MATH 142–511, 516, 517 Spring 2010 Solutions for the ...

1 MATH 142 511, 516, 517 Spring 2010 Solutions for the sample problems for the Final1. How many CDs would a record company have to make and sell to break even if the fixedcosts are $18000, variable costs are $ per CD, and the CDsare sold to retailers for$ each?SOLUTION. The cost function isC(x) = 18000 + , the revenue function isR(x) = The company breaks even ifR(x) =C(x)18000 + 75002. Trunsville Utilities uses the following rates to computethe monthly cost of natural gasfor residential customers. Write a piecewise definition forthe cost of consumingxCCF(cubic hundred feet) of natural per month$ per CCF for the first 50 CCF$ per CCF for the next 150 CCF$ per CCF for all over 200 CCFSOLUTION. If 0 x 50, thenC(x) = Note, thatC(50) = (50)( ) = 50 x 200, thenC(x) = + (x 50) = + (200) = + ( )(200) = >200, thenC(x) = + (x 200) = + (x) = ,if 0 x + ,if 50 x + ,ifx >2003.

2 Find the domain of the functionf(x) = x 5,if 5 x 1012x+5,if 10< x 205x,ifx > 5 0x 5and 2x+ 56= 0 orx6= Since <10, the domain offis [5, ).4. The revenue and cost functions for a particular product are given below. The cost andrevenue are given in millions of dollars, andxrepresents the number of units (in thou-sands).R(x) = 64x2+ 4400xC(x) = 250x+ 5200At what production level(s), rounded to the nearest whole unit, will the company breakeven on this product?SOLUTION. The company breaks even ifR(x) =C(x). 64x2+ 4400x= 250x+ 5200x1 1, x2 645. Solve the following equations forx:(a) 26x= 8x2+126x= 8x2+126x= (23)x2+126x= 23x2+36x= 3x2+ 32x=x2+ 1x2 2x+ 1 = 0(x 1)2= 0x= 1(b) (5 x)5= (2x 1)5(5 x)5= (2x 1)55 x= 2x 13x= 6x= 2(c) 5x= 145x= 14ln(5x) = ln(14)xln 5 = ln 14x=ln 14ln 5 (d) log2(x+ 2) + log2(2) = log221log2(x+ 2) + log2(2) = log221log22(x+ 2) = log2212(x+ 2) = 21x=21 42= (e) log4x= 3log4x= 3x= 43= 646.]

3 Suppose that $2500 is invested at 7% compounded quarterly. How much money will bein the account in 5 years?SOLUTION. A=$ (HINT: use TVM Solver forN= 4 5, I=7, PV=2500, P/Y=4and solve for FV).7. In its first 10 years the Gabelli Growth Fund produced an average annual return of that money invested in this fund continues to earn compounded long will it take money invested in this fund to double? 4(years) (HINT: use TVM Solver for I= , PV=1, FV=-2,P/Y=1 and solve for N).8. Evaluate each limit:(a) limx 2x2 3x+ 2x2+x 6= limx 2(x 2)(x 1)(x 2)(x+ 3)= limx 2x 1x+ 3=2 12 + 3=15(b) limx 2x3+x2+ 5x x3= limx 2x3 x3= limx 2 1= 2(c) limx 1 |x 1|x 1 Since|x 1|= x 1,ifx 1 (x 1),ifx <1thenlimx 1 |x 1|x 1= limx 1 (x 1)x 1= Determine where the functionf(x) = 1 +x,ifx 26 x,ifx >2is Ifx <2, thenf(x) = 1 +xandfis continuous on its domain ( ,2).Ifx >2 thenf(x) = 6 xandfis continuous on its domain (2, ).

4 Limx 2+f(x) = limx 2+(6 x) = 6 2 = 4limx 2 f(x) = limx 2 (1 +x) = 1 + 2 = 3 Since limx 2+f(x)6= limx 2 f(x), then limx 2f(x) does not exist. Thusf(x) is discontinuous atx= (x) is continuous on ( ,2) (2, ).10. Find all vertical and horizontal asymptotes for the functionf(x) =(2x 3)(x+ 2)(x2 1)(x 4)(x+ 3)x(x 1) (x) =(2x 3)(x+ 2)(x2 1)(x 4)(x+ 3)x(x 1)=(2x 3)(x+ 2)(x 1)(x+ 1)(x 4)(x+ 3)x(x 1)=(2x 3)(x+ 2)(x+ 1)(x 4)(x+ 3)xLinesx= 3,x= 0, andx= 4 are the vertical asymptotes limx f(x) = limx (2x 3)(x+ 2)(x+ 1)(x 4)(x+ 3)x= 2, the equation of the horizontal asymp-tote isy= Letp= 25 (x) = 2x+ 9000where 0 x 2500, be the price-demand equation and cost function, respectively, forthe manufacture of umbrellas.(a) Find the exact cost of producing the 31st umbrella. Use the marginal cost to ap-proximate the cost of producing the 31st The exact cost of producing the 31st umbrella isC(31) C(30) =2(31) + 9000 (2(30) + 9000) = $ marginal cost functionC (x) = (30) C(31) C(30).

