Properties of Derivatives

Transcription

1 6 CHAPTER Properties of Derivatives To investigate derivatives using first principles, we will look at the slope of f ( ) = at the point P (,9 ). Let Q1, Q, Q, Q4, be a sequence of points on the curve getting closer and closer to P. Let Q1 be ( 4,16 ), Q be (.5,1.5 ), Q be (.,10.4 ), Q4 be (.1,9.61 )

2 7 The idea is that the tangent to the curve at point P is the limiting line of the sequence of lines PQ1, PQ, PQ, PQ4 We say that the slope of the tangent at P is the limit of the slopes of lines PQ1, PQ, PQ slope PQ1 slope PQ slope PQ slope PQ4 16! 9 = = 7 4! 1.5! 9 = = 6.5.5! 10.4! 9 = = 6..! 9.61! 9 = = 6.1.1! It certainly appears that the slope of the tangent at P is close to 6. To show that the slope of f () = at (,9 ) is 6 ( ) Let P be (,9 ) and Q be a point h,( h) + + on the curve y =

3 8 The slope of PQ is ( ) ( h) + h! 9 6h + h = = 6 + h. * +! h In the limiting case i.e. where Q actually coincides with P then the line PQ becomes the tangent at P and h becomes zero. From the above * it can be deduced that when h = 0, the slope of the tangent i.e. the slope of the limiting case of PQ is 6+0 which is 6. To find the slope of f () = at ( a,a ) From the previous discussion we can say that the slope of the tangent at P is the limit of the slope of PQ as Q coincides with P i.e. as h approaches zero. The slope of the tangent at P is written: lim slope PQ = h! 0 ( ) ( ) a + h " a lim h! 0 a + h " a

4 9 = ah + h lim h! 0 h! 0 h = lim a + h When, in the limiting case, h does become zero, then the slope of the tangent at P = a + 0 = a. This means that if f ( ) f a = a. = then '( ) i.e. The slope of f ( ) = at an arbitrary point ( a, a ) is a. In general ( ) D = Definition of derivative from first principles Consider y = f ( ) f '( ) is the limit of QR PR as Q coincides with P. i.e. f '( ) = lim h! 0 f ( + h) " f ( ) h

5 0 This is called the first principles definition of the derivative of any function, f ( ). If we use the dy d notation then we say dy! y = lim d! " 0!. Eample 1 To find the derivative of 1 from first principles f ( ) 1 = f '( ) = lim h! 0 f ( + h) " f ( ) h = lim h! " + h h = lim h! 0 "( + ) ( + h) h h " h = lim h! 0 h h = lim ( + ) " 1 ( + ) h! 0 h! 1 = Note that from Chapter 1 we learned the rule " 1 #! 1!! 1 D$ % = D =! = & ' Therefore: ( ) ( 1) D( n ) n! n 1 =. as shown.

6 1 Eample To prove D( n )=n n-1 for the cases where n is a positive integer Let f ( ) = n Then f '() = lim h!0 f ( + h) " f ( ) h = lim h! 0 ( ) n + h " h n (By binomial theorem) # n$ # n$ + % & h + % & h +...! 1 = lim ' ( ' ( h" 0 h n n! 1 n! n # n$ # n$ # n$ = ' 1( ' ( ' ( n! 1 n! n! lim % & % & h % & h... h" 0 " n#! = $ % & 1' n 1 = n! n 1 CHAIN RULE (to be proven at the end of the chapter) D! " ( ) f g ( ) # $ = f ' g ( ( )) g '( ) This result is best illustrated by eamples. The graph of a function is said to be continuous if it is possible to draw the graph without lifting the pencil from the paper. ie if it has no jumps. It is not possible to find the slope or the derivative of a function at a discontinuity.

