AP Calculus BC Review Applications of Derivatives (Chapter 4) and f,

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1 AP alculus B Review Applicatios of Derivatives (hapter ) Thigs to Kow ad Be Able to Do Defiitios of the followig i terms of derivatives, ad how to fid them: critical poit, global miima/maima, local (relative) miima/maima, iflectio poit The Mea Value Theorem, icludig Rolle s Theorem, ad how to apply each of them The First Derivative Test ad the Secod Derivative Test Iterpretatios of cocavity i terms of derivatives Give ay oe of f, f, ad f, fid characteristics of the others ad be able to sketch a graph of y f Horizotal ad slat (oblique) asymptotes ad how to fid them, icludig polyomial log divisio Optimizatio problems icludig the use of trigoometry (be able to justify your aswers, usually by usig the Secod Derivative Test to show that the relevat poit is a miimum whe you wat a miimum or a maimum whe you wat a maimum) Newto s Method icludig iteratios thereof ad its derivatio Basic atiderivatives usig the Sum, Differece, ostat Multiple, Product, Quotiet, Power, ad hai Rules (do t forget!) Practice Problems Problems are to be doe without the use of a calculator Problems may be doe with a calculator Fid the most geeral atiderivative for each fuctio give a f sec ta cos csc c f ta sec 5 b f 5 d f cos si 5 7 (99AB) A particle, iitially at rest, moves alog the -ais so that its acceleratio at ay time t is give by () a t a Fid the value(s) of t for which the particle is at rest b Write a epressio for the positio ( t ) of the particle at ay time t c Fid the total distace traveled by the particle from t to t t The positio of the particle whe The fuctio f with domai [ 7,7] parts, justify your aswer t is is show at right For all a O what iterval(s) is f ( ) icreasig? b O what iterval(s) is f ( ) cocave dow? c O what iterval(s) is f positive? d At what value(s) of does f ( ) have a relative miimum? e At what value(s) of does f ( ) have a poit of iflectio? f If l is the taget lie to the graph of where, which is larger l or ( )? f at the poit f Eplai

2 (99B) Let f for ad f a The lie taget to the graph of f at the poit, b A isosceles triagle whose base is the iterval from (, ) to ( c,) k f k itercepts the -ais at What is the value of k? Show all of your work has its verte o the graph of f For what value of c does the triagle have maimum area? Justify your aswer 5 Fid a value of b for which the fuctio y b 8 has both a horizotal taget lie ad a poit of iflectio at the same value of Show all of your work 6 A crate, ope at the top, has four vertical sides, a square bottom with side legth, a height h, ad a volume of m If the crate has the least possible surface area, show use of calculus to fid its dimesios Justify your aswer does ot satisfy the coditios of the Mea Value Theorem because f does ot eist 7 The fuctio f o [ 8,8] a f ( ) is ot defied c f ( ) does ot eist e ( ) b f ( ) is ot cotiuous o [ 8,8 ] d f ( ) is ot defied for < 8 Let f be a fuctio that is differetiable o the ope iterval (, ) If f 5, f ( 5) 5, ad which of the followig must be true? I f has at least two zeros II The graph of f has at least oe horizotal taget III For some c, < c < 5, f () c a Noe b I oly c I ad II oly d I ad III oly e I, II, ad III 9 If Newto s Method is used to approimate the positive root of f value of is a b 5 c 9 d e Let f I usig Newto s Method to approimate the root of f i [ ] will fid this root (with iitial estimate 5)? a c b d Which of the followig represets the oblique asymptote for f e f 9 5, 5, startig with, the the,, which equatio below 5? a y b y c y 7 d y 5 e y 7

