5 + x x 5 = 1. x 0 cos(5x+ π ). Since cos(5 0 + π) = 2 cos(π) = 0, 2

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Math Aswers for December eam. Fid the followig its a) b) = = = 4 3. c) cos ) cos5+ π ) = = =

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Math Aswers for December eam. Fid the followig its a) b) = = = 4 3. c) cos ) cos5+ π ) = = = = ) = Let s first try ad uderstad the it cos5+ π ). Sice cos5 + π) = cosπ) =, this it is of the idetermiate form. Sice the fuctio i the umerator ad deomiator are differetiable, we ca use L Hôpital s rule to see that the it is cos5 + π ) = si5 + π ) 5 = 5 si π ) = 5. Fially, sice cos is a cotiuous fuctio, we ca use the theorem o its ad cotiuity to echage the process of takig the it ad evaluatig cos ad coclude that ) cos cos5 + π) = cos ) cos5 + π) = cos ). 5 Page of . a) State the ǫ-δ defiitio for the meaig of a f) = L. We say that a f) = L if, for every ǫ there is a δ such that if a δ the f) L ǫ. b) Show usig the defiitio) that 6 4 =. Iitial Ivestigatio: 4 = 6) + ) = 6 +. We will be able to cotrol the size of 6 sice 6 δ. O the other had, if 6 δ, the + 8 + δ. This follows from either the picture below: 8 + δ δ δ 6 R or the triagle iequality: + = 8 + 6) 8 + δ. Puttig these estimates together, if 6 δ the 4 = 6 + δ 8 + δ). We wat to figure out how to choose δ so that δ8 + δ) ǫ. We ca impose ay coditios o δ that we like, so we ca for istace) suppose that δ . If δ the 8 + δ 9 ad so δ8 + δ) 9δ. If we simultaeously choose δ ǫ the δ8 + δ) ǫ. 9 Give ǫ choose δ mi, ǫ ). The if 6 δ, 9 4 = 6 + ǫ δ 8 + δ) 9δ 9 9 = ǫ. Sice this works for ay ǫ , 6 4 = by the defiitio of it. 3. Let f) = { si ) if if =. Page of a) Use the defiitio of the derivative to fid f ). By the defiitio of the derivative, f f + h) f) fh) f) h si ) = = = ) h h h h h h h = h h si h ). I order to evaluate this it, we ca use the squeeze theorem. Sice si is always betwee ad +, h h si ) h for all h. Sice h h h = = h h we coclude that h h si ) = by the squeeze theorem. h Therefore f ) =. b) Is f ) cotiuous at =? Eplai your aswer. I order for f ) to be cotiuous at = we must have by defiitio of cotiuity) f ) = f ) =. I.e., the it must eist, ad must be equal to zero. Whe we compute by usig the product ad chai rules) that f ) = si ) + cos ) = si ) cos ). As the term si ) by usig the same kid of squeeze theorem argumet as above. However as cos ) does ot eist the similar it si ) has bee a frequet eample i class). Therefore f ) = si ) cos ) does ot eist, ad so f ) is ot cotiuous at =. For compariso, here are graphs of f) ad of f ) scale adjusted to show detail): f) f ) O the graph of f) we ca see that the slopes of the secat lies coectig, ) to other poits o the graph approach zero as the poit approaches, ), i.e., the Page 3 of derivative f ) =. O the other had, o the graph of the f ) it appears that for istace) f ) has output close to + ad to ifiitely may times as, ad therefore the it ca ot eist. 4. Fid the derivatives of the followig fuctios. You do ot have to simplify your aswers. a) f) = 4 + 5) ) f ) = ) ) ) 4 3 ) ) 3 6 ) 5 + 8) 4 + 5) ) ) ) b) g) = + si 5 ) )6. g ) = 6 + si 5 ) )5 + 5 si 4 ) cos)). c) h) = arcta7 ) + π l5 + 3) h ) = + 7 ) l π ) a) State the mea value theorem for a fuctio f o o iterval [a, b] icludig the hypotheses ecessary for the theorem to hold). If f) is a fuctio which is