Test Bank for Calculus For Biology and Medicine , 4th Edition

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INSTRUCTOR’SSOLUTIONSMANUALROGERLIPSETTTO ACCOMPANYCALCULUS FORBIOLOGY ANDMEDICINEFOURTHEDITIONClaudia NeuhauserUniversity of MinnesotaMarcus L. RoperUniversity of California—Los Angeles

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Contents1Preview and Review51.1Precalculus Skills Diagnostic Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .51.2Preliminaries. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .91.3Elementary Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .231.4Graphing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .50Chapter 1 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .782Discrete-Time Models, Sequences, and Difference Equations872.1Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .872.2Sequences. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .972.3Modeling with Recursion Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .113Chapter 2 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1263Limits and Continuity1333.1Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1333.2Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1473.3Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1563.4Trigonometric Limits and the Sandwich Theorem . . . . . . . . . . . . . . . . . . . . . . .1593.5Properties of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1643.6A Formal Definition of Limits (Optional). . . . . . . . . . . . . . . . . . . . . . . . . . .169Chapter 3 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1734Differentiation1854.1Formal Definition of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1854.2Properties of the Derivative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1914.3The Power Rule, the Basic Rules of Differentiation, and the Derivatives of Polynomials . .2014.4The Product and Quotient Rules, and the Derivatives of Rational and Power Functions.2144.5The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2244.6Implicit Functions and Implicit Differentiation. . . . . . . . . . . . . . . . . . . . . . . .2324.7Higher Derivatives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2394.8Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2424.9Derivatives of Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2474.10 Derivatives of Inverse Functions, Logarithmic Functions, and the Inverse Tangent Function 2544.11 Linear Approximation and Error Propagation . . . . . . . . . . . . . . . . . . . . . . . . .262Chapter 4 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .266

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2CONTENTS5Applications of Differentiation2755.1Extrema and the Mean-Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2755.2Monotonicity and Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2885.3Extrema and Inflection Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3025.4Optimization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3085.5L’Hˆopital’s Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3205.6Graphing and Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3315.7Recurrence Equations: Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3515.8Numerical Methods: The Newton-Raphson Method . . . . . . . . . . . . . . . . . . . . . .3625.9Modeling Biological Systems Using Differential Equations. . . . . . . . . . . . . . . . . .3695.10 Antiderivatives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .375Chapter 5 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3866Integration3996.1The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3996.2The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4186.3Applications of Integration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .430Chapter 6 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4577Integration Techniques and Computational Methods4657.1The Substitution Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4657.2Integration by Parts and Practicing Integration . . . . . . . . . . . . . . . . . . . . . . . .4717.3Rational Functions and Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . .4867.4Improper Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5097.5Numerical Integration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5227.6The Taylor Approximation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5397.7Tables of Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .546Chapter 7 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5528Differential Equations5658.1Solving Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5658.2Equilibria and Their Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5878.3Differential Equation Models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6158.4Integrating Factors and Two-Compartment Models . . . . . . . . . . . . . . . . . . . . . .634Chapter 8 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6489Linear Algebra and Analytic Geometry6559.1Linear Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6559.2Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6679.3Linear Maps, Eigenvectors, and Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . .6839.4Demographic Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7079.5Analytic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .714Chapter 9 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .72410 Multivariable Calculus73310.1 Functions of Two or More Independent Variables . . . . . . . . . . . . . . . . . . . . . . .73310.2 Limits and Continuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .74310.3 Partial Derivatives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .74810.4 Tangent Planes, Differentiability, and Linearization . . . . . . . . . . . . . . . . . . . . . .75710.5 The Chain Rule and Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . .766

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CONTENTS310.6 Directional Derivatives and Gradient Vectors. . . . . . . . . . . . . . . . . . . . . . . . .77010.7 Maximization and Minimization of Functions. . . . . . . . . . . . . . . . . . . . . . . . .77710.8 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .80510.9 Systems of Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .808Chapter 10 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .82411 Systems of Differential Equations83111.1 Linear Systems: Theory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83111.2 Linear Systems: Applications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .86311.3 Nonlinear Autonomous Systems: Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . .87311.4 Nonlinear Systems: Lotka-Volterra Model for Interspecific Interactions . . . . . . . . . . .89511.5 More Mathematical Models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .908Chapter 11 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .92412 Probability and Statistics93512.1 Counting. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .93512.2 What is Probability? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .94012.3 Conditional Probability and Independence . . . . . . . . . . . . . . . . . . . . . . . . . . .94812.4 Discrete Random Variables and Discrete Distributions. . . . . . . . . . . . . . . . . . . .95612.5 Continuous Distributions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .97912.6 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .100012.7 Statistical Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1011Chapter 12 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1020

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Chapter 1Preview and Review1.1 Precalculus Skills Diagnostic Test1.(a)y=x+ 5(b)y= 2x−1(c)y= 3x−5(Review 1.2.2)2.(a)59(80−32) =2409≈26.7◦C(b)Inverting the given formula givesx=95y+ 32, so that95 (−10) + 32 =−18 + 32 = 14°F(c)Sety=xin the conversion formula and solve forx:x= 59 (x−32) = 59x−1609⇒49x=−1609⇒4x=−160⇒x=−40.(Review 1.2.2)3.A circle with center (x, y) = (−1,5) and with radius 3. (Review 1.2.3)4.(a)180π·π7=1807≈25.7°.(b)x=−π3+ 2nπandx=−2π3+ 2nπfor anyn∈Z.(c)Divide both sides of the identity sin2θ+ cos2θ= 1 by cos2θ:sin2θ+ cos2θcos2θ=1cos2θso tan2θ+ 1 = sec2θ.(d)π12,7π12,3π4(Review 1.2.4)5.(a)(i)27/3(ii)28/3(b)Take base 2 logarithms of both sides:xlog24 = log2 12, so 2x=−1 and thereforex=−12.