3 The secod derivative of the fuctio f is give by f ( a)( b) The graph of f is show at right For what values of does the graph of f have a poit of iflectio? a ad a b ad m oly c b ad j oly d, a, ad b e j ad k Aswers a b c F ta d F si F sec si cot F a t ad t b () t t t c a (,5) ( 5,7), b ( 7, 5) (,5) c ( 5,) ( 5,7), d e { 5,,5}, f f ( ) a k, b c 5 b 6 6 ad h 7 e, 8 e, 9 c, c, e, a Solutios a The first term is the derivative of sec ; si ; the third, the derivative of cot ; ad the simply eeds to be multiplied by Do t forget the costat F sec si cot the secod (by the hai Rule), the derivative of 75 b Rewrite the secod through fourth terms as 5 ad apply the Power Rule for Atiderivatives to F the whole thig to get c This is the output of the hai Rule applied to d This is the output of the Product Rule applied to ta Icludig the costat, ad si Therefore a Give a(), t we ca fid v() t by atidifferetiatig a( t ) We get v ( ), we ca solve t {, } for This meas F ta F si, ad sice we kow that v t t t Solvig v t t t gives t t b Sice we have v() t from part a, we ca fid ( t ) by atidifferetiatig v( t ) This gives Sice we ve bee told that, we ca solve for, meaig that ( t) t t a The fuctio is icreasig whe f >, which is o ( ),5 5,7 b The fuctio is cocave dow whe f is decreasig, which is o ( ) 7, 5,5 t t t

4 c The secod derivative is positive whe f is icreasig, which is o ( ) d Relative miima occur at values of such that f ad f situatio occurs oly at e Poits of iflectio occur at values of such that f, ad 5 5, 5,7 is chagig from egative to positive This has a relative etreme This situatio occurs at 5, f The fuctio is cocave up from to f is icreasig o that iterval Therefore the taget lie at a value will, at a higher value of still o that iterval, have a smaller y-value tha the fuctio itself will f > l Therefore sice a f ad -itercepts occur at values of such that f ( ) The taget lie to f at, y f ( k) k( k), ad pluggig i ( y, ) (, ), we have f ( k) k( k) f ( k) k, the equatio is k k( k), which is solved for k {,6 } However, ( 6) the fuctio is oly defied whe its output is oegative, so k k f k is Sice we kow f ad b The isosceles triagle has a base of legth c ad the -coordiate of the third poit is halfway betwee ad c; that c c c c is, Therefore the triagle has area A c f ( ) Pluggig i to f gives A c 6 c 8c We wish to maimize this, so 6 8 c gives c ±, of which oly c is relevat We justify this by fidig c ad pluggig i c to fid, ad sice is egative, this is a relative maimum c 5 The horizotal taget meas that at some, y, 8 ad y Sice y b solves to ( y, ) (, ), ad more importatly, b 6 ad the poit of iflectio meas that at (the same), y b 8 y b, we ca set up the system b 8 which b 6 The volume is V h ad sice V, we kow h The surface area is A h, but usig our kow 6 h, A 6 We wish to miimize this, so gives We justify this by d fidig Pluggig i gives 6 ad sice 6>, this is a relative miimum d d 7 Optio a is ivalid because f ( ) eists (ad is equal to ) Optio b is ot true The derivative is c is ivalid; f ( ) Optio d is also ot true This leaves optio e, which is true: fidig ( ) f would require dividig by zero Therefore e is correct 8 I is true because, sice f 5 ad ( 5) 5, Value Theorem; likewise, sice f ( 5) 5 ad f so, f f must be zero somewhere betwee ad 5 by the Itermediate f 9 5, f must be zero somewhere betwee 5 ad 9 (Note that f beig differetiable meas that f is cotiuous ad thus meets the coditio of the Itermediate Value Theo- f f 9, there eists some c betwee ad 9 such that f ( c) rem) II is true by Rolle s Theorem: sice

5 III is true by the Itermediate Value Theorem: sice f 5 ad f 5 5, betwee ad 5, f assumes every value betwee 5 ad 5 Sice I, II, ad III are all true, the aswer is e 9 Newto s Method gives f 5 f ( ) f ( ) Sice f ( ), 9 f Newto s Method gives arry out the log divisio as follows: 7 R 5 ( ) 7 ( 7 8) f ( ) f ( ) Sice The quotiet is 7, which correspods to optio e This is choice c f Poits of iflectio occur at values of such that f, 5, choice c ad is chagig sig This situatio is at ad a oly ad

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