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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Document preview page 1

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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition

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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 1 preview imageDIAGNOSTIC TESTSTest AAlgebra1.(a)(−3)4= (−3)(−3)(−3)(−3) = 81(b)−34=−(3)(3)(3)(3) =−81(c)3−4=134=181(d)523521= 523−21= 52= 25(e)23−2=322=94(f )16−34=11634=14√163=123= 182.(a) Note that√200 =√100·2 = 10√2and√32 =√16·2 = 4√2. Thus√200− √32 = 10√2−4√2 = 6√2.(b)(333)(42)2= 3331624= 4857(c)33232−12−2=2−1233232= (2−12)2(3323)2=4−1936=4936=973.(a)3(+ 6) + 4(2−5) = 3+ 18 + 8−20 = 11−2(b)(+ 3)(4−5) = 42−5+ 12−15 = 42+ 7−15(c)√+√ √− √=√2− √√+√√−√2=−Or:Use the formula for the difference of two squares to see that√+√√− √=√2−√2=−.(d)(2+ 3)2= (2+ 3)(2+ 3) = 42+ 6+ 6+ 9 = 42+ 12+ 9.Note:A quicker way to expand this binomial is to use the formula(+)2=2+ 2+2with= 2and= 3:(2+ 3)2= (2)2+ 2(2)(3) + 32= 42+ 12+ 9(e) See Reference Page 1 for the binomial formula(+)3=3+ 32+ 32+3. Using it, we get(+ 2)3=3+ 32(2) + 3(22) + 23=3+ 62+ 12+ 8.4.(a) Using the difference of two squares formula,2−2= (+)(−), we have42−25 = (2)2−52= (2+ 5)(2−5).(b) Factoring by trial and error, we get22+ 5−12 = (2−3)(+ 4).(c) Using factoring by grouping and the difference of two squares formula, we have3−32−4+ 12 =2(−3)−4(−3) = (2−4)(−3) = (−2)(+ 2)(−3).(d)4+ 27=(3+ 27) =(+ 3)(2−3+ 9)This last expression was obtained using the sum of two cubes formula,3+3= (+)(2−+2)with=and= 3. [See Reference Page 1 in the textbook.](e) The smallest exponent onis−12, so we will factor out−12.332−912+ 6−12= 3−12(2−3+ 2) = 3−12(−1)(−2)(f )3−4=(2−4) =(−2)(+ 2)
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 2 preview image
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 3 preview image2¤DIAGNOSTIC TESTS5.(a)2+ 3+ 22−−2= (+ 1)(+ 2)(+ 1)(−2) =+ 2−2(b)22−−12−9·+ 32+ 1 = (2+ 1)(−1)(−3)(+ 3)·+ 32+ 1 =−1−3(c)22−4−+ 1+ 2 =2(−2)(+ 2)−+ 1+ 2 =2(−2)(+ 2)−+ 1+ 2·−2−2 =2−(+ 1)(−2)(−2)(+ 2)=2−(2−−2)(+ 2)(−2)=+ 2(+ 2)(−2) =1−2(d)−1−1=−1−1·=2−2−= (−)(+)−(−)=+−1=−(+)6.(a)√10√5−2 =√10√5−2·√5 + 2√5 + 2 =√50 + 2√10√52−22= 5√2 + 2√105−4= 5√2 + 2√10(b)√4 +−2=√4 +−2·√4 ++ 2√4 ++ 2 =4 +−4√4 ++ 2=√4 ++ 2=1√4 ++ 27.(a)2++ 1 =2++14+ 1−14=+122+34(b)22−12+ 11 = 2(2−6) + 11 = 2(2−6+ 9−9) + 11 = 2(2−6+ 9)−18 + 11 = 2(−3)2−78.(a)+ 5 = 14−12⇔+12= 14−5⇔32= 9⇔=23·9⇔= 6(b)2+ 1 = 2−1⇒22= (2−1)(+ 1)⇔22= 22+−1⇔= 1(c)2−−12 = 0⇔(+ 3)(−4) = 0⇔+ 3 = 0or−4 = 0⇔=−3or= 4(d) By the quadratic formula,22+ 4+ 1 = 0⇔=−4±42−4(2)(1)2(2)=−4±√84=−4±2√24= 2−2±√24=−2±√22=−1±12√2.(e)4−32+ 2 = 0⇔(2−1)(2−2) = 0⇔2−1 = 0or2−2 = 0⇔2= 1or2= 2⇔=±1or=±√2(f )3|−4|= 10⇔|−4|=103⇔−4 =−103or−4 =103⇔=23or=223(g) Multiplying through2(4−)−12−3√4−= 0by(4−)12gives2−3(4−) = 0⇔2−12 + 3= 0⇔5−12 = 0⇔5= 12⇔=125.9.(a)−45−3≤17⇔−9−3≤12⇔3 ≥ −4or−4≤ 3.In interval notation, the answer is[−43).(b)22+ 8⇔2−2−80⇔(+ 2)(−4)0. Now,(+ 2)(−4)will change sign at the criticalvalues=−2and= 4. Thus the possible intervals of solution are(−∞−2),(−24), and(4∞). By choosing asingle test value from each interval, we see that(−24)is the only interval that satisfies the inequality.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 4 preview imageTEST BANALYTIC GEOMETRY¤3(c) The inequality(−1)(+ 2)0has critical values of−20and1. The corresponding possible intervals of solutionare(−∞−2),(−20),(01)and(1∞). By choosing a single test value from each interval, we see that both intervals(−20)and(1∞)satisfy the inequality. Thus, the solution is the union of these two intervals:(−20)∪(1∞).(d)|−4|3⇔−3 −43⇔1  7. In interval notation, the answer is(17).(e)2−3+ 1≤1⇔2−3+ 1−1≤0⇔2−3+ 1−+ 1+ 1≤0⇔2−3−−1+ 1≤0⇔−4+ 1≤0.Now, the expression−4+ 1may change signs at the critical values=−1and= 4, so the possible intervals of solutionare(−∞−1),(−14], and[4∞). By choosing a single test value from each interval, we see that(−14]is the onlyinterval that satisfies the inequality.10.(a) False. In order for the statement to be true, it must hold for all real numbers, so, to show that the statement is false, pick= 1and= 2and observe that(1 + 2)26= 12+ 22. In general,(+)2=2+ 2+2.(b) True as long asandare nonnegative real numbers. To see this, think in terms of the laws of exponents:√= ()12=1212=√√.