7 Eample Consider ( 1) 5 +. This function can be decomposed by letting ( ) 5 ( ) = + 1. Then ( ) g ( ) ( 1) 5 4 Note that f '( ) = 5 and ( ) Eample 4! ( ) 5 D + 1 "# f g = + as required. g ' =. $ %& = D! " ( ) f g ( ) $ % = f '( g( ) ) g '( ) ( ) 4 = 5! " g # $ () ( ) 4 = f = and ( )! D # + " ( ) 1 ( + ) $ & = % + ( + ) = + In layman terms, the Chain Rule can be thought of as when differentiating a complicated epression, differentiate the most outside function, writing down the bracket as it is, then multiply by the derivative of what s inside the bracket.

8 Eample 5 Eample 6 ( ) 1 D! 5" $ # % = D # 5 1 ( ) 1 5! =! i =! 5 Eample 7 " # 1 D $ % = D! $ (! 7) % & '! ( 7)! ( ) 4 =!! 7 i =! 6 (! 7) 4

9 4 In a longer chain of functions we have ( ( ( ))) D! f g h " ( ( ( ))) g ' h( ) # $ = f ' g h ( )h' ( ). This can be etended to any length chain of functions. In dy, d notation the Chain Rule can be epressed as dy dy dt =! d dt d OR dy dy d = d dt dt In a longer chain we can say dy dy du dv = i i d du dv d i.e. We can think of these as though they were fractions where we can cancel. In fact, they are limits of fractions to be studied in Chapter 4. We have seen earlier that ( ) + ( ) = '( ) + '( ) D!# f g " $ f g but it is NOT TRUE that we can etend this result to the derivative of the product of two functions. PRODUCT RULE (to be proven at the end of the chapter) Consider f ( ) = g ( ) h( ) then f '( ) = h( ) g '( ) + g ( )h' ( )

10 5 Eample 8 If f ( ) = + 1i ( + ) then f '( ) = ( + ) 1 ( + 1 )! 1 ( ) Eample ( + 5) i ( + 6) D! # % " $ & ( ) ( ) ( ) ( ) = + 6 i4 + 5 i i7 + 6 i A useful mnemonic is the following: If u and v functions of then [ i ] D u v = vdu + udv Where du, dv denote the derivatives of u and v respectively with respect to. Similarly, the derivative of a quotient can be obtained by the following formula: QUOTIENT RULE (to be proven at the end of the chapter) ( ) ( )! D f $ # & "# g %& = g A mnemonic for this is ( ) f '( ) ' f ( ) g '( )! g ( ) $ " % " u # vdu! udv D $ v % = v & ' Eample 10! + 1 " D # & 1% $ ' = ( 1! )! + 1 ( )(!6) = ( 1! ) ( 1! )

11 6 Worksheet 1 PRODUCT RULE, QUOTIENT RULE, CHAIN RULE, FIRST PRINCIPLES 1. Find dy d for each of the following: a) = 16 + b) y ( 8 ) y = + c) y = (! 1)(! ) d) y ( 1)( ) =!! e)! 1 y = f) y =! 1! g) y 1 y =! 1 i) =! h) ( ) 1 4 y = j) y = Find f '( ) for each of the following: 1 a) f ( ) 4 6 =!! b) f ( ) = ( + 5) ( 1! ) 5 c) f ( ) =! d) f ( ) =! 4(! ). Find dy d if + y + y =. 4. f ( ) =! 1. Find '( ) f. 5. Find two values of for which the tangent to y = is parallel to + 1 y! + 8 = 0. n n 1 6. Given that D ( ) n! n ( ) n 1 D = n!. =, show by using the Product Rule that 1 7. Find the derivative of using first principles.

