PRACTICE FINAL/STUDY GUIDE SOLUTIONS

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1 Last edited December 9, 03 at 4:33pm) Feel free to sed me ay feedback, icludig commets, typos, ad mathematical errors Problem Give the precise meaig of the followig statemets i) a f) L ii) a + f) L iii) + f) L iv) + f) v) a f) Solutio i) For every ɛ > 0, there eists δ > 0 such that if 0 < a < δ the f) L < ɛ ii) For every ɛ > 0, there eists δ > 0 such that if 0 < a < δ the f) L < ɛ iii) For every ɛ > 0, there eists N > 0 such that if > N the f) L < ɛ iv) For every M < 0, there eists N > 0 such that if > N the f) < M v) For every M < 0, there eists δ > 0 such that if 0 < a < δ the f) < M Problem Prove the followig statemets usig the it defiitios i) 0 + ii) e iii) + e + Scratch work i) We wat to prove the followig statemet: for every ɛ > 0, there eists δ > 0 such that 0 < 0 < δ implies + < ɛ We have + + ii) We wat to prove the followig statemet: for every ɛ > 0, there eists δ > 0 such that 0 < < δ implies < ɛ We have ) ) ; we re goig to work with the last epressio If is small ie ), the 5 7 ad this is a ieact statemet which eeds to be made precise) I ca esure that 3 < 5 < if < 5, which is satisfied wheever δ 5 Coicidetally, < 5 also esures that 9 < < Thus < 5 5 implies that 5+0 < 3 9 check this) If δ ɛ is less tha 3/9 i additio to beig less tha or equal to 5 ) ) 3 ), the 5+0 < ɛ Notice that the coditio δ < ɛ 3/9 ad δ < ɛ 5 is equivalet to the coditio δ < mi{ 3/9, 5 } iii) This tured out to be harder tha I epected) We wat to prove the followig statemet: for every ɛ > 0, there eists N > 0 such that > N implies e e + < ɛ We have e as log as > 0, which we ca force by takig N 0) If e < ɛ, the e + look for a N such that > N implies e < ɛ We will show that e that the fuctio f) e Typo: the prited versio said 5 e + < e e + < ɛ, so we re goig to 0 We will also show is decreasig o the iterval, ) This will show that e 0

2 Solutio i) Let ɛ > 0 Set δ > ɛ Assume 0 < 0 < δ The 0 < δ ɛ Sice +, we have + < ɛ Thus + + < ɛ ii) Let ɛ > 0 Set δ < mi{ ɛ 3/9, 5 } Assume 0 < < δ The < 5, which implies 3 < 5 < ad 9 < < Thus < 3 ɛ 9 Also, we have < 3/9 Thus ) ) < 3 9 ɛ 3/9 ɛ iii) Let s first show that e 0 We have 0 by Problem i berkeleyedu/~shims/fa3-a/practice-problems-0pdf We have e < for all sice < e Let ɛ > 0 Sice 0, there eists N > 0 such that > N implies 0 < ɛ Assume > N The e 0 e < 0 < ɛ Thus e 0 Let s show that if < y, the f) > fy) We have ) y e > + y ) + y y where the iequality marked ) follows from the fact that e > + for all positive sice e + +! + 3 3! + ) The ey > y implies f) > fy) We ow show that e 0 Let ɛ > 0 Sice e 0, there eists a positive iteger N > 0 such that if is a positive iteger such that N, the e e 0 < ɛ Assume that is a real umber such that > N Let be a iteger such that > N The e 0 e < e < ɛ Thus e 0 e We ow prove that e + Let ɛ > 0 Sice e 0, there eists N > 0 such that > N implies e e e 0 < ɛ The e + e + < e e < ɛ Thus e + Problem 3 i) State ad prove the Squeeze Theorem ii) Use the Squeeze Theorem to compute Justify your aswer carefully 0 si ) 4 cos ) 3 Solutio i) If f, g, h are real-valued fuctios such that f) g) h) ) whe is ear a ecept possibly at a) ad if a f) a h) L, the a g) L The precise meaig of ear a ecept possibly at a) is: there eists some δ > 0 such that ) holds for all satisfyig 0 < a < δ Proof: Suppose f, g, h satisfy ) whe 0 < a < δ for some fied δ > 0 Also suppose that a f) a h) L Let ε > 0 Sice a f) L, there eists δ > 0 such that if 0 < a < δ, the f) L < ε Also, sice a h) L, there eists δ 3 > 0 such that if 0 < a < δ 3, the h) L < ε We ca also use L Hospital s rule, but the poit was to compute the it usig the defiitios