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6Preview and Review(c)Since 10 000 = 104, we have log1010 000 = 4.(d)log103x+ log105x= log10(3x·5x) = log10(15x2).(e)ln(x2)+ lnx= 2 lnx+ lnx= 3 lnx. Solving 3 lnx= 2 gives lnx=23, so thatx=e2/3.(f )If lnx= 3, thenx=e3, so that log10x= log10(e3)= 3 log10e.(Review 1.2.5)6.(a)Using the quadratic formula, the roots are−1± √12−4·1·12·1=−12±i√32.(b)(1 +i)(2−i) = 1·2 + 1·(−i) +i·2−i2= 3 +i.(c)z+z= (a+bi) + (a−bi) = 2a∈R(Review 1.2.6)7.(a)(i)f(x)∈[0,1](ii)f(x)∈[0,∞)(b)(i)f(3) =√3(ii)(f◦g)(x) =f((x+ 1)2)=√(x+ 1)2=|x+ 1|.(iii)(g◦f)(4) =g(√4) =g(2) = (2 + 1)2= 9.(c)Sincef(x) =|x| ≥0, then (f◦f)(x) =|f(x)|=f(x)(Review 1.3.1)8.(a)For large values ofx, (iii) grows much faster due to the presence of thex3term. For example,whenx= 6,p1(x) = 6,p2(x) = 18, andp3(x) = 72.(b)(i)The highest power ofrthat appears is 2, so this is a degree 2 polynomial inr.(ii)Since the velocity should not be negative we must have 1−r2a2≥0, so thatr≤a. Sinceris a distance, it must be nonnegative.Putting these together gives 0≤r≤afor thedomain.(iii)Since 1−r2a2≤1 for the allowable values ofr, the range is 0≤u(r)≤u0.(iv)The maximum value occurs whenr= 0; that value isu0.u(r) =12u0when 1−r2a2=12,which happens whenr=a√2.(Review 1.3.2)9.(a)For the three values ofc, we haver(1) =11 + 10≈0.091,r(2) =22 + 10≈0.167,r(3) =33 + 10≈0.231.The rate of metabolization is largest whenc= 3.(b)Sincer(5) =55 + 10 = 13,r(10) =1010 + 10 = 12,the rate of metabolization increases by less than double.

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1.1 Precalculus Skills Diagnostic Test7(c)From part (b),r(5) =13; to getr(c) =23we needr(c) =cc+ 10 = 23,so that 3c= 2(c+ 10). Solving givesc= 20.(d)No.r(10) =12.Doubling the rate of metabolization would increase it to 1.But since thedenominator ofr(c) is always strictly greater than the numerator, its value can never be 1.(Review 1.3.3)10.The relationship isN=kA1/5.(a)We want to findAso thatkA1/5=12kA1/50. Cancelling thek’s givesA1/5=12A1/50. Now raiseboth sides to the fifth power, givingA=A032. The new habitat area isA032.(b)We want to findAso thatkA1/5=13kA1/50. Cancelling thek’s givesA1/5=13A1/50. Now raiseboth sides to the fifth power, givingA=A0243. The new habitat area isA0243.(c)We want to findAso thatkA1/5= 2kA1/50. Cancelling thek’s givesA1/5= 2A1/50. Now raiseboth sides to the fifth power, givingA= 32A0. The new habitat area is 32A0.(Review 1.3.4)11.(a)Since 1000 =N(0) =N0er·0=N0we getN0= 1000. Since 2000 =N(2) = 1000er·2, we have2 =e2r. Take natural logarithms, giving 2r= ln 2 so thatr=ln 22. The formula is thereforeN(t) = 1000etln 2/2.(b)SolvingN(t) = 3000 givesetln 2/2= 3, so thattln 22= ln 3, and thent=2 ln 3ln 2≈3.17.(c)SolvingN(t) = 4000 givesetln 2/2= 4, so thattln 22= ln 4, and thent=2 ln 4ln 2= 4.(Review 1.3.5)12.(a)f−1(x) =√x−1(b)f−1(x) =−1 +ex/2(c)f−1(x) =x1/5(Review 1.3.6)13.(a)(i)lnx+ ln(x2+ 1) = ln(x·(x2+ 1)).(ii)log(x1/3)−log((x+ 1)1/3)=13logx−13log(x+ 1) =13(logx−log(x+ 1)) =13logxx+1.(iii)2 + log2x= log24 + log2x= log2(4x).(b)(i)ln 7ln 2(ii)ln 6ln 10(iii)ln 2lnx(Review 1.3.7)14.(a)(i)The amplitude is 2, and the period is 2π.(ii)The amplitude is 2, and the period is2π3.(iii)The amplitude is 3, and the period is2ππ/2= 4.(b)(i)The range is all ofR.The maximum domain is all ofRexcept for the points(2n+1)2πwherenis an integer (since tangent is undefined at odd multiples ofπ/2).

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8Preview and Review(ii)The range is [−1,1], since−1≤cosx≤1 for allx. The domain isR.(c)Compress cosxby a factor of 2 in thex-direction and stretch it by a factor of 3 in they-directionto get 3 cos 2x.(Review 1.3.8)15.-6-4-22x-0.50.5y(a)-4-224x0.51.1.5y(b)-4-224x-0.50.5y(c)-4-224x-0.50.5y(d)-8-448x-0.50.5y(e)(Review 1.4.1)16.(a)(i)2 kg(ii)30 kg(iii)100 kg(b)Adult weightPuppy weight=20 kg0.4 kg= 50(c)The point representing the cat will lie one tick below the Jack Russell Terrier.(Review 1.4.2)17.(a)Since both axes are log scales, this is a power relationship; since the slope of the line is 0.41,the relationship isD=k A0.41.(b)Since only the vertical axis is a log scale, this is an exponential relationship; since the slope ofthe line is−1, the relationship isN=k10−m.

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1.2 Preliminaries9(c)Since only the vertical axis is a log scale, this is an exponential relationship; since the slope ofthe line is 0.86, the relationship isN=k100.86t≈k7.24t.(d)Since only the vertical axis is a log scale, this is an exponential relationship; since the slope ofthe line is−0.09, the relationship isN=k10−0.09t≈k0.813t.(Review 1.4.3)18.(a)This is the graph of patient 1 shifted two hours to the right, so it is graph 4.(b)This graph should start from zero at 8AM just as for patient 1, but it should have a longertail since it takes longer to enter the bloodstream. This is graph 3.(c)This graph should have the same shape as the graph of patient A, which is graph 4, but is halfas high. This is graph 2.(Review 1.4.4)1.2 Preliminaries1.2.11.(a)Walking 4 units to the right and to the left from−1 we get the numbers 3 and−5, respectively.(b)|x−(−1)|= 4, sox+ 1 =±4, yieldingx= 3 orx=−5.2.With three numbers there will be three pairwise distances:|−5−2|=|−7|= 7,|2−7|=|−5|= 5and|−5−7|=|−12|= 12.3.(a)2x+ 4 =±6, so either 2x= 2 giving the solutionx= 1 or 2x=−10 and the other solution isx=−5.(b)x−3 =±2, from which we see thatx= 5 orx= 1.(c)2x−3 =±5, so 2x= 8 givingx= 4 or 2x=−2 andx=−1.(d)1−5x=±6, so that 5x= 1±6. Then 5x=−5, sox=−1, or 5x= 7, sox=75.4.(a)The equation implies±(2x+ 4) =±(5x−2), which reduces to two equations 2x+ 4 = 5x−2and 2x+ 4 =−(5x−2), so the solutions are:x= 2 andx=−27.(b)As in part (a) we get 1 + 2u=±(5−u), and solving both gives 1 + 2u= 5−uwith solutionu=43, and 1 + 2u=−(5−u) =u−5 with solutionu=−6.(c)As in part (a) we get the equations 4 +t2=±(32t−2) that solves tot= 6 andt=−1.(d)As in part (a) 2s−6 =±(3−s); solving both equations givess= 3 ands= 3. Sos= 3 isthe only solution.5.(a)We can rewrite the absolute value as two inequalities−4≤5x−2≤4⇒−4 + 2≤5x≤4 + 2⇒−2≤5x≤6⇒−25≤x≤65.(b)Rewriting as a pair of inequalities gives 3−4x <−8 or 3−4x >8. Solving the first we get:4x >11, orx >114, and solving the second we get 4x <−5, orx <−54. Sox <−54orx >114.(c)Rewriting as a pair of inequalities gives 7x+ 4≥3, which solves asx≥ −17, and 7x+ 4≤ −3which gives 7x≤ −7 orx≤ −1. Sox≤ −1 orx≥ −17.