12 7 8. If f '( ) = 6 + and f ( 1) = 7 find ( ) f. 9. Find where 10. Prove that y =! + is decreasing. = + + is always increasing. y Find the equation of the tangent to y = 5! at the point ( 4, ). Answers to Worksheet 1 1. a) ! ( + 1) b) c)! 4 5. = 0 OR! d) e) 4 5 9!! 8. 19! (! ) 4. 9.! 1 < < 1 f) g) 1! 1! 1! y + 4 = 5 1 4! h) (! ) 4 i) j) a) 8 6! + b) ( ) ( ) ! + 5! ( + 5) 5 ( 1! ) c) + 1! 4 ( + 6) d)! + +! 4 8 8! 4

13 8 Worksheet 1. Find dy d for each of the following: 1 a) y = b) ( ) n + y = + 1 c) y = y + 1 d) y = + 9 e) y = (! 1)(! )(! ). Differentiate ( 1)( 4 ) +! + with respect to.. Find! + 1" D # $. % + 1 & 4. Find dy d for each of the following: a) y ( 1) 5 = + b) y = + 1 c)! 1 y = + 1 d) y + y = 5. Find the equation of the tangent to the curve y = 5! at ( 0,0 ). Note that ( 0,0 ) is on the graph. 6. Find the equation of the tangent to + y = 5 at (,4 ). 7. Differentiate 5! y = + k is a tangent to y =! 1. Find k. 9. Find the maimum value of Find the minimum value of

14 9 Answers to Worksheet! 8 1. a) ( ) b) ( n + )( + ) 1 n 8. k = 9. c) d) e) ( 1! ) ! ! ! +! 1 ( + 1) 4. a) 5( + 1) b) c) ( + 1)!! 6 d) ( ) + 5. y = y = 5 7. ( )! 4i 5! 6!

15 40 Relative Maimum Minimum Points A relative maimum point occurs at a point on a curve where the y value at that point is greater than the y values of points in its neighbourhood. Usually this will occur when the slope is zero. A is a relative maimum point (on a smooth continuous curve). A relative maimum point occurs when the slope changes from + to! Eample 11 To find a relative maimum point on the graph of y =! : dy d =! ( ) =! 4 + ( )( ) =! 1! Note that when = 1 or = the slope equals zero. On a signed line diagram we have: Note that f '( ) changes from + to! as passes through the value 1.

16 41! = 1 yields a relative maimum. A ( 1,8 ) is a relative maimum. Similarly, note that B (,4 ) is a relative minimum point where f '( )! to + as passes through. changes from Summary A relative minimum occurs when f '( ) changes from! to +. A relative maimum occurs when f '( ) changes from + to!. An absolute maimum point is a point which is higher than all other points in the domain of the function, regardless of whether the function is discontinuous or has a slope of zero at the point. ie (a, f(a)) is an absolute maimum point if f(a)>f( ) for all values of in the domain of f.

17 4 If a curve is discontinuous however and f '( ) 1 relative ma/min point. For eample, y = looks like: changes sign then there will not be a When 0 =, '( ) f changes from + to! but there is not a local maimum point because of the discontinuity. A critical value of a function is a value such that f '( ) = 0 or is undefined changing from +! to! " or vice versa. For eample consider f ( ) = + 1 (! 1) which looks like:

18 4 On a sign diagram we have: and f '( ) changes sign at both =! and = 1 but =! yields a relative minimum whereas = 1 does not yield either a maimum or a minimum because of the discontinuity. =! and = 1 however are critical values. Note that further notes on relative ma/min points are given in Chapter. The Second Derivative Let s consider the significance of the derivative of f ', written f ".

19 44 Note that in going from A to E along the curve the slope of the curve is increasing everywhere. The slope of the curve is increasing everywhere i.e. f '( ) is increasing. From Chapter 1 we know that if a function is increasing then its derivative is positive. It follows that if f '( ) is increasing then f "( ) is positive. It can therefore be deduced that, for the curve shown, f "( ) is positive everywhere. is called concave up (note the useful mnemonic cup ) i.e. f "( ) is positive means the graph is part of a cup. Similarly, f "( ) is negative implies the graph is concave down. i.e. like an umbrella: f " is negative umbrella rain frown down f " is positive cup up smile happy We say that f "( ) is a measure of the curvature (or concavity) although the numerical relationship is not easy to define. Suffice it to say, usually, the larger the numerical value of f "( ) at a particular point, the more bendy the curve.