(c) False. To see this, let= 1and= 2, then√12+ 226= 1 + 2.(d) False. To see this, let= 1and= 2, then1 + 1(2)26= 1 + 1.(e) False. To see this, let= 2and= 3, then12−36= 12−13.(f ) True since1−·=1−, as long as6= 0and−6= 0.Test BAnalytic Geometry1.(a) Using the point(2−5)and=−3in the point-slope equation of a line,−1=(−1), we get−(−5) =−3(−2)⇒+ 5 =−3+ 6⇒=−3+ 1.(b) A line parallel to the-axis must be horizontal and thus have a slope of0. Since the line passes through the point(2−5),the-coordinate of every point on the line is−5, so the equation is=−5.(c) A line parallel to the-axis is vertical with undefined slope. So the-coordinate of every point on the line is 2 and so theequation is= 2.(d) Note that2−4= 3⇒−4=−2+ 3⇒=12−34. Thus the slope of the given line is=12. Hence, theslope of the line we’re looking for is also12(since the line we’re looking for is required to be parallel to the given line).So the equation of the line is−(−5) =12(−2)⇒+ 5 =12−1⇒=12−6.2.First we’llfind the distance between the two given points in order to obtain the radius,, of the circle:=[3−(−1)]2+ (−2−4)2=42+ (−6)2=√52. Next use the standard equation of a circle,(−)2+ (−)2=2, where( )is the center, to get(+ 1)2+ (−4)2= 52.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 5 preview image4¤DIAGNOSTIC TESTS3.We must rewrite the equation in standard form in order to identify the center and radius. Note that2+2−6+ 10+ 9 = 0⇒2−6+ 9 +2+ 10= 0. For the left-hand side of the latter equation, wefactor thefirst three terms and complete the square on the last two terms as follows:2−6+ 9 +2+ 10= 0⇒(−3)2+2+ 10+ 25 = 25⇒(−3)2+ (+ 5)2= 25. Thus, the center of the circle is(3−5)and the radius is5.4.(a)(−74)and(5−12)⇒=−12−45−(−7) =−1612=−43(b)−4 =−43[−(−7)]⇒−4 =−43−283⇒3−12 =−4−28⇒4+ 3+ 16 = 0. Putting= 0,we get4+ 16 = 0, so the-intercept is−4, and substituting0forresults in a-intercept of−163.(c) The midpoint is obtained by averaging the corresponding coordinates of both points:−7+524+(−12)2= (−1−4).(d)=[5−(−7)]2+ (−12−4)2=122+ (−16)2=√144 + 256 =√400 = 20(e) The perpendicular bisector is the line that intersects the line segmentat a right angle through its midpoint. Thus theperpendicular bisector passes through(−1−4)and has slope34[the slope is obtained by taking the negative reciprocal ofthe answer from part (a)]. So the perpendicular bisector is given by+ 4 =34[−(−1)]or3−4= 13.(f ) The center of the required circle is the midpoint of, and the radius is half the length of, which is10. Thus, theequation is(+ 1)2+ (+ 4)2= 100.5.(a) Graph the corresponding horizontal lines (given by the equations=−1and= 3) as solid lines. The inequality≥ −1describes the points( )that lieon orabovethe line=−1. The inequality≤3describes the points( )that lie on orbelowthe line= 3. So the pair of inequalities−1≤≤3describes the points that lie on orbetweenthe lines=−1and= 3.(b) Note that the given inequalities can be written as−4  4and−2  2,respectively. So the region lies between the vertical lines=−4and= 4andbetween the horizontal lines=−2and= 2. As shown in the graph, theregion common to both graphs is a rectangle (minus its edges) centered at theorigin.(c) Wefirst graph= 1−12as a dotted line. Since 1−12, the points in theregion liebelowthis line.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 6 preview imageTEST CFUNCTIONS¤5(d) Wefirst graph the parabola=2−1using a solid curve. Since≥2−1,the points in the region lie on orabovethe parabola.(e) We graph the circle2+2= 4using a dotted curve. Since2+22, theregion consists of points whose distance from the origin is less than 2, that is,the points that lieinsidethe circle.(f ) The equation92+ 162= 144is an ellipse centered at(00). We put it instandard form by dividing by144and get216 +29= 1. The-intercepts arelocated at a distance of√16 = 4from the center while the-intercepts are adistance of√9 = 3from the center (see the graph).Test CFunctions1.(a) Locate−1on the-axis and then go down to the point on the graph with an-coordinate of−1. The corresponding-coordinate is the value of the function at=−1, which is−2. So,(−1) =−2.(b) Using the same technique as in part (a), we get(2)≈28.(c) Locate2on the-axis and then go left and right tofind all points on the graph with a-coordinate of2. The corresponding-coordinates are the-values we are searching for. So=−3and= 1.