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10Preview and Review(d)Rewriting gives−7<3 + 2x <7⇒−7−3<2x <7−3⇒−10<2x <4⇒−5< x <2.6.(a)−6<2x+ 3<6; subtracting 3 gives−9<2x <3, so that−92< x <32.(b)Rewriting as a pair of inequalities gives 3−4x≥2 or 3−4x≤ −2, so thatx≤14orx≥54.(c)−1≤x+ 5≤1, which is the same as−6≤x≤ −4.(d)Since no absolute value can be negative, there are no values ofxthat will satisfy the equation.1.2.27.Use the point slope formulay−y0=m(x−x0). In this case we gety−2 =−2(x−3)⇒y−2 =−2x+ 6⇒2x+y−8 = 0.8.Use the point slope formulay−y0=m(x−x0). In this case we gety−(−1) = 14 (x−2)⇒y+ 1 = 14x−12⇒−x+ 4y+ 6 = 0.9.Use the point slope formulay−y0=m(x−x0). In this case we gety−(−2) =−3(x−0)⇒y+ 2 =−3x⇒3x+y+ 2 = 0.10.Use the point slope formulay−y0=m(x−x0). In this case we gety−5 = 12 (x−(−3))⇒2y−10 =x+ 3⇒−x+ 2y−13 = 0.11.First compute the slope using the two given points:m=y2−y1x2−x1= 4−(−3)1−(−2) = 73.Now use the point slope formulay−y0=m(x−x0) with (x0, y0) = (1,4) (we could equally welluse the other point, (−2,−3)):y−4 = 73 (x−1)⇒3(y−4) = 7(x−1)⇒3y−12 = 7x−7⇒−7x+ 3y−5 = 0.12.First compute the slope using the two given points:m=y2−y1x2−x1=−4−41−(−1) =−4.Now use the point slope formulay−y0=m(x−x0) with (x0, y0) = (−1,4) (we could equally welluse the other point, (1,−4)):y−4 =−4(x−(−1))⇒y−4 =−4x−4⇒4x+y= 0.

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1.2 Preliminaries1113.First compute the slope using the two given points:m=y2−y1x2−x1= 1−32−0 =−1.Now use the point slope formulay−y0=m(x−x0) with (x0, y0) = (0,3) (we could equally welluse the other point, (2,1)):y−3 =−1(x−0)⇒y−3 =−x⇒x+y−3 = 0.14.First compute the slope using the two given points:m=y2−y1x2−x1= 5−(−1)4−1= 63 = 2.Now use the point slope formulay−y0=m(x−x0) with (x0, y0) = (1,−1) (we could equally welluse the other point, (4,5)):y−(−1) = 2(x−1)⇒y+ 1 = 2x−2⇒2x−y−3 = 0.15.Horizontal lines are always of the formy=k; since this line goes through(4,14), its equation isy=14. The standard form is 4y−1 = 0.16.Horizontal lines are always of the formy=k; since this line goes through (0,−1), its equation isy=−1. The standard form isy+ 1 = 0.17.Vertical lines are always of the formx=k; since this line goes through (−2,0), its equation isx=−2. The standard form isx+ 2 = 0.18.Vertical lines are always of the formx=k; since this line goes through (2,−3), its equation isx= 2. The standard form isx−2 = 0.19.Use the slope-intercept formy=mx+b.We are givenm= 3 andb= 2, so the equation isy= 3x+ 2, which in standard form is 3x−y+ 2 = 0.20.Use the slope-intercept formy=mx+b.We are givenm=−1 andb= 5, so the equation isy=−x+ 5, which in standard form isx+y−5 = 0.21.Use the slope-intercept formy=mx+b.We are givenm=12andb= 2, so the equation isy=12x+ 2, or 2y=x+ 4, which in standard form isx−2y+ 4 = 0.22.Use the slope-intercept formy=mx+b.We are givenm=−13andb=13, so the equation isy=−13x+13, which in standard form isx+ 3y−1 = 0.23.Use the point slope formulay−y0=m(x−x0) where (x0, y0) = (1,0) is the given point. In thiscase we gety−0 =−2(x−1)⇒y=−2x+ 2⇒2x+y−2 = 0.24.Use the point slope formulay−y0=m(x−x0) where (x0, y0) = (−2,0) is the given point. In thiscase we gety−0 = 1(x−(−2))⇒y=x+ 2⇒x−y+ 2 = 0.25.Use the point slope formulay−y0=m(x−x0) where (x0, y0) =(−12,0)is the given point. In thiscase we gety−0 =−12(x−(−12))⇒y=−12x−14⇒2x+ 4y+ 1 = 0.

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12Preview and Review26.Use the point slope formulay−y0=m(x−x0) where (x0, y0) =(−12,0)is the given point. In thiscase we gety−0 = 15(x−(−12))⇒y= 15x+ 110⇒2x−10y+ 1 = 0.27.Since the line we want is parallel tox+ 2y−4 = 0, it must have the same slope. Now,x+ 2y−4 = 0implies that 2y=−x+ 4, ory=−12x+ 2, so the line has slopem=−12. Now use the point slopeformulay−y0=m(x−x0) where (x0, y0) = (2,−3) is the given point. In this case we gety−(−3) =−12 (x−2)⇒y+ 3 =−12x+ 1⇒2y+ 6 =−x+ 2⇒x+ 2y+ 4 = 0.28.Since the line we want is parallel tox−2y+ 4 = 0, it must have the same slope. Now,x−2y+ 4 = 0implies that 2y=x+ 4, ory=12x+ 2, so the line has slopem=12.Now use the point slopeformulay−y0=m(x−x0) where (x0, y0) = (1,2) is the given point. In this case we gety−2 = 12 (x−1)⇒y−2 = 12x−12⇒x−2y+ 3 = 0.29.The line passing through (0,2) and (3,0) has slopem= 0−23−0 =−23,so the line we want has the same slope. Now use the point slope formulay−y0=m(x−x0) where(x0, y0) = (−1,−1) is the given point. In this case we gety−(−1) =−23 (x−(−1))⇒y+ 1 =−23x−23⇒2x+ 3y+ 5 = 0.30.The line passing through (0,−4) and (2,1) has slopem= 1−(−4)2−0= 52,so the line we want has the same slope. Now use the point slope formulay−y0=m(x−x0) where(x0, y0) = (2,−1) is the given point. In this case we gety−(−1) = 52 (x−2)⇒y+ 1 = 52x−5⇒5x−2y−12 = 0.31.The line 2y−5x+ 7 = 0 can be written as 2y= 5x−7, ory=52x−72, so it has slope52. Since theline we want is perpendicular to it, it must have slopem⊥=−1m=−25. It passes through (1,4),so using the point slope formulay−y0=m(x−x0) we gety−4 =−25 (x−1)⇒5y−20 =−2x+ 2⇒2x+ 5y−22 = 0.32.The linex−2y+ 3 = 0 can be written as 2y=x+ 3, ory=12x+32, so it has slopem0=12.Since the line we want is perpendicular to it, it must have slopem=−2 (since for perpendicularlines we must havem·m0=−1).It passes through (1,−1), so using the point slope formulay−y0=m(x−x0) we gety−(−1) =−2(x−1)⇒y+ 1 =−2x+ 2⇒2x+y−1 = 0.