5 C (30) = 2.(b) Find the marginal revenue and the marginal average revenue The revenueR(x) =xp(x) = 25x marginal revenue function isR (x) = 25 average revenue is R=R(x)x=25x 25 marginal average revenue is function R (x) = (c) Find the average profit per umbrella if 50 umbrellas is produced. Find the marginalaverage profit at a production level of 50 umbrellas. Estimate the average profit perumbrella if 51 umbrella is The profit isP(x) =R(x) C(x) = 25x (2x+ 9000) = + 23x average profit is P(x) =P(x)x= + 23 marginal average profit is P (x) = + average profit per umbrella if 50 umbrellas is produced is P(50) = (50) +23 900050= marginal average profit at a production level of 50 umbrellas is P (50) = +9000(50)2= average profit per umbrella if 51 umbrella is produced is P(50) + P (50) = + = A bank offers a 10-year certificate of deposit (CD) that earns compounded contin-uously.

6 (a) If $10000 is invested in this CD, how much will it be worth in 10 years? ert, whereP= principal,r= annual nominal interest ratecompounded continuously,t= time in years,A= amount at 10000,r= ,t= 10,A=?.A= 10= $ (b) How long will it take for the account to be worth $18000?A= 18000,P= 10000,r= ,t=?A=P ertAP=ertlnAP= ln ert lnAP=rtt=1rlnAP= 1413. A note will pay $25000 at maturity 10 years from now. How much should you willing topay for the note now if money is worth 5% compounded continuously? 25000,t= 10,r= ,P=?.A=P ertP=Ae rt= 25000e 10 $ At what nominal rate compounded continuously must moneybe invested to double in 8years? 2P,t= 8,r=?.2P=P e8r2 =e8rln 2 = ln (e8r)ln 2 = 8rr=ln 28 Find the equation of the tangent line to the graph of the functionf(x) = ln(1 x2+ 2x4)at the point wherex= The equation of the tangent line to the graph of thefunctionf(x) = ln(1 x2+ 2x4) at the point wherex= 1 isy f(1) =f (1)(x 1)f(1) = ln(1 12+ 2(14)) = ln (x) =(1 x2+ 2x4) 1 x2+ 2x4= 2x+ 8x31 x2+ 2x4,f (1) = 2 + 81 1 + 2=62= 3 Thus, the equation of the tangent line isy ln 2 = 3(x 1) ory= 3x 3 + ln Find the value(s) ofxwhere the tangent line to the graph of the functiony= 5ex2 4x+1is The tangent line is horizontal means that its slope is (x) = 5ex2 4x+1(x2 4x+ 1) = (2x 4)5ex2 4x+1= 02x 4 = 0x= 217.

7 Find each derivative(a)ddxlog3(4 4x3+ 5x+ 7) =ddxlog3(4x3+ 5x+ 71/4) =ddx14log3(4x3+5x+7) =14ddxlog3(4x3+5x+ 7) =14(4x3+ 5x+ 7) (4x3+ 5x+ 7) ln 3=1412x2+ 5(4x3+ 5x+ 7) ln 3(b)ddx81 2x3=81 2x3(1 2x3) ln 8 = 6x281 2x3ln 8.(c)ddx3x2(x2+ 5)3=6x(x2+ 5)3 3(x2+ 5)2(x2+ 5) (3x2)(x2+ 5)6=6x(x2+ 5)3 3(x2+ 5)2(2x)(3x2)(x2+ 5)6=6x(x2+ 5)2[x2+ 5 3x2](x2+ 5)6=6x(5 2x2)(x2+ 5)4(d)ddx[(x2+x 3)e2x+3] = (2x+ 1)e2x+3+e2x+3(2x+ 3) (x2+x 3) = (2x+ 1)e2x+3+e2x+3(2)(x2+x 3) = (2x2+ 4x 5)e2x+318. Given the price-demand +p= 60(a) Find the elasticity of demandE(p).SOLUTION. First, let s solve the equation +p= 60 3000 (p) = pf (p)f(p),wheref(p) = 3000 (p) = pf (p)f(p)= p( 50)3000 50p=p60 p(b) For which values ofpis demand elastic?SOLUTION. The demand is elastic ifE(p)> p>1p60 p 1>0p60 p 60 p60 p>0p 60 +p60 p>02p 6060 p>0p 3060 p>030< p <60(c) Ifp= $10 and the price is increased by 5%, what is the approximatechange indemand?