(d) Using the same technique as in part (c), we get≈ −25and≈03.(e) The domain is all the-values for which the graph exists, and the range is all the-values for which the graph exists.Thus, the domain is[−33], and the range is[−23].2.Note that(2 +) = (2 +)3and(2) = 23= 8. So the difference quotient becomes(2 +)−(2)= (2 +)3−8= 8 + 12+ 62+3−8= 12+ 62+3=(12 + 6+2)= 12 + 6+2.3.(a) Set the denominator equal to 0 and solve tofind restrictions on the domain:2+−2 = 0⇒(−1)(+ 2) = 0⇒= 1or=−2. Thus, the domain is all real numbers except1or−2or, in intervalnotation,(−∞−2)∪(−21)∪(1∞).(b) Note that the denominator is always greater than or equal to1, and the numerator is defined for all real numbers. Thus, thedomain is(−∞∞).(c) Note that the functionis the sum of two root functions. Sois defined on the intersection of the domains of these tworoot functions. The domain of a square root function is found by setting its radicand greater than or equal to0. Now,
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 7 preview image6¤DIAGNOSTIC TESTS4−≥0⇒≤4and2−1≥0⇒(−1)(+ 1)≥0⇒≤ −1or≥1. Thus, the domain ofis(−∞−1]∪[14].4.(a) Reflect the graph ofabout the-axis.(b) Stretch the graph ofvertically by a factor of2, then shift1unit downward.(c) Shift the graph ofright3units, then up2units.5.(a) Make a table and then connect the points with a smooth curve:−2−1012−8−1018(b) Shift the graph from part (a) left1unit.(c) Shift the graph from part (a) right2units and up3units.(d) First plot=2. Next, to get the graph of() = 4−2,reflectabout thex-axis and then shift it upward4units.(e) Make a table and then connect the points with a smooth curve:01490123(f ) Stretch the graph from part (e) vertically by a factor of two.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 8 preview imageTEST DTRIGONOMETRY¤7(g) First plot= 2. Next, get the graph of=−2by reflecting the graph of= 2about thex-axis.(h) Note that= 1 +−1= 1 + 1. Sofirst plot= 1and then shift itupward1unit.6.(a)(−2) = 1−(−2)2=−3and(1) = 2(1) + 1 = 3(b) For≤0plot() = 1−2and, on the same plane, for 0plot the graphof() = 2+ 1.7.(a)(◦)() =(()) =(2−3) = (2−3)2+ 2(2−3)−1 = 42−12+ 9 + 4−6−1 = 42−8+ 2(b)(◦)() =(()) =(2+ 2−1) = 2(2+ 2−1)−3 = 22+ 4−2−3 = 22+ 4−5(c)(◦◦)() =((())) =((2−3)) =(2(2−3)−3) =(4−9) = 2(4−9)−3= 8−18−3 = 8−21Test DTrigonometry1.(a)300◦= 300◦180◦= 300180= 53(b)−18◦=−18◦180◦=−18180 =−102.(a)56= 56180◦= 150◦(b)2 = 2180◦=360◦≈1146◦3.We will use the arc length formula,=, whereis arc length,is the radius of the circle, andis the measure of thecentral angle in radians. First, note that30◦= 30◦180◦=6. So= (12)6= 2cm.4.(a)tan(3) =√3You can read the value from a right triangle with sides 1, 2, and√3.(b) Note that76can be thought of as an angle in the third quadrant with reference angle6. Thus,sin(76) =−12,since the sine function is negative in the third quadrant.(c) Note that53can be thought of as an angle in the fourth quadrant with reference angle3. Thus,sec(53) =1cos(53) =112 = 2, since the cosine function is positive in the fourth quadrant.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 9 preview image8¤DIAGNOSTIC TESTS5.sin=24⇒= 24 sinandcos=24⇒= 24 cos6.sin=13andsin2+ cos2= 1⇒cos=1−19= 2√23. Also,cos=45⇒sin=1−1625=35.So, using the sum identity for the sine, we havesin(+) = sincos+ cossin= 13·45 + 2√23·35 = 4 + 6√215=1154 + 6√27.(a)tansin+ cos= sincossin+ cos= sin2cos+ cos2cos=1cos= sec(b)2 tan1 + tan2= 2 sin(cos)sec2= 2 sincoscos2= 2 sincos= sin 28.sin 2= sin⇔2 sincos= sin⇔2 sincos−sin= 0⇔sin(2 cos−1) = 0⇔sin= 0orcos=12⇒= 0,3,,53,2.9.Wefirst graph= sin 2(by compressing the graph ofsinby a factor of 2) and then shift it upward1unit.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 10 preview image1FUNCTIONS AND SEQUENCES1.1Four Ways to Represent a Function1.The functions() =+√2−and() =+√2−give exactly the same output values for every input value, soandare equal.2.() =2−−1=(−1)−1=for−16= 0, soand[where() =] are not equal because(1)is undefined and(1) = 1.3.(a) The point(13)is on the graph of, so(1) = 3.(b) When=−1,is about−02, so(−1)≈ −02.(c)() = 1is equivalent to= 1When= 1, we have= 0and= 3.(d) A reasonable estimate forwhen= 0is=−08.(e) The domain ofconsists of all-values on the graph of. For this function, the domain is−2≤≤4, or[−24].The range ofconsists of all-values on the graph of. For this function, the range is−1≤≤3, or[−13].