Please put away all books, calculators, cell phoes ad other devices. You may cosult a sigle two-sided sheet of otes. Please write carefully ad clearly, USING WORDS (ot just symbols). Remember that the paper you had i will be your oly represetative whe your work is graded. Please write your ame clearly o each page of your exam. Your paper will be scaed ad will be processed usig Gradescope. It is essetial that you had i all pages that you have received (icludig this cover sheet) ad that the order of the pages be preserved. Poit couts: Problem 3 4 5 6 7 8 9 Total Poits 7 8 7 6 7 8 7 7 6 7 7 These quick ad dirty solutios were writte by me (Ribet) just before the exam. These are ot ecessarily perfect, model solutios, but rather my attempt to explai what s goig o. a. Fid all poits o the iterval [, ] where the istataeous rate of chage of f(x) = x 3 + x is equal to the average rate of chage of f(x) o the iterval. The average rate of chage is defied o page 8 of Schreiber i a highlighted f(b) f(a) gree box: it s if we re dealig with the iterval [a, b]. Here the b a iterval is [, ], so it s f() f() =. The istataeous rate of chage at c is f (c) = 3c +. We are asked to fid all c i the iterval such that 3c + =, i.e., such that c = 3. There s oly oe such c, amely 3 =. 3 b. If the derivative of f(x) is x +, what is the derivative of f(x )? d By the Chai Rule, dx f(x ) = f (x ) d dx (x ) = x + x = x +. Thus the derivatives of f(x) ad f(x ) are egatives of each other. This may seem strage at first. The explaatio is that f(x) is arcta x (plus a costat), so that f(x ) is the agle whose taget is x. π f(x), so f(x ) is really a costat plus f(x). You acted with hoesty, itegrity, ad respect for others. That agle is
a. Suppose that is a positive iteger. Calculate the itegral l x dx. The fact that the upper limit of itegratio is a iteger is a red herrig it could be ay real umber. To do this problem, you eed to fid a atiderivative of l x. This is a stadard itegratio by parts situatio, ad the atiderivative will tur out to be x l x x; see Example o page 395 of Schreiber. (For the test, my itetio was for you to do the itegratio by parts o your paper.) The defiite itegral is the (x l x x) ] = ( l ) ( l ) = l +. b. For what values of x is the series = (x + 7) ( + ) coverget? Let a = (x + 7) ( + ). If c = x + 7, the a = ( + ) c. Sice ( + ) (as ), a if ad oly if c. We kow that c if ad oly if c <. Hece the series diverges wheever c. This coditio traslates to x + 7 ; it holds precisely whe x 3 or x 7. It remais to see what happes whe 7 < x < 3, or equivaletly whe c <. By the ratio test, the series a is coverget if lim a + is less tha. This limit is c, so we re i good shape: we lear that the series coverges exactly i those situatios where we did t kow that it diverges. To summarize: for all values of x, either the series diverges because its th term does t approach or it coverges by the ratio test. A priori there might have bee cases where the ratio a + a approached ad the th term approached. The we would have to do further aalysis to see whether or ot the series coverged i those cases. That would have bee a added difficulty ( tricky problem ); maybe ext year. a
3. What approximatio to (.) / is provided by the quadratic Taylor polyomial for f(x) = x / at the poit a =? (Leave your aswer as a usimplified umerical expressio.) Basically this problem was a exercise i locatig the relevat formula o your cheat sheet: the quadratic Taylor polyomial associated with f(x) at x = a is f(a) + f (a)(x a) + f (a) (x a). We have a =, x =., x a =.. Now f(x) = x, so f() = ; also f (x) = x /, so f () =. Similarly, f () = =. Hece the 4 approximatio i questio is + (.) (.4) =.995. 8 This approximatio is correct to four decimal places. 4. Determie the volume of the solid obtaied by revolvig the area uder the curve y = x + from x = to x = about the x-axis. This is aother problem where the mai exercise is to cosult your cheat sheet ad fid the right formula. The right formula here is V = πy dx. I this specific case, we have π (x + ) dx = π (x 4 + x + ) dx. A atiderivative of the itegrad is x5 5 + 3 x3 + x, so the value of the itegral is 5 5 + 3 3 + = 3 5 + 6 96 + 8 + 3 + = 3 5 is 6π 5. = 6. Accordigly, the volume 5 ( 5a. If a is a real umber, calculate lim + a. (As for all problems o ) this exam, be sure to explai your reasoig with care.) 3