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1.2 Preliminaries1333.The given line, passing through (−2,1) and (1,−2), has slopem=y2−y1x2−x1= (−2)−11−(−2) =−33 =−1.Therefore the slope of the perpendicular line ism⊥=−1m= 1. It passes through (5,−1), so usingthe point slope formulay−y0=m(x−x0) we gety−(−1) = 1(x−5)⇒y+ 1 =x−5⇒x−y−6 = 0.34.The given line, passing through (−2,0) and (1,1), has slopem=y2−y1x2−x1=1−01−(−2) = 13.Therefore the slope of the perpendicular line ism⊥=−1m=−3.It passes through (4,−1), sousing the point slope formulay−y0=m(x−x0) we gety−(−1) =−3(x−4)⇒y+ 1 =−3x+ 12⇒3x+y−11 = 0.35.The line we want is horizontal and passes through (1,3). Since all horizontal lines are of the formy=k, here we must havey= 3, ory−3 = 0.36.The line we want is horizontal and passes through (1,5). Since all horizontal lines are of the formy=k, here we must havey= 5, ory−5 = 0.37.The line we want is vertical and passes through (−2,3).Since all vertical lines are of the formx=k, here we must havex=−2, orx+ 2 = 0.38.The line we want is vertical and passes through (3,1). Since all vertical lines are of the formx=k,here we must havex= 3, orx−3 = 0.39.A line perpendicular to a horizontal line is a vertical line. Since all vertical lines are of the formx=k, and the line passes through (1,−3), we must havex= 1, orx−1 = 0.40.A line perpendicular to a horizontal line is a vertical line. Since all vertical lines are of the formx=k, and the line passes through (1,3), we must havex= 1, orx−1 = 0.41.A line perpendicular to a vertical line is a horizontal line. Since all horizontal lines are of the formy=k, and the line passes through (7,3), we must havey= 3, ory−3 = 0.42.A line perpendicular to a vertical line is a horizontal line. Since all horizontal lines are of the formy=k, and the line passes through (−2,5), we must havey= 5, ory−5 = 0.43.(a)We convert each of these into centimeters, remembering that the formulay= 30.5xrequiresthatxbe measured in feet:(i)y= 30.5·6 ft = 183 cm(ii)y= 30.5·412ft≈30.5·0.1667 ft≈10.2 cm(iii)1 ft,7 in = 1 +712ft≈1.583 ft, soy≈30.5·1.583 ft≈48.3 cm.(iv)y= 30.5·20.512ft≈30.5·1.708 ft≈52.1 cm(b)Divide both sides ofy= 30.5xby 30.5, givingy30.5=x, orx=y30.5. Sincexis measured infeet andyin centimeters, this gives a way to convert from centimeters to feet.

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14Preview and Review(c)(i)x=19530.5≈6.39 ft(ii)x=1230.5≈0.39 ft(iii)x=4830.5≈1.57 ft44.(a)Since 1 pound equals 2.20 kg, lety= 2.20x, wherexis kilograms andyis pounds.(b)(i)y= 2.20x, so 63 = 2.20x, orx≈28.6 kg.(ii)y= 2.20x, so 5 = 2.20x, orx≈2.27 kg.(iii)y= 2.20·2.5 = 5.5 pounds.(iv)y= 2.20·76≈167 pounds.45.Distance = rate·time; here time = 15 min =14h = 0.25 h, and the distance is 10 mi. The constantof proportionality is miles per hour or “mph”. Using the given values gives10 mi = speed·0.25 h⇒speed =100.25 = 40 mph.The speed (constant of proportionality) is 40 mph.46.Letybe the number of seeds produced, andxbe the biomass. Sinceyis proportional tox, we havey=mxwheremis the constant of proportionality. Then 13 =m·213, so thatm=13213≈0.061.47.1 ft = 0.305 m, so 3.279 ft = 1 m. Then1 m2= (1 m)(1 m) = (3.279 ft)(3.279 ft)≈10.75 ft2.48.1 ha = 10 000 m2, and 1 acre = 4046.86 m2. Therefore the area ratio of a hectare to an acre (thatis, the number of acres in a hectare) is1 ha1 acre =10 000 m24046.86 m2≈2.471 acre.49.(a)Since the relationship is linear, if we letybe the number of liters andxthe number of ounces,we havex=mywherem= 33.81. So the relationship isx= 33.81y, ory≈0.0296x.(b)Withx= 12 ounces, we gety= 0.0296·12≈0.355 l.50.(a)Since the relationship is linear, if we letybe the number of kilometers andxthe number ofmiles, we haveykm =mxmi wherem= 1.609. So the relationship isy= 1.609x.(b)We have 434 = 1.609x, so thatx=4341.609≈270 mi.51.(a)Since one cup of flour weighs 120 g, it follows that 2.5 cups of flour weighs 2.5·120 = 300 g.(b)Since one cup of flour weighs 120 g, one gram of flour is1120cups.Thus 225 g of flour is225·1120= 1.875 cups.(c)Letybe the number of cups andxthe weight in grams. Then from part (b),y=1120x.52.(a)Letxbe the temperature in◦C and letybe the temperature in°F. Then we have the points(0,32) and (100,212). First find the slope between these two pts:m= 212−32100−0= 180100 = 1810 = 95They-intercept is 32. A linear equation that relates Celsius and Fahrenheit then isy=95x+32,orF=95C+ 32