(f) Asincreases from−2to1,increases from−1to3. Thus,is increasing on the interval[−21].4.(a) The point(−4−2)is on the graph of, so(−4) =−2. The point(34)is on the graph of, so(3) = 4.(b) We are looking for the values offor which the-values are equal. The-values forandare equal at the points(−21)and(22), so the desired values ofare−2and2.(c)() =−1is equivalent to=−1. When=−1, we have=−3and= 4.(d) Asincreases from0to4,decreases from3to−1. Thus,is decreasing on the interval[04].(e) The domain ofconsists of all-values on the graph of. For this function, the domain is−4≤≤4, or[−44].The range ofconsists of all-values on the graph of. For this function, the range is−2≤≤3, or[−23].(f) The domain ofis[−43]and the range is[054].5.No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve failsthe Vertical Line Test.6.Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is[−22]and the rangeis[−12].7.Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is[−32]and the rangeis[−3−2)∪[−13].8.No, the curve is not the graph of a function since for= 0,±1, and±2, there are infinitely many points on the curve.9.(a) The graph shows that the global average temperature in 1950 was(1950)≈138◦C(b) By drawing the horizontal line= 142to the curve and then drawing the vertical line down to the horizontal axis, we seethat≈1992(c) The temperature was smallest in1910and largest in2006(d) The range is{|135≤≤145}= [135145]
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 11 preview image10¤CHAPTER 1FUNCTIONS AND SEQUENCES10.(a) The range is{Width|0Width≤16}= (016](b) The graph shows an overall decline in global temperatures from 1500 to 1700, followed by an overall rise in temperatures.Thefluctuations in temperature in the mid and late 19th century are reflective of the cooling effects caused by several largevolcanic eruptions.11.If we draw the horizontal linepH = 40we can see that the pH curve is less than 4.0 between 12:23AMand 12:52AM.Therefore, a clinical acid reflux episode occurred approximately between 12:23AMand 12:52AMat which time the esophagealpH was less than4012.The graphs indicate that tadpoles raised in densely populated regions take longer to put on weight. This is sensible since morecrowding leads to fewer resources available for each tadpole.13.(a) At30◦S and20◦N, we expect approximately 100 and 134 ant species respectively.(b) By drawing the horizontal line at a species richness of 100, we see there are two points of intersection with the curve, eachhaving latitude values of roughly30◦N and30◦S.(c) The function is even since its graph is symmetric with respect to the y-axis.14.Example 1:A car is driven at60mih for2hours. The distancetraveled by the car is a function of the time. The domain of thefunction is{|0≤≤2}, whereis measured in hours. The rangeof the function is{|0≤≤120}, whereis measured in miles.Example 2:At a certain university, the number of studentsoncampus at any time on a particular day is a function of the timeaftermidnight. The domain of the function is{|0≤≤24}, whereismeasured in hours. The range of the function is{|0≤≤},whereis an integer andis the largest number of students oncampus at once.Example 3:A certain employee is paid$800per hour and works amaximum of30hours per week. The number of hours worked isrounded down to the nearest quarter of an hour. This employee’sgross weekly payis a function of the number of hours worked.The domain of the function is[030]and the range of the function is{0200400    2380024000}.240payhours0.250.500.75029.50 29.75302423823615.The person’s weight increased to about160pounds at age20and stayed fairly steady for10years. The person’s weightdropped to about120pounds for the next5years, then increased rapidly to about170pounds. The next30years saw a gradualincrease to190pounds. Possible reasons for the drop in weight at30years of age: diet, exercise, health problems.16.Initially, the person’s forward moving heel contacts the ground resulting in a ground reaction force in the opposite or negativedirection. In moving from heel-strike to toe-off, the foot transitions from a forward push to a backward push. Hence, theground reaction force switches from a negative value to a positive value, becoming zero at some point in between.