1.3 c. inc: x! 2, 3 dec: x! "3, 0 d. CU: x! "3, 0 CD: x! 0, 1.3 # ( 0, 2) # ( 1.3, 3) 742

71 4a. max x = ±2 min x = 0 b. x = ±1.5 c. inc: x! "3, " 2 dec: x! "2, 0 # ( 0, 2) # ( 2, 3) d. CU: x! "1.5, 1.5 # ( 1.5, 3) CD: x! "3, "1.5 6a. min x = 2 max x =!2 b. x = 0 c. inc: x! "3, " 2 # ( 2, 3) dec: x! "2, 2 CD: x!("3, 0) d. CU: x! 0, 3 5a. max x = ±2 min x = 0 b. x = ±1.3 c. inc: x! "3, " 2 dec: x! "2, 0 # ( 0, 2) # ( 2, 3) CD: x! "3, "1.3 d. CU: x! "1.3, 1.3 # ( 1.3, 3) 743

72 The Second Derivative Practice Test Answer Key Multiple Choice 1. E 2. A 3. D 4. D 5. B 6. D Free Response Zeros: None 1. Domain: x! "#, 0 2. Increasing: x! "#,"3 Decreasing: x! "3, 0 Extreme Point:!3,! CD: x! "#," CU: x! "6.175, 0 POI:!6.175,!

73 Practice Group Test Answer Key Page 1 1. Extreme Points: ( 0, 0),!4, 32e!2 2. No POIs 3. x y y! y!! Conclusion (!",! 6 ) + below, decreasing, CD x =! 6! POI (! 6,! 2) + below, decreasing, CU x =! 2! min pt. (! 2, 0) + + below, increasing, CU x = zero, POI ( 0, 2) + + above, increasing, CD x = maximum pt. ( 2, 6) + above, decreasing, CD 1 x = POI 6,! + + above, decreasing, CU

74 Page 2 1. Extreme Points: (!2, 0), (!1.372,! 2.952), ( 4.372, ), 6, 0 2. POIs: (!1.172,!1.845), (!6.828, 0.635) 3. x y y! y!! Conclusion x =!2 DNE DNE DNE VA!2,! below, increasing, CD x =! zero "!1.861,! 2 % ' # 3& + + above, increasing, CD x =! max. pt. " #! 2 3,! % ' & + above, decreasing, CD x =! zero!0.254, 0 below, decreasing, CD x = 0 DNE DNE DNE VA x = 2 DNE DNE DNE VA 2, below, increasing, CD x = zero 2.115,! + + above, increasing, CD

75 Page 3 1. Extreme Points: ( 2, 2),! 2,! 2 2. POIs: ( 1.320, 5.203) 3. x y y! y!! Conclusion (!",! 6.828) above, increasing, CU x =! POI (!6.828,! 4) + + above, increasing, CD x =! max pt. (!4,!1.472) + above, decreasing, CD x =! POI (!1.472, 0) + + above, decreasing, CU x = zero, min pt. ( 0,!) above, increasing, CU

76 Page 4 1. Extreme Points: "!2 # 3, % & ' 2. POIs: ( 0, 0), (! 6,!1.225), 6, x y y! y!! Conclusion x =!2 0 DNE DNE zero!2,! below, decreasing, CU x =! min pt.!1.372, below, increasing, CU x = zero 0, above, increasing, CU x = POI 1.320, above, increasing, CD x = max pt , 6 + above, decreasing, CD x = 6 0 DNE DNE zero