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1.2 Preliminaries15(b)Solving the equation from part (a) forCgivesC= 59 (F−32).Then the lower and upper normal body temperatures in humans are97.6°F corresponds to 59 (97.6−32) = 59·65.6≈36.44◦C99.6°F corresponds to 59 (99.6−32) = 59·67.6≈37.56◦C(c)SettingC=Fin the equation from part (a) givesF= 95F+ 32⇒−45F= 32⇒F=−40.The two scales read the same at a temperature of−40°.53.(a)Since the relationship is linear, these two points, (0.16,0.52) and (1.0,1.0), must lie on a line.The slope of that line must bem=y2−y1x2−x1= 1.0−0.521.0−0.16 = 0.480.84≈0.5714.The equation of the line is then (using the point slope form with the point (1.0,1.0))y−1.0 = 0.5714(x−1.0)⇒y= 0.5714x+ 0.4286.(b)Substitute each value ofxinto the equation from part (a):(i)y= 0.5714·0.5 + 0.4286 = 0.7143(ii)y= 0.5714·0.9 + 0.4286 = 0.9429(iii)y= 0.5714·0 + 0.4286 = 0.4286(c)Solving the equation in part (a) forx, we gety= 0.5714x+ 0.4286⇒y−0.4286 = 0.5714x⇒x=10.5714y−0.42860.5714⇒x= 1.75y−0.75.54.(a)A plot of two perpendicular lines, with their inclinations labeled, is below.`2has also beentranslated so we can see what the relative angles of inclination are:θ2θ1xyl2l1From the graph, comparing`1with the dashed line, we see thatθ1=θ2+π2.

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16Preview and Review(b)We see that the slopes of the two lines arem2=4y4x= tanθ2m1=4y4x=−tan(π−θ1) = tanθ1(c)Now using the fact from (a) that−θ2=π2−θ1, we havem1= tanθ1=1cotθ1=1tan(π2−θ1) =1tan(−θ2) = cot(−θ2) =−cotθ2.(d)m1m2=−cotθ2·tanθ2=−1.1.2.355.Using the general equationr2= (x−x0)2+ (y−y0)2withr= 2 and (x0, y0) = (1,−2) gives4 = (x−1)2+ (y−(−2))2⇒4 = (x−1)2+ (y+ 2)2.56.Using the general equationr2= (x−x0)2+ (y−y0)2withr= 4 and (x0, y0) = (2,3) gives42= (x−2)2+ (y−3)2⇒16 = (x−2)2+ (y−3)2.57.(a)Using the general equationr2= (x−x0)2+ (y−y0)2withr= 4 and (x0, y0) = (2,5) gives42= (x−2)2+ (y−5)2⇒16 = (x−2)2+ (y−5)2.(b)Whenx= 0 the circle is on they-axis. Settingx= 0 gives16 = (0−2)2+(y−5)2⇒16 = 4+(y−5)2⇒12 = (y−5)2⇒ ±2√3 =y−5⇒y= 5±2√3.The points of intersection are (0,5±2√3).(c)Wheny= 0 the circle is on thex-axis. Settingy= 0 gives16 = (x−2)2+ (0−5)2⇒16 = (x−2)2+ 25⇒−9 = (x−2)2.But the right hand side is the square of a real number, which cannot be negative, so this isimpossible and the circle does not intersect thex-axis.58.(a)If we start with a point at (2,−5), and think of increasing the radius of a circle centered atthat point, it will touch one axis when the radius becomes 2, and will touch the second axiswhen the radius becomes 5, so 2≤r <5.(b)As in part (a), it will intersect both axes whenr≥5.59.(x+ 2)2+y2= 25, so (x−(−2))2+ (y−0)2= 52. This is a circle with center (−2,0) and radius 5.60.(x+ 1)2+ (y−3)2= 9, so (x−(−1))2+ (y−3)2= 32.This is a circle with center (−1,3) andradius 3.

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1.2 Preliminaries1761.Completing the square, we get0 =x2+y2+ 6x+ 2y−120 = (x2+ 6x+ 9) + (y2+ 2y+ 1)−12−9−122 = (x+ 3)2+ (y+ 1)2= (x−(−3))2+ (y−(−1))2,which is a circle with center (−3,−1) and radius√22.62.Completing the square, we get0 =x2+y2+ 2x−4y+ 10 = (x2+ 2x+ 1) + (y2−4y+ 4) + 1−1−44 = (x+ 1)2+ (y−2)2,which is a circle with center (−1,2) and radius 2.1.2.463.(a)Convert to radian measure: 65°·π180°=1336πrad.(b)Convert to degree measure:1112π·180°π=11·180°12= 165°.64.(a)−15◦·π180◦=−π12radiansor2312πradians.(b)7π4·180◦π=1260◦4= 315◦.65.(a)sin(−π4)=−√22.(b)cos(5π6)=−√32.(c)tan(π6)=1√3.66.(a)sin(54π)=√22.(b)cos(−11π6)=√32.(c)tan(π3)=√3.67.(a)From the definition of cosine as adjacent divided by hypotenuse, it is clear that reflectingαacross they-axis negates the cosine; that is, cos(π−α) =−cosα. Since cosπ3=12√3, we getcos 2π3= cos(π−π3)=−cosπ3 =−12√3.Then, since cos(−α) = cosα, we also getcos 4π3= cos(2π−2π3)= cos(−2π3)= cos 2π3=−12√3.So the two solutions areα=2π3,4π3.(b)From the table in the text, we see that tanα=sinαcosαis equal to1√3whenα=π6. The othersolution isα=π+π6=7π6, since for that angle both sine and cosine have the same numericalvalues as forπ6, but they are both negative.

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18Preview and Review68.(a)From the definition of cosine as adjacent divided by hypotenuse, it is clear that reflectingαacross they-axis negates the cosine; that is, cos(π−α) =−cosα. Since cosπ4=12√2, we getcos 3π4= cos(π−π4)=−cosπ4 =−12√2.Then, since cos(−α) = cosα, we also getcos 5π4= cos(2π−3π4)= cos(−3π4)= cos 3π4=−12√2.So the two solutions areα=3π4,5π4.(b)Since secα=1cosα, we are looking for solutions to cosα=1√2=12√2. From the table in thetext, one solution isα=π4; since cos(−α) = cosα,α=−π4is another solution. Adding 2πtoget it into the required range givesα=7π4. The two solutions areα=π4,7π4.69.Using the facts that tanθ=sinθcosθand that secθ=1cosθ, we getsin2θ+ cos2θ= 1sin2θ+ cos2θcos2θ=1cos2θ(Divide both sides by cos2θ)sin2θcos2θ+ cos2θcos2θ=1cos2θ(Distribute)tan2θ+ 1 = sec2θ(Make substitution)70.Using the facts that cotθ=cosθsinθand that cscθ=1sinθ, we getsin2θ+ cos2θ= 1sin2θ+ cos2θsin2θ=1sin2θ(Divide both sides by sin2θ)sin2θsin2θ+ cos2θsin2θ=1sin2θ(Distribute)1 + cot2θ= csc2θ(Make substitution)71.Using the identity sin2θ+cos2θ= 1, the equation cos2θ−2 = 2 sinθis the same as (1−sin2θ)−2 =2 sinθ, or sin2θ+ 2 sinθ+ 1 = (sinθ+ 1)2= 0. Thus sinθ=−1, so thatθ=3π2.72.Use the identity 1 + tan2x= sec2xto rewrite the equation:sec2x=√3 tanx+ 1⇒1 + tan2x=√3 tanx+ 1⇒tanx(tanx− √3) = 0.The solutions to this equation are the solutions to tanx= 0 or tanx=√3 for 0≤x < π, whicharex= 0 (tanx= 0),x=π3 (tanx=√3).