3 Put δ mi{δ, δ, δ 3 } Suppose 0 < a < δ The ), L ε < f) < L + ε ad L ε < h) < L + ε hold Hece g) h) < L + ε Also L ε < f) g) Thus L ε < g) < L + ε, so g) L < ε Hece a g) L ii) Let g) si ) 4 cos ) 3, f), h) Sice cos ad si for all, we have si ) 4 cos ) 3 for all Thus ) holds I additio, we have 0 f) 0 h) 0 Thus 0 g) 0 Problem 4 Let f ad g be fuctios, ad suppose a f) L ad a g) K Prove that a f) + g)) L + K Solutio Let f ad g be fuctios, ad suppose a f) L ad a g) K Let ɛ > 0 Sice a f) L, there eists δ > 0 such that if 0 < a < δ, the f) L < ɛ Also, sice a g) K, there eists δ > 0 such that if 0 < a < δ, the g) K < ɛ Put δ mi{δ, δ } Suppose 0 < a < δ The f) L < ɛ ad g) K < ɛ Hece f) + g)) L + K) f) L) + g) K) f) L + g) K ɛ + ɛ ɛ Hece a f) + g)) L + K Problem 5 Evaluate ) You should show your reasoig carefully; however you may use ay of the it laws without eplaatio or proof Solutio We have ) ) 0 0 )) )) + ) + 0 ) ) ) 0 Problem 6 Idicate true if the statemet is always true; idicate false if there eists a coutereample

4 i) If a f) L, the a + f) L ii) If a + f) L, the a f) L iii) If f) 0, the f)e 0 iv) If a f)), the a f) Solutio i) True Ituitio: If the two-sided it is L, the a oe-sided it is also L) Proof: Suppose a f) L Let ɛ > 0 The there eists δ > 0 such that 0 < a < δ implies f) L < ɛ Thus if 0 < a < δ, the f) L < ɛ Thus a + f) L ii) False Cosider the fuctio if > 0 f) if < 0 The 0 + f) but 0 f) does ot eist iii) False We have e 0 but e e 0 iv) False Cosider the fuctio f i the solutio to ii) agai; the f)) is idetically o, 0) Problem 7 0, ) so 0 f)), but, as oted above, 0 f) does ot eist i) Give the precise meaig of the statemet f is cotiuous at a ii) Usig the defiitio i i), show that f) is cotiuous at Solutio i) The statemet f is cotiuous at a meas f is defied i some iterval cotaiig a ad a f) fa) ii) The fuctio f) is defied everywhere, so it is defied i a iterval cotaiig a We have, so f) is cotiuous at Problem 8 i) State the Itermediate Value Theorem 3 ii) Prove that e si 40 has a solutio i 0, ) Solutio i) Let f be a cotiuous fuctio o the iterval [a, b] Suppose fa) fb) ad let N be a value strictly betwee fa) ad fb) The there eists some c a, b) such that fc) N Remark: f eeds to be cotiuous o the closed iterval [a, b], ot just o the ope iterval a, b), ad we ca guaratee that such c eists i the ope iterval a, b), ot just the closed iterval [a, b]) ii) Let f) e si Note that f) is cotiuous o R We have f0) e 0 si 0 0 < 40 We have Problem 9 f 03π ) e 03π si 03π e 03π si006π + π ) e 03π > 03π > 03 > 40 Sice f) is cotiuous o the closed iterval [0, 03π ], the IVT implies that there eists c 0, 03π ) such that fc) 40 Thus f) 40 has a solutio i 0, 03π ) Thus f) 40 has a solutio i 0, ) i) Give the precise meaig of the statemet f is differetiable at a ii) Usig the defiitio i i), show that f) is differetiable at Solutio i) The statemet f is differetiable at a meas f is defied i some iterval cotaiig fa+h) fa) a ad the it h 0 h eists which we deote f a) ) 4 3 The paper versio asked you to prove the IVT, too; it was a typo 4 f) fa) It s equivalet to say that the it a eists a