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 12 preview imageSECTION 1.1FOUR WAYS TO REPRESENT A FUNCTION¤1117.The water will cool down almost to freezing as the ice melts. Then, whenthe ice has melted, the water will slowly warm up to room temperature.18.Runner A won the race, reaching thefinish line at100meters in about15seconds, followed by runner B with a time of about19seconds, and then by runner C whofinished in around23seconds. B initially led the race, followed by C, and then A.C then passed B to lead for a while. Then A passedfirst B, and then passed C to take the lead andfinishfirst. Finally,B passed C tofinish in second place. All three runners completed the race.19.Initially, the bacteria population size remains constant during which nutrients are consumed in preparation for reproduction. Inthe second phase, the population size increases rapidly as the bacteria replicate. The population size plateaus in phase three atwhich point the "carrying capacity" has been reached and the available resources and space cannot support a larger population.Finally, the bacteria die due to starvation and waste toxicity and the population declines.20.The summer solstice (the longest day of the year) isaround June 21, and the winter solstice (the shortest day)is around December 22. (Exchange the dates for thesouthern hemisphere.)21.Of course, this graph depends strongly on thegeographical location!22.The temperature of the pie would increase rapidly, leveloff to oven temperature, decrease rapidly, and then leveloff to room temperature.23.As the price increases, the amount solddecreases.24.The value of the car decreases fairly rapidly initially, then somewhat less rapidly.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 13 preview image12¤CHAPTER 1FUNCTIONS AND SEQUENCES25.(a)198019902000201020406080100120xy0YearCount(b)198019902000201020406080100120xy0YearCountWe see from the graph that there were approximately92,000 birds in 1997.26.(a)012340.2tC0.4(b) Alcohol concentration increases rapidly within thefirsthour of consumption and then slowly decreases over thefollowing three hours.27.() = 32−+ 2(2) = 3(2)2−2 + 2 = 12−2 + 2 = 12(−2) = 3(−2)2−(−2) + 2 = 12 + 2 + 2 = 16() = 32−+ 2(−) = 3(−)2−(−) + 2 = 32++ 2(+ 1) = 3(+ 1)2−(+ 1) + 2 = 3(2+ 2+ 1)−−1 + 2 = 32+ 6+ 3−+ 1 = 32+ 5+ 42() = 2·() = 2(32−+ 2) = 62−2+ 4(2) = 3(2)2−(2) + 2 = 3(42)−2+ 2 = 122−2+ 2(2) = 3(2)2−(2) + 2 = 3(4)−2+ 2 = 34−2+ 2[()]2=32−+ 22=32−+ 232−+ 2= 94−33+ 62−33+2−2+ 62−2+ 4 = 94−63+ 132−4+ 4(+) = 3(+)2−(+) + 2 = 3(2+ 2+2)−−+ 2 = 32+ 6+ 32−−+ 228.A spherical balloon with radius+ 1has volume(+ 1) =43(+ 1)3=43(3+ 32+ 3+ 1). We wish tofind theamount of air needed to inflate the balloon from a radius ofto+ 1. Hence, we need tofind the difference(+ 1)−() =43(3+ 32+ 3+ 1)−433=43(32+ 3+ 1).29.() = 4 + 3−2, so(3 +) = 4 + 3(3 +)−(3 +)2= 4 + 9 + 3−(9 + 6+2) = 4−3−2,and(3 +)−(3)= (4−3−2)−4=(−3−)=−3−.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 14 preview imageSECTION 1.1FOUR WAYS TO REPRESENT A FUNCTION¤1330.() =3, so(+) = (+)3=3+ 32+ 32+3,and(+)−()= (3+ 32+ 32+3)−3=(32+ 3+2)= 32+ 3+2.31.()−()−=1−1−=−−=−(−) =−1(−)(−) =−132.()−(1)−1=+ 3+ 1−2−1=+ 3−2(+ 1)+ 1−1=+ 3−2−2(+ 1)(−1)=−+ 1(+ 1)(−1) =−(−1)(+ 1)(−1) =−1+ 133.() = (+ 4)(2−9)is defined for allexcept when0 =2−9⇔0 = (+ 3)(−3)⇔=−3or3, so thedomain is{∈R|6=−33}= (−∞−3)∪(−33)∪(3∞).34.() = (23−5)(2+−6)is defined for allexcept when0 =2+−6⇔0 = (+ 3)(−2)⇔=−3or2, so the domain is{∈R|6=−32}= (−∞−3)∪(−32)∪(2∞).35.() =3√2−1is defined for all real numbers. In fact3(), where()is a polynomial, is defined for all real numbers.Thus, the domain isRor(−∞∞).36.() =√3−− √2 +is defined when3−≥0⇔≤3and2 +≥0⇔≥ −2. Thus, the domain is−2≤≤3, or[−23].37.() = 14√2−5is defined when2−5 0⇔(−5)0. Note that2−56= 0since that would result indivision by zero. The expression(−5)is positive if 0or 5. (See Appendix A for methods for solvinginequalities.) Thus, the domain is(−∞0)∪(5∞).38.() =+ 11 +1+ 1is defined when+ 16= 0[6=−1] and1 +1+ 16= 0. Since1 +1+ 1 = 0⇒1+ 1 =−1⇒1 =−−1⇒=−2, the domain is{|6=−2,6=−1}= (−∞−2)∪(−2−1)∪(−1∞).39.() =2− √is defined when≥0and2− √≥0. Since2− √≥0⇒2≥ √⇒√≤2⇒0≤≤4, the domain is[04].40.() =√4−2. Now=√4−2⇒2= 4−2⇔2+2= 4, sothe graph is the top half of a circle of radius2with center at the origin. The domainis|4−2≥0=|4≥2={|2≥||}= [−22]. From the graph,the range is0≤≤2, or[02].