20 45 Consider the graph shown below: Convince yourself that the table below represents the values of f, f ' and f " for the different intervals of. 0 < < 1 1< < < < < < 4 4 < < 5 5 < < 6 f f ' + + f " Note that in particular when = the graph is changing concavity from down to up i.e. f " is changing from negative to positive i.e. f "( ) = 0. We say that when = the graph has an inflection point.

21 46 Inflection Points Points P and Q illustrate eamples of inflection points. An inflection point occurs when f "( ) a tangent at that point. changes sign and when it is possible to draw Most often, but not always, this will occur when f "( ) = 0. A good method for locating inflection points is to find where f "( ) = 0 and check to see if f " changes sign at those values of. It is possible for an inflection point to occur when f " is infinite. For eample, consider ( ) 1 f =. At ( 0,0 ) the graph looks like! f " is infinite but ( 0,0 ) is an inflection point since f "( 0 ) is + positive and f "( 0 ) is negative.

22 47 It is not true however that f "( ) = 0 guarantees an inflection point. For eample, on the graph of y 4 = at ( 0,0 ) the second derivative is 0 but ( 0,0 ) is not an inflection point because "( ) + positive and f "( 0 )! f does not change sign at ( 0,0 ) since f "( 0 ) is positive also. is Note also that if f ( ) is discontinuous when = a then it is not an inflection point even if f ( ) changes concavity because it is not possible to draw a tangent when = a. For eample, f ( ) inflection point. 1 = changes concavity when = 0 but = 0 does not yield an In summary, remember that An inflection point occurs when f "( ) changes sign AND it is possible to draw a tangent.

23 48 It is quite often helpful to think of an inflection point as a point where the slope is a maimum or a minimum. A is an inflection point where slope is a minimum. B is an inflection point where slope is a maimum. An even function is one in which An eample is f ( ) = f (! ) for all f ( ) = 4 +! 1 An odd function is one in which An eample is f (! ) =! f ( ) ( ) f = + 4

24 49 It follows that an even function is symmetric about the y ais and an odd function is symmetric about the origin. An even function An odd function

25 50 Worksheet 1. Find the equations of the tangents, with slope 9, to the curve y ( ). Show that the function ( ) f =! is always increasing. =!.. The curve ( )! b " f = # a + $ % & passes through the point ( 4,8 ) at which the slope of the tangent is. Find a and b. 4. y = is a tangent to y 4 k = + +. Find k. 5. Prove by first principles that if f ( ) = then f '( ) 1 =. 6. Find the equation of the curve whose slope at the point (, y ) is + if the curve passes through ( 1,6 ). 7. Find, for what values of, f ( ) is increasing if f ( ) ( 1) ( ) =! Find where y =! + is increasing. 9. If f "( ) > 0 for all for a certain function f, what can be said about the graph of f? 10. Find the inflection points of ( ) 4 f =! + 5! Find the minimum points on the graph of f ( ) =! For what values of k will the graph of y k = have positive slope for all values of? 1. If f ( ) =!, find a) f ( ) b) f '( ) c) f '(! 1)

26 Find the equation of the tangent to y = + at the point where = Find the equation of the tangent to y 4 = ! at the point ( -,15 ). Find another point on the curve which has eactly the same equation for the tangent at the point. 16. Find a point P on y = such that the tangent at P intersects the original curve again at point Q so that the slope of the tangent at Q is 4 times the slope of the tangent at P. 17. Shown below is a graph of y = f '( ), the derivative of ( ) f. Describe what signifance, if any, points A, B, C, D, E, F, G, H tell us about the graph of y = f ( ), the original function.