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1.2 Preliminaries191.2.573.(a)248−2/3= 24(23)−2/3= 2423(−2/3)= 242−2= 22= 4.(b)333−1/231/2= 333−1/23−1/2= 33−1/2−1/2= 32= 9.(c)5k(25)k−152−k= 5k(52)k−15−(2−k)= 5k52(k−1)5k−2= 5k+(2k−2)+(k−2)= 54k−4.74.(a)(24·2−3/2)2= (28/2·2−3/2)2= (28/2−3/2)2= (25/2)2= 25= 32.(b)(65/2·62/361/3)3=(65/2+2/361/3)3=(615/6+4/662/6)3=(619/662/6)3=(6(19/6)−(2/6))3=(617/6)3= 6(17/6)·3= 617/2.(c)(3−2k+334+k)3=(3−2k+3−4−k)3=(3−3k−1)3= 3(−3k−1)(3)= 3−9k−3.75.(a)log4x=−2, so thatx= 4−2=142=116.(b)log1/3x=−3, so thatx=(13)−3=1(13)3=1127= 27.(c)log10x=−2, so thatx= 10−2=1102=1100.76.(a)log2x=−3, so thatx= 2−3=18.(b)log1/4x=−12, so thatx=(14)−1/2=1(1/4)1/2= 41/2= 2.(c)log3x= 0, so thatx= 30= 1.77.(a)log1/232 =x, so that 32 =(12)x= 1x2x= 2−x. Therefore 2−x= 32 = 25, so thatx=−5.(b)log1/381 =x, so that 81 =(13)x= 1x3x= 3−x. Therefore 3−x= 81 = 34, so thatx=−4.(c)log100.001 =x, so that 10x= 0.001 = 10−3. It follows thatx=−3.78.(a)log381 =x, so that 81 = 34= 3x. It follows thatx= 4.(b)log5125=x, so that125= 5−2= 5x. It follows thatx=−2.(c)log101000 =x, so that 1000 = 103= 10x. It follows thatx= 3.79.(a)−ln(13)= ln[(13)−1]= ln 3.(b)log4(x2−4) = log4[(x−2)(x+ 2)] = log4(x−2) + log4(x+ 2).(c)log243x−1= (3x−1) log2(22) = (3x−1)(2·log22) = (3x−1)(2·1) = 6x−2.80.(a)−log3 14= log311/4= log34.(b)log(x3−xx−1)= log(x(x2−1)x−1)= log(x(x−1)(x+1)x−1)= log(x(x+ 1)) = log(x2+x).(c)ln(ex−2)= (x−2) lne=x−2.

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20Preview and Review81.(a)Sincee3x−1= 2, we have (3x−1) lne= ln 2, so that 3x−1 = ln 2 and thereforex=1+ln 23.(b)Sincee−2x= 10, we have−2xlne= ln 10, so that−2x= ln 10 and thereforex=−ln 102.(c)Sinceex2−1= 10, we have (x2−1) lne= ln 10, so thatx2−1 = ln 10 and thereforex=±√1 + ln 10.82.(a)Since 625 = 54, we have 5x= 54so thatx= 4.(b)Since 256 = 44, we have 44x= 44, so that 4x= 4 and thusx= 1.(c)Since 0.0001 = 10−4, we have 102x= 10−4, so that 2x=−4 and thusx=−2.83.(a)Since ln(x−3) = 5, we haveeln(x−3)=e5, so thatx−3 =e5. Thereforex=e5+ 3.(b)Since ln(x+ 2) + ln(x−2) = ln ((x+ 2)(x−2)) ln(x2−4), we have ln(x2−4) = 1, so thateln(x2−4)=e1=e.Thereforex2−4 =eso thatx=±√e+ 4.However,x=−√e+ 4 isinvalid, since thenx−2<0 so that ln(x−2) is undefined. The only solution isx=√e+ 4.(c)We have log3x2−log32x= log3x22x= log3x2, so that log3x2= 2. Therefore 3log3(x/2)= 32= 9,so thatx2= 9 and thenx= 18.84.(a)log3(2x−1) = 2⇒2x−1 = 32= 9, so that 2x= 10 and thenx= 5.(b)ln(2−3x) = 0⇒2−3x=e0= 1, so that 3x= 1 and thenx=13.(c)log(x)−log(x+ 1) = logxx+1.Therefore logxx+1= log23, so thatxx+1=23.It follows thatx= 2.1.2.685.(3−2i)−(−5 + 2i) = 3−2i+ 5−2i= 8−4i86.(6 +i)−4i= 6 +i−4i= 6−3i87.(4−2i) + (9 + 4i) = 4 + 9−2i+ 4i= 13 + 2i88.(6−4i) + (2 + 5i) = 6 + 2−4i+ 5i= 8 +i89.4(5 + 3i) = 20 + 12i90.(2−3i)(3 + 2i) = 6 + 4i−9i−6i2= 6−5i+ 6 = 12−5isincei2=−1.91.(6−i)(6 +i) = 36−6i+ 6i−i2= 36−(−1) = 37.92.(−4−3i)(4 + 3i) =−4·4−4·3i−3i·4−9i2=−16−12i−12i−9(−1) =−16−24i+ 9 =−7−24i.93.Since conjugation replacesiby−i, we getz= 1 + 2i= 1−2i.94.z+u= (1 + 2i) + (2−3i) = 1 + 2i+ 2−3i= 3−i.95.z+v= (1 + 2i) + (1−5i) = 1 + 2i+ 1−5i= 2−3i= 2 + 3i.96.v−w= (1−5i)−(1 +i) = 1−5i−1−i=−6i= 6i.97.vw= (1−5i)(1 +i) = 6−4i= 6 + 4i.98.uz= (2−3i)(1 + 2i) =8 +i= 8−i.