cotiuous o [a, b] ad differetiable o a, b) the there is some poit c a, b) such that f fb) fa) ) =. b a b) If f satisfies the hypotheses of the mea value theorem o a iterval [a, b], ad if f ) = for all [a, b] show that f is a costat fuctio o [a, b]. Proof by cotradictio: Suppose that f is ot the costat fuctio. The there would be two poits ad i [a, b] where f ) f ). i.e., if the values of fuctio are equal for ay pair of poits, the fuctio must be costat). Page 4 of By switchig the order of the poits if ecessary we ca assume that . Sice [, ] is cotaied i [a, b], f) is differetiable o, ) ad cotiuous o [, ]. By the mea value theorem there is a poit c [, ] such that f c) = f ) f ). Sice f ) f ), f ) f ), therefore f c). This cotradicts the coditio that f ) = for all [a, b]. Therefore f) is a costat fuctio. Alterate solutio: Istead of arguig by cotradictio, we ca reverse the argumet to show that f ) = f ) for all poits, [a, b], ad therefore coclude that f is costat. Give ay two poits, [a, b] we ca assume that be switchig them if ecessary. The [, ] is cotaied i [a, b] so f is cotiuous o [, ] ad differetiable o, ). Therefore by the mea value theorem there is a poit c, ) so that f c) = f ) f ). Sice by hypothesis) f c) =, this meas that f ) f ) =, i.e., that f ) = f ). Sice the value of f at ay pair of poits is equal, f must be the costat fuctio. 6. a) If , show that ) ) e 3 + si ) e 3 + si ). Let f) = e 3 +si). The f ) = +3 e 3 +cos) = +cos))+3 e 3. Sice cos) for all, + cos) for all. The fuctios ad e 3 are also for all. Therefore f ) for all R, ad so the fuctio f is a icreasig fuctio. By the defiitio of icreasig ) fuctio, this meas ) that if the f ) f ), i.e., e 3 + si ) e 3 + si ). ) 4 + b) Show that for all real umbers,. Let f) = 4. Sice f ) =, f ) for all R. By the secod derivative test for cocavity, this meas that f) is cocave up. If r =, r 3 =, the Jese s iequality applied to a cocave-up fuctio says that 3 for ay, R, f 3 + ) 3 3 f ) + 3 f ), ) 4 + i.e., that = ) for all real umbers,. Page 5 of 7. a) A object is travellig alog the curve y =. At a poit i time t, t ) = ad t ) = 3. Fid the rate of chage of the distace from the object to the poit, ) at time t. Remider: The distace betwee poits, y ) ad, y ) is ) + y y ). ) The distace from the poit, ) to the poit, ) o the curve is D = ) + ) = + ). Differetiatig with respect to t ad usig implicit differetiatio, sice is also chagig with t) we get dd dt = + ) ) + ) = + ) + ) = ) + ). At time t = ad = 3. At time t the rate of chage of D is therefore ) 3 + ) = 4 3. b) Fid the poit o the graph of the parabola y = closest to the poit 54, ). The distace betwee the poit 54, ) ad a poit, ) o the parabola is give by D) = 54) + ). As or the fuctio D) also goes to ifiity. Therefore there must be a overall miimum distace. To fid it, we look for critical poits. Differetiatig, dd d = 56) + ) = 54) + ) ) + ). This is zero whe the umerator is zero, i.e., whe 3 56 = or 3 = 7, which oly occurs whe = 3. Sice there is a absolute miimum distace, ad sice ay absolute miimum must also be a local miimum, ad sice there is oly oe critical poit, we coclude that the absolute miimum distace occurs whe = 3. The poit o the parabola y = closest to the poit 54, ) is therefore the poit 3, 9). Page 6 of 8. Below is a graph of a fuctio f, made up of straight lies ad a piece of a circle y = f) 8 π