27 5 Answers to Worksheet 1. y = or y = 9! a = 6, b =! 8 4. k = y = <! 1 or > 1 8.! 1 " " 1 9. graph is always concave upwards like a parabola 10. ( 0,- ) ( 1,1 ) 11. ( -,-1 ) (,-1 ) 1. k > 4 1. a) b)9 c)0 14. y =! y =! + 6 ( 1, 1 4 ) 16. Any point on the curve satisfies the conditions stated.! P can be any point on y =. 17. A tells us f ( ) is increasing. B tells us f ( ) has an inflection point like C tells us f ( ) is increasing. D tells us f ( ) has a relative maimum point. E tells us that f ( ) has an inflection point like F tells us that f ( ) has a relative minimum point. G tells us that f ( ) has an inflection point like H tells us that f ( ) has an inflection point like

28 5 Worksheet 4 SECOND DERIVATIVES, MAX/MIN POINTS, INFLECTION POINTS 4 1. Find the inflection points for f ( ) =! ! 7.. If f ( ) =! find a) f ( ) b) f '( ) c) "( ) f.. Find the minimum value of y on the graph of y 4 =!. 4. a) Does b) Does c) Does 1 y = have an inflection point? Or a ma/min point? + 1 y = have an inflection point? Or a ma/min point? + 1 y = have an inflection point? Or a ma/min point? *5. Find the relative ma/min points and inflection points on the graph of y 4 =! 4! If ( ) 1 f = find the following: a) f ( 4) b) f '( 4) c) f "( 4) *7. Find the local ma/min points and inflection point(s) on the graph of 4 y =! +!. 8. Find derivative of 1 1+ from first principles. 9. In which quadrants is dy d positive for + y = 5 (Do not differentiate). In which quadrants is d y d positive for the same graph?

29 54 *10. Given f '( ) =! 5 + 4! 5 a) For what values of is f ( ) increasing? b) For what values of is f '( ) increasing? c) Does the graph of f change concavity? If so, where? 11. Does y 8 = have an inflection point? If so, where? If not, why not? 1. f '( ) = + 1! a) on which intervals is f increasing? b) on which intervals is f ' increasing? c) does f have a local maimum? Answers to Worksheet 4 1. (,5 ) (,108 ). a) b) 9 c) 1. 1! 8 4. a) No, No b) No, No c) No, Rel. Min at ( 1, ), Rel. Ma at ( -1,- ) 5. Rel. Min ( -1,-5 ) and (,- ) Rel. Ma ( 0,0 ) Inflection points ( ,-.68 ) ( 1.15,-18.5 ) 6. a) b) 1 4 c) 1! 7. Min point ( 4,- ), Inflection point ( 1,-5 ) (,-1 ) 9. Q and Q4. Q and Q a) 1< < 4 or > 5 b) < or > 7 c) at = and = No because d y d is always positive. 1. a) <! 1 or > b) Nowhere c) Yes, when =! 1

30 55 Worksheet 5 1. If f ( ) = then '( ) f = (A) 4 (B) (C) 1 4 (D) 1 (E) 1!. If the line y = 4 + is tangent to the curve y = + c, then c is (A) (B) 4 (C) 7 (D) 11 (E) 15. Suppose that the domain of the function of f is all real numbers and its derivative is given by: f ( ) ' = (! 1)(! 4) 1+ Which of the following is true about the original function of f? I. f is decreasing on the interval (!",1 ). II. f has a local minimum at = 4. III. f is concave up at = 8. (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, III *4. What are all values of for which the graph of 6 y = is concave down? a) <! 1 b) < 0 c)! 1 < < 0 d) 0 < < 1 e) >! 1

31 56 5. A graph of the function f is shown at the right. Which of the following is true? (A) f ( 4)! f ( ) 4! = 4 (B) f ( ) > f "( ) (C) f '( ) = f "( ) (D) f '( 1 ) f '( ) < (E) None of these Note: This is the graph of f "( ), NOT the graph of f ( ). 6. The figure above shows the graph of f "( ) function f ( ). The function f ( ) following statements about f is true?, the second derivative of a is continuous for all. Which of the I. f is concave down for < 0 and for b < < c. II. f has a relative minimum in the open interval b < < c.