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1.2 Preliminaries2199.Using the quadratic formula witha= 2,b=−3,c= 2, we getx1,2=−b± √b2−4ac2a=−(−3)±√(−3)2−4·2·22·2= 3± √−74= 3± √7i4.100.Using the quadratic formula witha= 1,b= 1,c= 1, we getx1,2=−b± √b2−4ac2a=−1± √12−4·1·12·1=−1± √−32=−1± √3i2.101.Using the quadratic formula witha=−1,b= 1,c= 2, we getx1,2=−b± √b2−4ac2a=−1±√12−4·(−1)·22·(−1)=−1± √9−2=−1±3−2,so that the two roots arex1=−1+3−2=−1 andx2=−1−3−2= 2.102.Using the quadratic formula witha= 1,b= 2,c= 3, we getx1,2=−b± √b2−4ac2a=−2± √22−4·1·32·1=−2± √−82=−2±2√−22=−1± √2i.103.Using the quadratic formula witha= 1,b= 1,c= 6, we getx1,2=−b± √b2−4ac2a=−1± √12−4·1·62·1=−1± √−232=−1± √23i2.104.Using the quadratic formula witha=−2,b= 4,c=−3, we getx1,2=−b± √b2−4ac2a=−4±√42−4·(−2)·(−3)2·(−2)=−4± √−84=−4±2i√24=−1±√22i.105.The discriminant isb2−4ac= (−4)2−4·3·(−7) = 16 + 84 = 100,so there are two real roots, which arex1,2= 4± √1006= 4±106=−1,73.106.The discriminant isb2−4ac= (−1)2−4·1·(−1) = 1 + 4 = 5,so there are two real roots, which arex1,2= 1± √52.107.The discriminant isb2−4ac= (−1)2−4·3·(−4) = 49,so there are two real roots, which arex1,2= 1± √496= 1±76=−1,43.

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22Preview and Review108.The discriminant isb2−4ac= (−1)2−4·4·1 =−15,so there are two complex roots, which are complex conjugates. They arex1,2= 1± √−158= 1± √15i8.109.The discriminant isb2−4ac= (−5)2−4·3·6 =−47,so there are two complex roots, which are complex conjugates. They arex1,2= 5± √−476= 5± √47i6.110.The discriminant isb2−4ac= (−1)2−4·(−3)·(−4) =−47,so there are two complex roots, which are complex conjugates. They arex1,2= 1± √−47−6=−1± √47i6.111.Ifz=a+bi, thenz=a−bi, so thatz+z= (a+bi) + (a−bi) =a+bi+a−bi= 2az−z= (a+bi)−(a−bi) =a+bi−a+bi= 2bi.112.Ifz=a+bi, thenz=a−bi. Since taking complex conjugates negates the imaginary portion ofthe number we also have(z) =a−bi=a+bi=z.113.Ifz=a+bi, then(z) =a−bi=a+bi=z.114.Letz=a+biandw=c+di. Thenz+w= (a+bi) + (c+di) = (a+c) + (b+d)i= (a+c)−(b+d)i= (a−bi) + (c−di) =z+w.115.Letz=a+biandw=c+di. Thenzw= (a+bi)(c+di) = (ac−bd) + (ad+bc)i= (ac−bd)−(ad+bc)i= (a−bi)(c−di) =z·w.

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1.3 Elementary Functions231.3 Elementary Functions1.3.11.The range isy≥0.-2-112x1234y2.The range is 0≤y≤4.12x1234y3.The range is 0≤y <4 (note thaty= 4 is excluded since the domain does not includex=−2).-2-1x1234y

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24Preview and Review4.The range is 0≤x <(32)2, or 0≤x <94.-1212132x12y5.(a)Factoring the numerator, we getx2−1x−1= (x+ 1)(x−1)x−1=x+ 1 ifx6= 1 (so that the function is defined).(b)No the functions are not equal; they have different domains. The functionfis not defined atthe pointx= 1 whilegis defined for all ofR.6.(a)Forx≥2 the quantity inside the absolute value is non-negative, so it is equal to 2(x−2) .Forx <2, the quantity inside the absolute value is negative, and so the absolute value will beequal to−2(x−2) = 2(2−x). This shows the equality. (Note that the casex= 2 is given inboth branches on the right. However, the value is in any case 0, so the two definitions agree.)(b)Indeedfandgare the same function because they are defined on the same domain andf(x) =g(x) for allxin the domain.This can be seen as follows:for 0≤x≤2, g(x) =2|x−2|=−2(x−2) = 4−2x, and for 2≤x≤3, g(x) = 2|x−2|= 2(x−2) = 2x−4.7.The function is odd, as the graph below indicates:-2-112x-6-4-2246yTo see this algebraically, notice thatf(−x) = 3(−x) =−3x=−f(x).

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1.3 Elementary Functions258.The function is even, as the graph below indicates:-2-112x4812yTo see this algebraically, notice thatf(−x) = 3(−x)2= 3x2=f(x).9.The function is even, as the graph below indicates:-2-112x246yTo see this algebraically, notice thatf(−x) =|3(−x)|=|−3x|=|3x|=f(x).10.The function is neither even nor odd, as the graph below implies:-2-112x-5-3-113yTo see this algebraically, notice that for examplef(−1) = 2(−1)−1 =−3, whilef(1) = 2·1−1 = 1,so thatf(−1)6=f(1) andf(−1)6=−f(1).11.The function is even, as the graph below implies.To see this algebraically, notice thatf(−x) =− |−x|=− |x|=f(x).-2-112x-2-1y

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26Preview and Review12.The function is odd, as the graph below implies:-2-112x-24-16-881624yTo see this algebraically, notice thatf(−x) = 3(−x)3=−3x3=−f(x).13.(a)(f◦g)(x) =f(g(x)) =f(3 +x) = (3 +x)2.(b)(g◦f)(x) =g(f(x)) =g(x2) = 3 +x2.14.(a)(f◦g)(x) =f(g(x)) =f(1−2x) =√1−2x.(b)(g◦f)(x) =g(f(x)) =g(√x) = 1−2√x.15.(a)(f◦g)(x) =f(g(x)) =f(√x) = 1− √x.The domain of this function is{x∈R:x≥0},since otherwise√xis undefined.(b)(g◦f)(x) =g(f(x)) =g(1−x) =√1−x. The domain of this function is{x∈R:x≤1},since otherwise√1−xis undefined.16.(a)(f◦g)(x) =f(g(x)) =f(2x2) =12x2+1. The domain of this function isR, since the denominatoris never zero and is therefore defined everywhere, and sinceg(x) is defined for allx.(b)(g◦f)(x) =g(f(x)) =g(1x+1)= 2·(1x+1)2=2(x+1)2.The domain of this function is{x∈R:x6=−1}, since the denominator vanishes forx=−1.17.(a)(f◦g)(x) =f(g(x)) =f(√x) =1√x. The domain of this function is{x∈R:x >0}, since√xis undefined forx <0, and the fraction is undefined forx= 0.(b)(g◦f)(x) =g(f(x)) =g(1x)=√1x=1√x.This is the same function as in part (a), so itsdomain is also{x∈R:x >0}.18.(f◦g)(x) =f(g(x)) =f(√x+ 1) = (√x+ 1)4= (x+ 1)2.In order for the composition to bedefined, we must havex≥3 (since this is the domain ofg), and also√x+ 1≥3 (since this is thedomain off). These two taken together givex≥8. Since (x+ 1)2is defined forx≥8, the domainis in factx≥8.19.(f◦g)(x) =f(g(x)) =f(√x) = (√x)2=x(g◦f)(x) =g(f(x)) =g(x2) =√x2=|x|.In addition, both compositions have the same domain, namely{x∈R:x≥0}. As a result,|x|=xon that domain, so thatf◦g=g◦f.

