Math 10A final exam, December 16, 2016

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1 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 Total Poits 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

2 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 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) = 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 x x3 + x, so the value of the itegral is = = 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

4 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

5 (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

6 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

7 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 l. If we average these two, we get this approximatio ( trapezoidal rule ): l x dx (l + l ) + l + l l( ). We ca write the right-had side as l + l + l 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

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