8 (10) =1060 10= (change is demand)=E(10) (change in price)=( )(5%)=1%(d) Ifp= $40 and the price is decreased, will revenue increase or decrease? (40) =4060 40= 2>1, the demand is elastic. If the price isdecreased then the revenue will Findf (x) for the functions(a)f(x) =x2(2x3 5) (x) = 2x(2x3 5)4+ (x2)(4)(2x3 5)3(2x3 5) = 2x(2x3 5)4+(x2)(4)(2x3 5)3(6x2) = 2x(2x3 5)4+ 24x4(2x3 5)3f (x) = 2(2x3 5)4+(2x)(4)(2x3 5)3(2x3 5) +(24)(4x3)(2x3 5)3+24x4(3)(2x3 5)2(2x3 5) = 2(2x3 5)4+(2x)(4)(2x3 5)3(6x2)+(24)(4x3)(2x3 5)3+24x4(3)(2x3 5)2(6x2) = 2(2x3 5)4+ 48x3(2x3 5)3+ 96x3(2x3 5)3+ 432x6(2x3 5)2=2(2x3 5)4+ 144x3(2x3 5)3+ 432x6(2x3 5)2(b)f(x) =2x 6x3= 2x 1 6x (x) = 2( 1)x 2 6( 3)x 4= 2x 2+ 18x 4f (x) = 2( 2)x 3+ 18( 4)x 5= 4x 3 72x Given the graph of the derivativef (x) of the functiony=f(x).(a) Find the intervals on whichfis increasing, increasing on the interval wheref (x)>0 and decreasing on theinterval wheref (x)< increasing on (a, e) (g, ).

9 Fis decreasing on (e, g).(b) Findx-coordinates of the critical points for the , andx=g.(c) Find the intervals on whichfis concave upward, concave CU on the interval wheref is increasing and CD on the intervalwheref is CU on (b, c) (f, )fis CD on (a, b) (c, f).(d) Findx-coordinates of the inflection points ,x=c, andx= Given the functionf(x) =13x3+12x2 3x+ 4.(a) Find critical values off(x).f (x) =x2+x 3 = 0x1= , x2= (b) Find intervals on whichf(x) is increasing and find intervals we have to constract the sigh chart forf (x).The sign chart forf (x)fis increasing on ( , ) ( , ) and decreasing on ( , ).(c) Find local extrema forf(x).It follows from the sign chart thatfhas the local minimum atx= and the localmaximum atx= (d) Find intervals on whichf(x) is concave upward and concave (x) = 2x 1 = 0x= find intervals we have to construct the sign chart forf (x).The sign chart forf (x)fis CU on ( , ) and CD on ( , ).

10 (e) Find all inflection points off(x).fhas the inflection point atx= Find the absolute maximum and absolute minimum for the functionf(x) = 9 x2onthe interval [ 1,2]. (x) =12(9 x2)1/2 1(9 x2) =12(9 x2) 1/2( 2x) = x(9 x2) 1/2= 0x= 0f( 1) = 9 1 = 8 (0) = 9 0 = 9 = 3 absolute maximum valuef(2) = 9 4 = 5 absolute minimum value23. Find the absolute maximum and minimum for the functionf(x) =x2 1x2+ (x) =2x(x2+ 1) 2x(x2 1)(x2+ 1)2=4x(x2+ 1)2= 0x= 0To classify the critical valuex= 0 we have to use the Second Derivative (x) =4(x2+ 1)2 (4x)(2)(x2+ 1)(x2+ 1) (x2+ 1)4=4(x2+ 1)2 8x(x2+ 1)(2x)(x2+ 1)4f (0) = 4>0 Functionfhas the absolute minimum atx= What are the dimensions of the rectangular field of 20000 square feet that will minimizethe cost of fencing if one side costs three times as much per unit length as the other three?SOLUTION. Letxbe the length of the rectangle andybe the width of the area of the rectangle isA=xy= 20000.