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1.3 Elementary Functions2720.(a)(f◦g)(x) =f(g(x)) =f(x+ 1) = (x+ 1)4(g◦f)(x) =g(f(x)) =g(x4) =x4+ 1.Since these expressions are not equal, (f◦g)(x)6= (g◦f)(x).(b)(f◦g)(x) =f(g(x)) =f(√x) = (√x)4=x2(g◦f)(x) =g(f(x)) =g(x4) =√x4=x2.While these expressions are equal, note that (f◦g)(x) is defined only forx≥0, while (g◦f)(x)is defined for anyx∈R. Since the domains off◦gandg◦fare unequal, the functions areunequal.(c)(f◦g)(x) =f(g(x)) =f(1x)=(1x)4=1x4(g◦f)(x) =g(f(x)) =g(x4) =1x4.Since these expressions are equal, (f◦g)(x) = (g◦f)(x).(d)(f◦g)(x) =f(g(x)) =f(−x) = (−x)4(g◦f)(x) =g(f(x)) =g(x4) =−x4.Since (−x)4=x46=−x4, these expressions are not equal, so (f◦g)(x)6= (g◦f)(x).(e)(f◦g)(x) =f(g(x)) =f(|x|) =|x|4(g◦f)(x) =g(f(x)) =g(x4) =∣∣x4∣∣.Since|x|4=x4=∣∣x4∣∣, these expressions are equal, so (f◦g)(x) = (g◦f)(x).1.3.221.x2x30.51.1.5x123yf(x)> g(x) for 0< x <1, andf(x)< g(x) forx >1.

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28Preview and Review22.x3x50.51.1.5x12345yf(x)> g(x) for 0< x <1, andf(x)< g(x) forx >1.23.0.51.1.5x12345yxx2x3x4The curves intersect atx= 0 and again atx= 1.24.(a)xx20.51.1.52.x1234y(b)f(x)≥g(x) for 0≤x≤1, andf(x)≤g(x) forx= 0 andx≥1.25.(a)x2x30.51.1.5x123y(b)Multiply both sides ofx≤1 byxto getx2≤x. Note that we do not have to worry aboutreversing the sign of the inequality since we also knowx≥0.(c)Multiply both sides ofx≥1 byxto getx2≥x. Again note that we do not have to worryabout reversing the sign of the inequality since we must havex >0.26.First suppose 0≤x≤1. Multiplying byx, we have 0≤x2≤x≤1. Multiplying byxagain, wehave 0≤x3≤x2≤x≤1. Repeating this processn−1 times, we get the chain of inequalities0≤xn≤xn−1· · · ≤xm≤xm−1· · · ≤x≤1Thusxn≤xmifn≥m.

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1.3 Elementary Functions29Next supposex≥1.Multiplying byx, we havex2≥x≥1.Multiplying byxagain, we havex3≥x2≥x≥1. Repeating this processn−1 times, we get the chain of inequalitiesxn≥xn−1· · · ≥xm≥xm−1· · · ≥x≥1Thusxn≥xmifn≥m.27.(a)Letf(x) =y=x4; thenf(−x) = (−x)4=x4=f(x), so thatfis even.(b)Letf(x) =y=x3; thenf(−x) = (−x)3=−x3=−f(x), so thatfis odd.28.(a)Ifnis even, writen= 2kwherekis an integer. Then lettingf(x) =y=xn=x2kwe getf(−x) = (−x)n= (−x)2k=((−x)2)k=(x2)k=x2k=f(x),so thatfis even.(b)Ifnis odd, writen= 2k+ 1 wherekis an integer. Then lettingf(x) =y=xn=x2k+1wegetf(−x) = (−x)n= (−x)2k+1= (−x)2k·(−x) =−x((−x)2)k=−x(x2)k=−x·x2k=−x2k+1=−f(x),so thatfis odd.29.(a)I(p) is a probability, and all probabilities must be between 0 and 1. A graph ofIis below:-112p12345IFrom the graph we see thatI(p)>0 everywhere.To see whenI(p) = 1, solveI(p) =2p2−2p+ 1 = 1, giving 2p(p−1) = 0, so thatp= 0 orp= 1. Thus for 0≤p≤1, the valueofI(p)∈[0,1]; it follows that the domain ofIisp∈[0,1].(b)See part (a).(c)The minimum value ofI(p) occurs forp=12, where it is 2·(12)2−2·12+ 1 =12, so the rangeI([0,1]) =[12,1].30.(a)The reaction rateRis proportional to the amount of the two reactants that are left; that is,R∝AB. Denote bykthe constant of proportionality. Ifxis the concentration ofAB, thenwe have used upxunits of bothAandB, so thatR(x) =k([A]−x)([B]−x) =k(3−x)(1−x)since there were three units ofAand one unit ofBto start with.(b)We cannot use up more of either component than there was to start with; since there is onlyone unit ofB, and we use a nonnegative amount ofB, the domain isx∈[0,1]. Note that forx≤1 thatR(x)≤3k, andR(x)≥0 everywhere, so that the range ofR(x) is [0,3k].

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30Preview and Review(c)Substituting [A] = 6 and [B] = 3 in the equation in part (a) givesR(x) =k(6−x)(3−x).Here, by the same argument as in part (b), the domain ofR(x) isx∈[0,3], and its range is[0,18k], since the maximum reaction rate occurs whenx= 0, at which time it is 18k.31.The beetle has a constant speed of 1 m/h. Fromrate·time = distance,we see that in one hour the beetle goes(1 m/h)·(1 h) = 1 m.Similarly,In two hours the beetle goes (1 m/h)·(2 h) = 2 mIn three hours the beetle goes (1 m/h)·(3 h) = 3 mLet the distance in meters bedand the time in hours bet; then from the distance-rate-time equationwe get a polynomial equation of degree 1 int:d= rate·t⇒d= 1·t⇒d=t.32.The fungal disease spreads radially at the speed of 3 ft/day. To calculate the area of the affectedpart of the orchard, we need to know its radius. Using the fact that speed·time = distance, we getfor the radius(3 ft/day)(2 day) = 6 ft on day two.Thus, the area of infection on the 2ndday isA=π·62≈113.1 ft2.To find the radius on the fourth and eighth days, use the same idea:(3 ft/day)(4 day) = 12 ft on day four(3 ft/day)(8 day) = 24 ft on day eightTherefore the area on these days isA=π·122= 144π≈452.4 ft2on day fourA=π·242= 576π≈1809.6 ft2on day eightAn equation that expresses the area as a function of time: we know that the circular area isA=πr2wherer= 3t; substituting forrgives a polynomial equation of degree 2 int:A=πr2=π(3t)2= 9πt2,1.3.333.The domain is (−∞,1)∪(1,∞), and the range is (−∞,0)∪(0,∞)

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