32 57 III. f has points of inflection at = 0 and = b. (A) I only (B) II only (C) III only (D) I and III only (E) I,II, and III 7. Find the derivative of y = + at the point ( 1, ) using FIRST PRINCIPLES. 8 + k 8. For what value of k will have a relative maimum at = 4? (A) (B) 16 (C) 0 (D)16 (E) 9. a) If the graph of y a b = + +! 8 has a point of inflection at (,0 ), what is the value of b? b) For which values of is the function increasing at an increasing rate? 10. The tangent line to the graph of ( ) f =! at the point T(,1 ) intersects the graph of f at another point P. a) Write the equation of the line PT. b) Find the - and y - coordinates of point P. 11. The function f is defined on the interval [-4,4] and its graph is shown below.

33 58 a) Where does f have critical values? b) on what intervals is f ' negative? c) Where does f ' achieve its minimum value? Estimate this value of d) Sketch a graph of f '. e) Sketch a graph of f " Answers to Worksheet 5 f '. 1. C. C 11. d). D 4. C 5. B 6. D 8. B 9. a) +1 b) > 11. e) 10. a) 1 y = b) ( -6,- ) 11. a) =!, -1.4, 1.5 b) ( -4,- ) or ( -1.4,1.5 ) c) At = 0. Estimated value of f ' is.

34 59 Eample This is a graph of y f '( ) =. The domain of f and f ' is { :! 10 " " + 10}. Question a) Does f ( ) have a relative maimum or relative minimum point? Answer a) No because f '( ) is never zero. Question b) Does f ( ) have an inflection point? Answer b) Yes when = 0 because an inflection point of f occurs at a relative maimum point of f '.

35 60 Question c) If f ( 0) = find an approimate intercept for y f ( ) =. Answer c) f '( ) is always positive.! f is an increasing function. ( ) graph of y = f ( ) looks like: f 0 =! a f '( ) is less than 1 for all and therefore the slope of any tangent to y = f ( ) is always less than 1.! intercept of y f ( ) than. = must be less Question Answer d) How many intercepts does y = f ( ) have? d) Since y = f ( ) is always an increasing function it can only have one intercept.

36 61 Eample Shown is the graph of the derivative of a function i.e. the graph shown is y f '( ) The domain of f is the set { : }! " ". =. Question Answer a) For what values of does f have a relative minimum?, relative maimum? a) When f has a relative minimum point, f ' is changing from negative to positive.! = 0 yields a relative minimum point. When f has a relative maimum point, f ' is changing from positive to negative.! =! yields a relative maimum point.

37 6 Question b) For what values of is the graph of f concave up? Answer b) The graph of f is concave up when f " is positive i.e. f ' is increasing. In the graph shown of y f '( ) =, ( -1,1 ) and (, ) are two intervals of where f ' is increasing and so f is concave up when! 1 < < 1 OR < <. Question Answer Question c) Locate the co-ordinate of inflection points of f. c) An inflection point of f occurs when f ' has a relative maimum or a relative minimum value. i.e. =! 1 OR = 1 OR = d) If f ( 0) =! 1 sketch a possible graph of y f ( ) =.

38 6 Answer d) Note: A is a relative maimum point when =!. B is an inflection point when =! 1 C is a relative minimum point when = 0 D is an inflection point when = + 1 E is an inflection point where slope = 0 and when =.

