x \eu r^)l\t fee g$ca- Oeccstol5 f,-r*rufs OS f*dd* (tt.1\ t#(ns+,ns*$ l::i:ffi 'c^* * s""d F::*r2'#rHHHt\ A (, fh:?,-#':.""::t?l#.

Transcription

1 togtrurrb rro+ Corcrc4 on ShaS o( LrY$ ld,br lo,l, lo.-?l ll.lr ll'(4r ll'7' l3'{' lrl'o' l5'3 Math 202: Calculus for Business and Economics Fali 2013 Final Exam FridaY, December 13' 20L3 Name: Student ID:=--..- Malh 202 Section- Version A Directions ' Read all directions carefully,. you must show all work or jusfify your reasoning wlth rneaningful evidence to receive full credit, regarclless of whether it is explicitly asked for in the question'. Simplify completelywhereverplssible'. Clearly inclicate your final answer for each problem' ' Labql units and problems appropriately', The back page is ieft blank for you in case you need more space to work through a problem. Please indicate on the exam if you h;lge used the back page for a problem' Average Value o[ a Lontlnllous ar"rrgerareorchange (ro.q\ ^"'igii>^-irr>," tl+o A (, fh:?,-#':.""::t?l#.'{) fromx=et1x=q+his t t- b-al_f{*)o* Q rnstantr"":, Rate orchange(ro,nt)' 3:TJ*',':;:I?[i*(,-*s\ atx=ois lim*- F::*r2'#rHHHt\ (n,r) h+0 A - Dol't Relative Rate of Change (tt.1\ t#(ns+,ns*$ Elasticity of Demand (no,trffs*$ (tl,r) E(p) = - pf '(p) f (p) Error Bounds for APProximation of Area by Left or Right Sums error < lf Q) - f (a)l ry (r"s) (m+ ft\$""s) ff..}tr Jg f'. l::i:ffi 'c^* * s""d fv :,u' f (t)e*'t dt {o' "li::ff. consumers'surplus (rt,o) &m*ot$t'$i),, :,O, I o(x) - p) dx producers' surptus (t1,3) (no+crt$rgi) rx Ps = Jn I p.- s(x)] ax Integration by Parts Formula (fq 'a) :^"""'"1"0"'= ", - lr r" inot;sno$ x \eu r^)l\t fee g$ca- Oeccstol5 f,-r*rufs OS f*dd*

2 1". [5 points eachj Find the indicated derivatives. ll,e,ll,tla'f'(x)forf(x):(3xz+l)(x-2)z ll, b b. s'(x) for s(x) = H ll,4 c, h" (x) for h(x) -,x2+4x-3 2

3 2. [6 points each] Evaluate the limits. la.3 a. ex -t lim=..": x.6 2xt lo. b. x2+x-2 Iimx-l x-7 lb.df lo.b rim C, 10x-9x3 +1,7 6x3+2x*1 la.3 d. x-]. lim, x*1 lll X J

4 3. (2 points eachl Write TRUE if the statement is correct and FALSE if the statement is incorrect. lo, { a, The derivative can be interpreted as the average rate ofchange f,or any glven function. lo. b lq.& c, lo,a d. I l r-, e, f "(x) > o. A fnnction is concave upwards on the interval[s) where A Gini index of 1 indic{es ab;qlute equality - all people share equally in the income. (not Oh tfn$ll) f (x) =o? t, continuous at every value of x. &r*rton*'*$ When demand,is elastic, a nrice decrease will cause an increase in re'cnue. /ylo+ Un S-ne!) 16. I f. l&.5e lb, I h, lo.&, l{.d i. The criticai values of a functio n f (x) occurs where f '(x) = 0 or where f '(x) is undefined, The absolute maximum or minimum of a function can only occur at endpoints of a closed interval. A function F is an antiderivative of a function/ if F '(x) : f (x)' The value of x where the rate of changc of sales goes from increasing to decreasing is called the point of diminishing returns. The consumers' surplus represents the total savings to conslrmers who are willing to pay more than $p fo1(re product but are still able to buy the product for $p. (no# On +irfa.lt) le,l k. f'(x) : x2. lf f '(x) = 2r, then there is exactly one antiderivative, l4,o r' lb.6 m. ld. & n' Given a probability density lunctionl[xj, the probability that a product will last between a and b years can be represented by the integral, rb {' ru) or (no* on Stlq.[\ The average value of f (x) = 2x over 12, \ is L2. An inflection point on a graph of a function is the point where the slope of the tangent Iines to the graph change from positive to negative or from negative to positive,

5 l3.o la.6 4. [5 points each) Evaluate the integrals..1. ftz I ) W 2)o"' -,lt t?,b b l,'('-'*se*-lo' It. 3 c' lo'*,,* o* (no* on f, t',o.s) 5

6 lq,b d' {xtnx dx (nct on Snar) 16,& S"'(tnt)z o, Jn t le,5 p lb,t f, I (Zxz - x-t) dx

7 l.5 5. [3 points each) Circ]e the most appropriate answer to t]-re follolving questions given /(x) = 2x3 * 3xz - 1.Zx defined on the closed interval [-3,2] a. What is the critical value of the f,unctiort f (x)? i. x = 0,2,-3 ii. x: -2,3 iii. x = -1,,2 iv, x = 1,-2 v. x : -1,-3 b. What is the absolute maximum point of the function on [-3, 2]? i, (-3, -9) ii, (-2,20) iii. (0,0) iv. (1, -7) v, (2, +) c. What is the absolute minimum point of the function on [-3,2]? i. (-3, -9) ii. (*2,20) iii. (0,0) iv. (1, *7) v, (2, +) 16,rl 6. [3 points each) SuPPose /l ^- ' J, f t-> dx : -6, 1..' f G) a* a ll fe)a,= 'n the integrals are as defined belo',v, - 2, lf s@) d.x = 3, ana ff s(x) dx^= 5. (no+ an S"^$) b, Il f(i a, = d j,'(su1'ry' : e. Iief f*> - 3s(x)) dx :

8 . 7. [10 pointsj Find the area bounded by the graphs ofthe given equations over the Itl,l giveninterval. Y=x2+!, Y=*xz-1, -2<x<2 lll,& B. The price - demand and price - supply equation of a product are given by (""*, q**) P = Dt8;,1?; "X?fJ,ii3,'rl#; a. [5 points) Find the equilibrium quantity and price' b. [5 pointsj Calculate the consumers' surplus at the equilibrium price level. c. [5 pointsj Calcr,tlate the producers' surplus at the equilibrium price level.

9 la, ht g. Surnmarize the pertinent information obtained by applying the graphing strategy to analyze and sketch the function x _ Z ' x-5 y:f(x)::--; a, [2 points] Domain of f (x): b. [2 pointsj x - intercept[sj: r : c. [2 points] y - intercept; Y - d. [2 points) Vertlcal asymptote[sj: e. [2 pointsj Horizonta] asymptote[s): f. {2 pointsj I (x) is increasing on the interval: 9

10 o b' [2 points] f (x) is decreasing on the interval: h. [2 points] /(x) is concave upward on the interval: (2 points) /(x) is concave downward on the interval: ). (2 pointsj Does /(x) have an inflection Point? ICircle answerj YES NO I<. [6 pointsj Sl<etch the ft-rnction. 10