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 15 preview image14¤CHAPTER 1FUNCTIONS AND SEQUENCES41.() = 2−04is defined for all real numbers, so the domain isR,or(−∞∞)The graph ofis a line with slope−04and-intercept2.42.() =2−2+ 1 = (−1)2is defined for all real numbers, so thedomain isR, or(−∞∞). The graph ofis a parabola with vertex(10).43.() = 2+2is defined for all real numbers, so the domain isR, or(−∞∞). The graph ofis a parabola opening upward since thecoefficient of2is positive. Tofind the-intercepts, let= 0and solvefor.0 = 2+2=(2 +)⇒= 0or=−2. The-coordinate ofthe vertex is halfway between the-intercepts, that is, at=−1. Since(−1) = 2(−1) + (−1)2=−2 + 1 =−1, the vertex is(−1−1).44.() = 4−22−= (2 +)(2−)2−, so for6= 2,() = 2 +. The domainis{|6= 2}. So the graph ofis the same as the graph of the function() =+ 2(a line) except for the hole at(24).45.() =√−5is defined when−5≥0or≥5, so the domain is[5∞).Since=√−5⇒2=−5⇒=2+ 5, we see thatis thetop half of a parabola.46.() =|2+ 1|=2+ 1−(2+ 1)if2+ 1≥0if2+ 11=2+ 1−2−1if≥ −12if −12The domain isR, or(−∞∞).
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 16 preview imageSECTION 1.1FOUR WAYS TO REPRESENT A FUNCTION¤1547.() = 3+||. Since||=if≥0−if 0, we have() =3+if 03−if 0=4if 02if 0=4if 02if 0Note thatis not defined for= 0. The domain is(−∞0)∪(0∞).48.() =||−=−if≥0−−if 0 =0if≥0−2if 0.The domain isR, or(−∞∞).49.() =+ 2if 01−if≥0The domain isR.50.() =3−12if≤22−5if 2The domain isR.51.() =+ 2if≤ −12if −1Note that for=−1, both+ 2and2are equal to 1. The domain isR.52.() =+ 9if −3−2if||≤3−6if 3Note that for=−3, both+ 9and−2are equal to6; and for= 3, both−2and−6are equal to−6. The domain isR.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 17 preview image16¤CHAPTER 1FUNCTIONS AND SEQUENCES53.Let the length and width of the rectangle beand. Then the perimeter is2+ 2= 20and the area is=.Solving thefirst equation forin terms ofgives= 20−22= 10−. Thus,() =(10−) = 10−2. Sincelengths are positive, the domain ofis0  10. If we further restrictto be larger than, then5  10would bethe domain.54.Let the length and width of the rectangle beand. Then the area is= 16, so that= 16. The perimeter is= 2+ 2, so() = 2+ 2(16) = 2+ 32, and the domain ofis 0, since lengths must be positivequantities. If we further restrictto be larger than, then 4would be the domain.55.Let the length of a side of the equilateral triangle be. Then by the Pythagorean Theorem, the heightof the triangle satisfies2+122=2, so that2=2−142=342and=√32. Using the formula for the areaof a triangle,=12(base)(height), we obtain() =12()√32=√342, with domain 0.56.Let the volume of the cube beand the length of an edge be. Then=3so=3√, and the surface area is() = 62= 63√2= 623, with domain 0.57.Let each side of the base of the box have length, and let the height of the box be. Since the volume is2, we know that2 =2, so that= 22, and the surface area is=2+ 4. Thus,() =2+ 4(22) =2+ (8), withdomain 0.58.We can summarize the monthly cost with a piecewisedefined function.() =35if0≤≤40035 + 010(−400)if 40059.We can summarize the total cost with a piecewise defined function.() =75if0 ≤2150 + 50(−2)if 260.One example is the amount paid for cable or telephone system repair in the home, usually measured to the nearest quarter hour.Another example is the amount paid by a student in tuition fees, if the fees vary according to the number of credits for whichthe student has registered.61.The period can be estimated by measuring the peak-to-peak distance on the graph. This is approximately 77 hours. Note thatthe graph shown is for a single person’s temperature. The period for this species of malaria is, on average, 72 hours.62.The cycle of increased body temperature followed by a drop in temperature is indicative of a recurrent fever. This is typical ofaP. falciparuminfection. The period is approximately 48 hours, but the fever is also subsiding. This might be because theperson is being treated for infection.63.is an odd function because its graph is symmetric about the origin.is an even function because its graph is symmetric withrespect to the-axis.