39 64 Worksheet 6 1. If f ( ) = then '( ) f equals (A) 1 10 (B) 1 5 (C) 4 5 (D) 4! (E) none of these 5. The equation of the tangent line to the curve + 4 y = 4! at the point ( 1,7 ) is (A) y + 5 = (B) y! 1 =! 4 (C) y! 7 = 0 (D) y + 5 = 1 (E) y! 5 =! 18 k. Consider the function f ( ) = for which f ( ) + (A) 5 (B) 4 (C) (D) (E) 1 ' 0 = 1. The value of k is 4. The graph of the first derivative of a function f is shown at the right. Which of the following are true? I. The graph of f has an inflection point at = 1. graph of f ' II. The graph of f is concave down on the interval ( -,1 ). III. The graph of f has a relative maimum value at =. (A) I only (B) II only (C) III only (D) I and II only E) I, II, III

40 65 5. If f ( )! k = and f ' 0 + k ( ) = 1, then k = (A) 1 (B)! 1 (C) (D)! (E) 0 6. The composite function h is defined by h( ) f g ( ) are functions whose graphs are shown below. =! ", where f and g # $ The number of horizontal tangent lines to the graph of h is (A) (B) 4 (C) 5 (D) 6 (E) 7 7. If u + 1 y = u! 5 and u = + +, dy d at = is: (A)! 1 (B)! 5 (C)! 4 (D) 5 4 (E)! 15

41 66 8. The graph of the second derivative of f is shown below. Which of the following are true about the function f? f '' I. f ' is decreasing at = 0 II. f is concave up at = 5 III. f has a point of inflection at = 1 (A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, III Answers to Worksheet 6 1. C. A. C 4. D 5. C 6. E 7. B 8. B

42 67 Worksheet 7 1. If f ( )! 1 =, then the slope of the tangent at its intercept is: (A) (B) 0 (C) undefined (D) 1 (E)! 9. If h( ) f ( g ( ) ) =, and it is known that f ( 1) =! 4, f '( 1) = 5, g ( 6) = 1, g '( 6) = 9, then '( ) h equals: (A) 45 (B) 0 (C) 15 (D) 90 (E) 60. In the graph of f ( ) shown below, at which points listed is the slope increasing? (A) A only (B) B only (C) A, B (D) A,C (E) B,C

43 68 4. The equation of the tangent to y =! is: + y = when 1 (A)! 4y! = 0 (B) + y + 4 = 0 (C) + 4y + 6 = 0 (D) + 5y + 8 = 0 (E) y! =!4 5. g ( ) is an even function and f ( ) is an odd function. It is known that f ( 4) = 6, f '( 4) =, g ( 4) = 8, and ( ) f ( ) If h( ) =, then '( 4) g ( ) h! = g ' 4 =!. (A) 1 (B) (C) 1 (D) (E) cannot be found 6. If y =!, the minimum value of y is: (A) 1 (B) 1! (C) 0 (D) 1! (E) In the graph of f ( ) ( 1) ( 4) function is decreasing is: =!!, the intervals of for which the (A) < 1 only (B) 10 < < only (C) < 1 or 10 < < 7 7 (D) > or < 1 (E) > or 10 1< < 7

44 69 8. Consider the graph shown below. State which of the following has the least numerical value: (A) f ( 1) (B) f '( 1) (C) f "(! 1) (D) ( ( 1) ) f f (E) f ( 1)! f ( 0) 1! 0 Answers to Worksheet 7 1. A. D. D 4. E 5. A 6. D 7. B 8. D

45 70 Worksheet Let f be the function defined by f ( ) a) On what intervals is f increasing? =! 5 +. b) On what intervals is the graph of f concave upward? c) Write the equation of each horizontal tangent line to the graph. d) Sketch an approimate graph of y = f ( ).. Note: This is a graph of the derivative of f, not the graph of f. '( ) f is defined for! 4 " " 6. The figure above shows the graph of f ', the derivative of the function f. a) For what values of does the graph of f have a horizontal tangent? b) For what values of does f have a relative maimum?