Suppose that F is a fuctio defied o [, ] with F = f. No eplaatios are ecessary for your aswers to the questios below. a) O which itervals) is F icreasig? O [4, 8] ad [6, ] i.e, where F is positive). b) O which itervals) is F cocave dow? O [6, ] i.e., where F is decreasig). c) Which critical poits) of F are local maima? At = 8 The critical poits are where F =. The local maes are where F chages from positive to egative. Of the critical poits = 4, = 8, ad = 6, F chages from positive to egative oly at = 8). d) At which -value does F attai its absolute maimum o [, ]? At =. The absolute ma is foud amog the critical poits or at the edpoits = ad =. The oly possible local ma for the critical poits is = 8 by part d) above. We ca compare the values of F), F8), ad F) by lookig at the differeces F8) F) ad F) F). By the Fudametal Theorem of Calculus I these differeces are the same as areas uder the graph of F. Lookig at the graph of F, F8) F) = 8 + π which is egative, so F8) F). The differece Page 7 of F) F) = 6+π which is also egative, so F) F). This meas that F) is the largest of the three possible values for the global maimum, ad must therefore be the global maimum.) 9. The object of this questio is to prove, from the defiitio of the itegral, that for every b the fuctio f) = is itegrable o [, b]. + ) The formula k = may be useful. k= Let P be the partitio of [, b] ito equal itervals. a) Write dow the sums Uf, P ) ad Lf, P ). Let P = {,,..., } be the partitio of [, b] ito equal itervals. Sice f ) = 4 we see that f) is a icreasig fuctio. Therefore o ay subiterval [ k, k ] the miimum value m k of f o the iterval occurs at k ad the maimum value M k of f o the iterval occurs at k. Therefore m k = f k ) ad M k = f k ). Sice k = bk, this meas that m k = f k ) = 4 bk ) + 7 ad M k = f k ) = 4 bk + 7. The width of each iterval is = b. The upper ad lower sums are therefore Uf, P ) = 4 bk ) + 7 b 4b = k + 7b ), k= k= ad Lf, P ) = k= ) bk ) b 4bk ). = + 7b ). k= b) Fid formulas for Uf, P ) ad Lf, P ) which oly deped o b ad. 4b k Uf, P ) = + 7b ) = 4b k= = b + ) + 7b. k= k + 7b k= = 4b + ) + 7b Page 8 of Lf, P ) = 4bk ) k= = b ) + 7b ) = 4b + 7b. k= k ) + 7b k= = 4b ) + 7b c) Use your aswer from b) to show that is itegrable o [, b] ad to fid the value of the itegral b d. By defiitio f is itegrable o [, b] if, for ay ǫ we ca fid so that Uf, P ) Lf, P ) ǫ. Sice Uf, P ) Lf, P ) = b + ) + 7b b ) ) + 7b = 4b, we see that if we pick 4b the Uf, P ǫ ) Lf, P ) = 4b ǫ. Therefore f is itegrable o [, b]. The value of the itegral o [, b] is the uique umber betwee Uf, P ) ad Lf, P ) for all, i.e., the umber b + ) + 7b = b + 7b. Therefore b. Let A) = a) Fid A) d = b + 7b. + t4 dt. Sice + t 4, this implies that A) = A) Therefore as we have A) too ad so form ad we ca apply L Hôpital s rule. By L Hôpital s rule, A) 3 = A ) 3. + t4 dt dt = for all. is of the idetermiate 3 Page 9 of By the Fudametal Theorem of Calculus I, A ) = + 4. Therefore A) 3 = A ) 3 = = = + 3 = 3. b) Let g = A be the iverse fuctio for A. If c = A), compute g c). If g is the iverse fuctio of A, the the formula for the derivative of the iverse fuctio says that g c) =. Sice c = A), gc) = by defiitio of iverse fuctio. A gc)) We also kow that A ) = + 4 by Fudametal Theorem of Calculus I. Therefore g c) = A gc)) = = Page of
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