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 18 preview imageSECTION 1.1FOUR WAYS TO REPRESENT A FUNCTION¤1764.is not an even function since it is not symmetric with respect to the-axis.is not an odd function since it is not symmetricabout the origin. Hence,isneithereven nor odd.is an even function because its graph is symmetric with respect to the-axis.65.(a) Because an even function is symmetric with respect to the-axis, and the point(53)is on the graph of this even function,the point(−53)must also be on its graph.(b) Because an odd function is symmetric with respect to the origin, and the point(53)is on the graph of this odd function,the point(−5−3)must also be on its graph.66.(a) Ifis even, we get the rest of the graph by reflectingabout the-axis.(b) Ifis odd, we get the rest of the graph by rotating180◦about the origin.67.() =2+ 1.(−) =−(−)2+ 1 =−2+ 1 =−2+ 1 =−().Sois an odd function.68.() =24+ 1.(−) =(−)2(−)4+ 1 =24+ 1 =().Sois an even function.69.() =+ 1, so(−) =−−+ 1 =−1.Since this is neither()nor−(), the functionisneither even nor odd.70.() =||.(−) = (−)|−|= (−)||=−(||)=−()Sois an odd function.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 19 preview image18¤CHAPTER 1FUNCTIONS AND SEQUENCES71.() = 1 + 32−4.(−) = 1+3(−)2−(−)4= 1+32−4=().Sois an even function.72.() = 1 + 33−5, so(−) = 1 + 3(−)3−(−)5= 1 + 3(−3)−(−5)= 1−33+5Since this is neither()nor−(), the functionisneither even nor odd.73.(i) Ifandare both even functions, then(−) =()and(−) =(). Now(+)(−) =(−) +(−) =() +() = (+)(), so+is anevenfunction.(ii) Ifandare both odd functions, then(−) =−()and(−) =−(). Now(+)(−) =(−) +(−) =−() + [−()] =−[() +()] =−(+)(), so+is anoddfunction.(iii) Ifis an even function andis an odd function, then(+)(−) =(−) +(−) =() + [−()] =()−(),which is not(+)()nor−(+)(), s o+isneithereven nor odd. (Exception: ifis the zero function, then+will beodd. Ifis the zero function, then+will beeven.)74.(i) Ifandare both even functions, then(−) =()and(−) =(). Now( )(−) =(−)(−) =()() = ( )(), sois anevenfunction.(ii) Ifandare both odd functions, then(−) =−()and(−) =−(). Now( )(−) =(−)(−) = [−()][−()] =()() = ( )(), so is anevenfunction.(iii) Ifis an even function andis an odd function, then( )(−) =(−)(−) =()[−()] =−[()()] =−( )(), so is anoddfunction.1.2Mathematical Models: A Catalog of Essential Functions1.(a)() = log2is a logarithmic function.(b)() =4√is a root function with= 4.(c)() =231−2is a rational function because it is a ratio of polynomials.(d)() = 1−11+ 2542is a polynomial of degree2(also called aquadratic function).(e)() = 5is an exponential function.(f)() = sincos2is a trigonometric function.2.(a)=is an exponential function (notice thatis theexponent).(b)=is a power function (notice thatis thebase).
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 20 preview imageSECTION 1.2MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS¤19(c)=2(2−3) = 22−5is a polynomial of degree5.(d)= tan−cosis a trigonometric function.(e)=(1 +)is a rational function because it is a ratio of polynomials.(f)=√3−1(1 +3√)is an algebraic function because it involves polynomials and roots of polynomials.3.We notice from thefigure thatandare even functions (symmetric with respect to the-axis) and thatis an odd function(symmetric with respect to the origin). So (b)=5must be. Sinceisflatter thannear the origin, we must have(c)=8matched withand (a)=2matched with.4.(a) The graph of= 3is a line (choice).(b)= 3is an exponential function (choice).(c)=3is an odd polynomial function or power function (choice).(d)=3√=13is a root function (choice).5.(a) An equation for the family of linear functions with slope2is=() = 2+, whereis the-intercept.(b)(2) = 1means that the point(21)is on the graph of. We can use thepoint-slope form of a line to obtain an equation for the family of linearfunctions through the point(21).−1 =(−2), which is equivalentto=+ (1−2)in slope-intercept form.(c) To belong to both families, an equation must have slope= 2, so the equation in part (b),=+ (1−2),becomes= 2−3. It is theonlyfunction that belongs to both families.6.All members of the family of linear functions() = 1 +(+ 3)havegraphs that are lines passing through the point(−31).