5 ii) Let f) The f is defied o all of R, so it is defied o a iterval cotaiig We fa+h) fa) a+h) a) have h 0 h h 0 h h 0, which eists ad is equal to ), so f is differetiable at Problem 0 i) State Rolle s Theorem ii) State the Mea Value Theorem iii) Prove the Mea Value Theorem usig Rolle s Theorem Solutio i) Let f be a fuctio cotiuous o [a, b] ad differetiable o a, b) Suppose fa) fb) The there eists c a, b) such that f c) 0 ii) Let f be a fuctio cotiuous o [a, b] ad differetiable o a, b) The there eists c a, b) such that f c) fb) fa) b a iii) Let f be a real-valued fuctio that is cotiuous o [a, b] ad differetiable o a, b) Defie g) f) fa) fb) fa) b a a) Note that g is a real-valued fuctio that is cotiuous o [a, b] ad differetiable o a, b) Note that ga) gb) Hece, by Rolle s Theorem, there eists c a, b) such that g c) 0 But g ) f ) fb) fa) b a Hece f c) fb) fa) b a Problem I each of the followig cases, evaluate dy d i) y + ii) y arctasi ) ) iii) y + 3y + e cos iv) y Solutio i) Use the quotiet rule: ii) Use the chai rule: iii) Differetiate implicitly: Thus so dy d + ) )) + ) + + ) dy d + si ) ) si ) ) si )cos ) si cos + si ) ) + si ) 4 y dy d + 3y + 3 dy ) + e cos e si d y + 3) dy d e cos e si 3y dy d e cos e si 3y y + 3

6 iv) Let g) e l The g ) e l l + ) l + ) The dy d g) ) e g) l ) e g) l )g) l ) e g) l ) g ) l + g) ) l + )l ) + ) Problem Aleader Coward s youtube chael has subscribers at time t 0, ad the umber of subscribers grows epoetially with respect to time At time t 4, he has 03 subscribers After how log will Aleader have 0 6 subscribers? Solutio Let yt) be the umber of subscribers at time t Sice yt) grows epoetially, we have yt) y0)e Ct for some costat C We re give that y0) ad y4) 03 Thus 03 e 4C, so C Let t 0 be the time at which yt 0 ) 0 6 The 0 6 e Ct0 implies t 0 Problem 3 Which poit o the graph of y is closest to the poit 5, )? l 4 l C l 03 l 03 4 Solutio A arbitrary poit o the graph y is of the form, ) The distace betwee, ) ad 5, ) is d) 5) + + ) We have d ) / ) Thus d ) 0 if ad oly if Sice the coefficiets of sum to 0, a root of is Thus divides , ad ) + + 5) ) Sice ) + 4, it is always positive Thus has eactly oe root, amely Furthermore, by ), we have > 0 if > ad < 0 if < Thus d ) > 0 if > ad d ) < 0 if < Thus is a global miimum of d) Thus, ) is the poit o the graph of y which is closest to 5, ) Problem 4 The iterior of a bowl is a coic frustum, where the top surface is a disk of radius ad the bottom surface is a disk of radius ad the height of the cup is 3 A liquid is beig poured ito the bowl at a costat rate of 4 How fast is the height of the water icreasig whe the bowl is full? Solutio Let ht), rt), ad V t) be the height, radius of the surface), ad volume of the water i the bowl at time t, respectively The followig is the side view of the bowl:

7 4 3 rt) ht) 3 + ht) 3 By similar triagles, we have Thus We have 3 rt) 3 + ht) rt) ht) V t) 3 π rt)) )3 + ht)) 3 π /) )3) ) ) ht) + 3 π 3 + ht)) π π 7 ht) + 3)3 π 4 Thus V t) π 9 ht) + 3) h t)) Let t 0 be the time at which the bowl is full The ht 0 ) 3 Sice V t) 4 for all t by assumptio, we have h t 0 ) 36 πht 0 ) + 3) 36 π 6 π Problem 5 Showig your work carefully, evaluate the it Solutio We have + si ) cos ) 0 + si ) cos ) + si )cos ) cos ) si ) 0 0 by L Hospital s Rule The secod it does ot eist sice + si )cos ) cos ) si ) + 3) 0 +