46 71 c) For what values of is the graph of f concave downward? d) For what values of does f have an inflection point?. Note: This is a graph of the derivative of f, not the graph of f. The figure above shows the graph of f ', the derivative of a function f. The domain of f is the set of all such that 0!!. a) Does 1 = yield an inflection point on the graph of y f ( ) b) Is (1,) a maimum point on the graph of y = f ( )? c) Sketch a graph of y = f ( ) given that f ( ) function. =? 0 = 1 and f is a continuous

47 7 4. Let f be a function that is even and continuous on the closed interval [ -, ]. The function f and its derivatives have the properties indicated in the table below. 0 0 < < 1 1 1< < < < f ( ) 1 Positive 0 Negative -1 Negative f '( ) Undefined Negative 0 Negative Undefined Positive f "( ) Undefined Positive 0 Negative Undefined Negative a) Find the -coordinate of each point at which f attains an absolute maimum value or an absolute minimum value. For each -coordinate you give, state whether f attains an absolute maimum or an absolute minimum. b) Find the -coordinate of each point of inflection on the graph of f. Justify your answer. c) Sketch the graph of a function with all the given characteristics of f.

48 7 Answers to Worksheet 8 1. a) f '( ) = 15 4! 15 = 15 (! 1) f is increasing when! " 1 or! 1. b) f "( ) = 60! 0 = 0(! 1) f "( ) = 0 when = 0, 1, or 1! f concave up when 1! < < or 1 > 0 c) f '( ) = 0 when = 0, 1 or! 1. d) =!,,4 f ( 0) = f ( 1) = 0 ( ) f! 1 = 4 Horizontal tangents: y =, y = 0, y = 4

49 74. a) f has a horizontal tangent at points where f '( ) = 0. This occurs at =!, 0, and 4. b) f has a relative ma. at =!. c) f is concave down when! 4 < <! or < < 4.. a) Yes b) No c)

50 75 4. a) f is even f ( ) f ( ) f ( ) f ( ) f ( ) f ( )! = ", ' = " ' ; " 1< " = # 0 f has absol. ma. at = 0 ; absol. min. at = ± b) f has inflection points at = ± 1 c)

51 76 Proof of the Product Rule Let y = f ( ) g ( ). Then dy d = ( by definition) f ( + h) g ( + h) " f ( ) g ( ) lim h! 0 h = ( ) ( + )" ( ) ( ) + ( + ) ( + )" ( ) ( + ) h! 0 f lim g h f g f h g h f g h ( ) ( ) ( ) h ( ) ( ) f # g + h " g $ # f + h " f $ = lim % & + lim g ( + h % & ) h! 0 h h! 0 h ( ) ( ) ( ) ( ) g + h " g f + h " f = f ( ) lim + lim g ( + h) lim h! 0 h h! 0 h! 0 h ( ) '( ) ( ) '( ) = f g + g f Let y = f g ( ) ( ) = f ( )! " g( ) By the Product Rule and Chain Rule: dy d = g ( %1! " )# $ f ' '( ) ( ) Proof of the Quotient Rule # $ %1 ( ) + f ( ) (%1)! g( ) ( ) '( ) g ( ) = f f g g! " $ # % ( ) '( )! ( ) '( ) " g ( ) # g f = f g $ % " ( ) % # $ g '

52 77 Proof of the Chain Rule The Chain Rule is best shown by using dy and d notation (attributable to Liebnitz). Suppose y = f ( ) = g( h( ) ) then this can be decomposed as: f ( ) = g ( u) where u h( ) =. i.e. ( ) ( ) y = g u! "# u = h "$ We wish to find dy d. We wish to show dy dy d =. d du du dy d = (by definition)! y #! y! u $ lim = lim % i &! " 0!! " 0'! u! (! y! u = lim i lim! " 0! u! " 0! Since u is a continuous function of, it follows that as! " 0 then! u " 0. dy! y! u # = lim i lim d! u" 0! u! " 0! dy du dy d = i = du d du du