Whe I set up this problem at first, the real umber was called A ad I wrote + A i place of + a. I decided that the joke of havig A be part of the problem was ot worth the extra burde o you. ( We ca write + ) a as e l((+ ) a ). Hece if L = lim ( l + a ), the ( the limit to be calculated will be e L. Now l + a ) ( = l + a ). Set h = a, so h as. We have L = lim ( l + a ) l( + h) = lim a h h sice = a h. l( + h) The limit lim is because it s (by defiitio) the h h derivative of the fuctio l(x) at x =. (You ca see this also by l Hôpital s rule.) Hece L = a, ad the aswer to the questio is e a. Why was this problem o the exam? Look for the setece This is a good exercise (or exam problem) o the December slides. b. Let f(x) = l x x approaches? for x >. What happes to f(x) as the positive umber x The fuctio l x approaches as x +. You re dividig a large egative umber by x, which is tiy ad positive. The quotiet is, so to speak, eve more large ad egative. So f(x) approaches as the positive umber x approaches. (Although the aswer was more or less apparet, I hope that your aswer icluded a explaatio ad was ot simply the usubstatiated statemet that the fuctio approaches.) 6a. Let f(x) be the fuctio l x, defied for x positive. Fid lim x f(x). x Both x ad l x approach ifiity as x, so we ca use l Hôpital s rule to calculate the limit: the ratio of the derivatives is x / =, which approaches x as x. Hece the limit is. Takeaway: f(x) at the right ed of its domai of defiitio, ad f(x) at the left ed of its domai of defiitio 4
(by problem 5b). Hece the maximum value of f(x) occurs somewhere i the middle. b. Does f(x) have a global maximum value? If so, what is this value? Yes, it has a maximum, which you fid by settig the derivative equal to. The derivative is a fractio, whose deomiator is x. The derivative vaishes whe the umerator is ; the umerator of the derivative is x d dx (l x) (l x) d (x) = l x. dx The derivative is exactly whe l x =, i.e., whe x = e. The maximum value is the f(e) = e. 7. Which is more likely: gettig 6 or more heads i tosses of a fair coi or gettig 5 or more heads i 4 tosses of a fair coi? This is very similar to problems you ve ecoutered before. We have to compute the z-statistic for the two situatios. The statistic that s the furthest from the ceter is the less likely. The geeral formula is Z = (X µ) N. Here, µ = σ ad σ = as well. Of course, σ will be the same i both scearios, so we ca igore it ad just see which of the two expressios (X ) N is bigger i absolute value. I the first sceario, X is replaced by.6 ad N =, so (X µ) N =. =. I the secod sceario, we have 5 4 = 5 4. Sice 5 >, the secod sceario 4 is the less likely oe. Thus it s more likely to get 6 heads i tosses tha to get 5 heads i 4 tosses. 8. If the cotiuous radom variable X has PDF equal to f(x), the we have E[g(X)] = g(x)f(x) dx for all reasoable fuctios g. Use this iformatio to calculate the expected value of X whe X is a stadard ormal variable (with mea ad stadard deviatio equal to ). 5
Accordig to the HW3 solutios file, the aswer is. Let s see why. I π the geeral descriptio of the problem, the fuctio g is the absolute value fuctio; moreover, f(x) = e x /. Cosultig the descriptio at the π begiig of the statemet of the problem, we realize that the umber to be computed is x e x / dx. Because the fuctio beig itegrated is π a eve fuctio, the itegral i questio is twice the aalogous itegral from to. Sice =, we ow recogize that the aswer i the HW π π solutios is correct if ad oly if we have xe x / dx =. I the itegral, set u = x /, du = x dx. The itegral becomes e u du = e u ] =, as required. 9. Fid all values of a ad b such that p(t) = is a cumulative distributio fuctio. aebt + ae bt The fuctio p(t) is a CDF if it has the followig three properties: () it s at ; () it s at + ; (3) it s a o-decreasig fuctio. The last property meas that p(t) p(t ) if t t. To see how p(t) behaves, we might multiply umerator ad deomiator by e bt. The we see that p(t) = a a + e bt. If a =, the p(t) is idetically, so it ca t be a CDF. Let s assume that a is o-zero ad divide umerator ad deomiator by a. If c = /a, the p(t) = + ce bt. 6
I order to get property (), we eed the deomiator to get large (i absolute value) for t ; this requires that b be positive. If b is positive, the e bt as t +, so () will be satisfied as well. For (3), we wat ce bt to decrease as t icreases. For that, we d better have c positive because ce bt is decreasig i view of our assumptio that b is positive. Note that c is positive if ad oly if a is positive because c = /a. Coclusio: p(t) is a CDF if ad oly if a ad b are both positive.. Explai how the approximatio l x dx l(!) l ca be obtaied by averagig together left- ad right-edpoit approximatios to the itegral. (Recall that! = 3.) [Problems a ad lead to Stirlig s approximatio to!.] The itegral i questio is the area uder y = l x betwee x = ad x =. The segmet of the x-axis from to ca be divided ito segmets, the first from to ad the last oe from to. If you use left-edpoits to approximate the area, the l x dx is approximated by the sum of the areas of rectagles, all of which have width ; their heights are l, l,..., l( ). The left-had approximatio is thus Similarly, usig right edpoits, we get l x dx l + l + + l( ). l x dx l + l 3 + + l. If we average these two, we get this approximatio ( trapezoidal rule ): l x dx (l + l ) + l + l 3 + + l( ). We ca write the right-had side as l + l + l 3 + + l (l + l ). The positive terms sum to l(!), sice! = 3 ad sice the log of a product is the sum of the logs. Furthermore, the term l ca be igored, sice it s. Thus the right-had side is l(!) l. 7