8 ad + si )cos ) cos ) si ) 0 Notice that we caot apply L Hospital s Rule to 3) sice it is ot a idetermiate form: si )cos ) cos ) si ) while 0 + 0) Thus the desired it does ot eist Problem 6 i) Give the precise defiitio of the defiite itegral usig Riema sums ii) What s the differece betwee a defiite itegral ad a idefiite itegral? iii) Usig the defiitio i i), compute 0 d Solutio i) Let f : [a, b] R be a fuctio For each N recall that N is the set of positive itegers), pick a collectio of sample poits,, so that b a i lies i [a + i ), a + i b a ] The defiite itegral of f from a to b is defied as b ) b a f) d f i ), a i provided this it eists ad gives the same value for all possible choices of sample poits ii) A defiite itegral is a umber, while a idefiite itegral ie atiderivative) is a fuctio iii) Let f), a 0, ad b For coveiece, let s choose i computig the right-had Riema sums) We have d i ) Problem 7 0 i i i 8 + ) + ) / i) State the Fudametal Theorem of Calculus a + i b a for all i, we re ii) Let f : R R be a differetiable fuctio Prove that if g is a atiderivative of f, the there eists a costat C such that f) g) + C for all iii) Are all cotiuous fuctios differetiable? iv) Do all cotiuous fuctios have atiderivatives? Solutio i) Let f be a cotiuous real-valued fuctio o [a, b] Let g : [a, b] R be defied by g) a ft) dt The g) is cotiuous o [a, b] ad differetiable o a, b), ad g ) f) Furthermore, if F is ay atiderivative of f, the F b) F a) b a ft) dt By lettig b vary, this implies, i particular, that F ) F a) ft) dt for all [a, b]) a

9 ii) Notice that f ad g are fuctios defied o R Choose a iterval [a, b] Notice that both f ad g are atiderivatives of f The secod part of the FTC states that f) fa) g) ga) for all [a, b] This implies that f) g) fa) ga) for all [a, b] Let C fa) ga) If [a, b], the we ca repeat the above argumet to ay closed iterval I [a, b ] which cotais [a, b] ad which will yield the coclusio that f) g) fa ) ga ) for all [a, b ]) Thus f) g) C for all R iii) Not all cotiuous fuctios are differetiable For eample, f) is ot differetiable at 0 iv) All cotiuous fuctios have atiderivatives, by the first part of FTC: if f is cotiuous o [a, b], the the fuctio g) ft) dt is a atiderivative of f a Problem 8 Compute a atiderivative of the followig fuctios i) f) ii) f) 5 + ) iii) f) + iv) f) taarcsi)) v) f) 3 + Solutio i) We have ) d C C ii) We have 5 + ) d /5 + /5 + ) d 5 7 7/ /5 + + C iii) Use the substitutio u + The du d We have + du ) d u d u du d 3 u3/ + C 3 + ) 3/ + C where we have used the Substitutio Rule i the step marked ) iv) Write taarcsi)) ad use the substitutio u Thus taarcsi)) d d ) u du d d u du u + C + C v) Write We compute atiderivatives of + ad + separately A atiderivative of + is 3 + ) 3/ + C by iii) We use the same substitutio u + to compute a atiderivative of

10 + du The d We have + d du u d PRACTICE FINAL/STUDY GUIDE SOLUTIONS d ) du u + C + + C u where we have used the Substitutio Rule i the step marked ) Hece 3 + d 3 + ) 3/ + + C Problem 9 i) Fid the volume of the solid obtaied by rotatig the regio {, y) : 0 e y, y } about the y-ais ii) Fid the volume of the solid obtaied by rotatig about the y-ais the regio betwee y ad Solutio y i) The itersectio of the solid with the plae y y 0 is a circle of radius e y0 Thus the ifiitesimal cross-sectio of the solid has volume πe y0 ) dy where dy is the thickess of the crosssectio Thus the volume of the solid is πe y ) dy πe y dy π ey π e4 e ) ii) The curves y ad y itersect at 0, 0) ad, ) ad they are strictly icreasig, so the regio they boud is cotaied i the bo {, y) : 0, 0 y } The itersectio of the solid with the plae y y 0 is a aulus a disk with a smaller cocetric disk removed from it) of ier radius y 0 ad outer radius y 0, which has area π y 0 ) πy 0) Thus the volume of the solid is 0 π y) πy ) dy 0 πy y 4 ) dy π y ) 5 y5 3π 0 0 Problem 0 Simplify log log3 9log 4 ) Solutio We have log log3 9log 4 ) log /)