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 21 preview image20¤CHAPTER 1FUNCTIONS AND SEQUENCES7.All members of the family of linear functions() =−have graphsthat are lines with slope−1. The-intercept is.8.The vertex of the parabola on the left is(30), so an equation is=(−3)2+ 0. Since the point(42)is on theparabola, we’ll substitute4forand2fortofind.2 =(4−3)2⇒= 2, so an equation is() = 2(−3)2.The-intercept of the parabola on the right is(01), so an equation is=2++ 1. Since the points(−22)and(1−25)are on the parabola, we’ll substitute−2forand2foras well as1forand−25forto obtain two equationswith the unknownsand.(−22):2 = 4−2+ 1⇒4−2= 1(1)(1−25):−25 =++ 1⇒+=−35(2)2·(2)+(1)gives us6=−6⇒=−1. F rom(2),−1 +=−35⇒=−25, so an equationis() =−2−25+ 1.9.Since(−1) =(0) =(2) = 0,has zeros of−1,0, and2, so an equation foris() =[−(−1)](−0)(−2),or() =(+ 1)(−2). Because(1) = 6, we’ll substitute1forand6for().6 =(1)(2)(−1)⇒−2= 6⇒=−3, so an equation foris() =−3(+ 1)(−2).10.(a) For= 002+ 850, the slope is002, which means that the average surface temperature of the world is increasing at arate of002◦Cper year. The-intercept is850, which represents the average surface temperature in◦Cin the year 1900.(b)= 2100−1900 = 200⇒= 002(200) + 850 = 1250◦C11.(a)= 200, so= 00417(+ 1) = 00417(200)(+ 1) = 834+ 834. The slope is834, which represents thechange in mg of the dosage for a child for each change of 1 year in age.(b) For a newborn,= 0, so= 834mg.12.(a) We are givenchange in pressure10feet change in depth= 43410= 0434. Usingfor pressure andfor depth with the point( ) = (015), we have the slope-intercept form of the line,= 0434+ 15.(b) When= 100, then100 = 0434+ 15⇔0434= 85⇔=850434≈19585feet. Thus, the pressure is100 lbin2at a depth of approximately196feet.13.(a)(b) The slope of95means thatincreases95degrees for each increaseof1◦C. (Equivalently,increases by9whenincreases by5anddecreases by9whendecreases by5.) The-intercept of32is the Fahrenheit temperature corresponding to a Celsiustemperature of0.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 22 preview imageSECTION 1.2MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS¤2114.(a) Assumingis a linear function ofwe can sketchthe graph of()by plotting the points(150035)and(50014)and drawing the straight line that passesthrough both these points.50150V(150, 0.35)(50, 0.14)A10000.10.40.30.2(b) The slope is=(150)−(50)150−50= 035−014100= 00021min−1This represents the rate of change of absorptionrate with respect to volume. The slope of00021means thatincreases by00021mLmin for each1mL increase in(c) The-intercept of0035mLmin is the absorption rate corresponding to a cerebrospinalfluid volume of0mL.15.(a) Usingin place ofandin place of, wefind the slope to be2−12−1=80−70173−113 = 1060 = 16. So a linearequation is−80 =16(−173)⇔−80 =16−1736⇔=16+30763076= 5116.(b) The slope of16means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricketchirps per minute. Said differently, each increase of6cricket chirps per minute corresponds to an increase of1◦F.(c) When= 150, the temperature is given approximately by=16(150) +3076= 7616◦F≈76◦F.16.(a) Usingin place ofandin place of, wefind the slope to be2−12−1= 460−380800−480 =80320 = 14.So a linear equation is−460 =14(−800)⇔−460 =14−200⇔=14+ 260.(b) Letting= 1500we get=14(1500) + 260 = 635.The cost of driving 1500 miles is $635.(c)The slope of the line represents the cost permile,$025.(d) The-intercept represents thefixed cost, $260.(e) A linear function gives a suitable model in this situation because you havefixed monthly costs such as insurance and carpayments, as well as costs that increase as you drive, such as gasoline, oil, and tires, and the cost of these for eachadditional mile driven is a constant.17.(a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form() =cos() +seems appropriate.(b) The data appear to be decreasing in a linear fashion. A model of the form() =+seems appropriate.18.(a) The data appear to be increasing exponentially. A model of the form() =·or() =·+seems appropriate.(b) The data appear to be decreasing similarly to the values of the reciprocal function. A model of the form() =seemsappropriate.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 23 preview image22¤CHAPTER 1FUNCTIONS AND SEQUENCES19.(a)A linear model does seem appropriate.(b) Using the points(4000141)and(60,00082), we obtain−141 =82−14160,000−4000 (−4000)or, equivalently,≈ −0000105357+ 14521429.(c) Using a computing device, we obtain the least squares regression line=−00000997855+ 13950764.The following commands and screens illustrate how tofind the least squares regression line on a TI-84 Plus.Enter the data into list one (L1) and list two (L2). Pressto enter the editor.Find the regession line and store it in Y1. Press.Note from the lastfigure that the regression line has been stored in Y1and that Plot1 has been turned on (Plot1 ishighlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressingor bypressing.Now pressto produce a graph of the data and the regressionline. Note that choice 9 of the ZOOM menu automatically selects a windowthat displays all of the data.(d) When= 25,000,≈11456; or about115per100population.(e) When= 80,000,≈5968; or about a6%chance.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 24 preview imageSECTION 1.2MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS¤23(f) When= 200,000,is negative, so the model does not apply.20.(a)(b)Using a computing device, we obtain the least squaresregression line= 4856−22096.(c) When= 100◦F,= 2647≈265chirpsmin.21.(a)3555140185LH(b) Using a calculator to perform a linear regression gives= 18807+ 826497whereis the height in centimeters andis the femur length in centimeters. This line, having slope 1.88 and-intercept 82.65, is plotted below.3555140185LH(c) The height of a person with= 53is(53) = (18807)(53) + 826497≈1823 cm.22.(a) Using a calculator to perform a linear regression gives= 00188+ 03048(b) The plot shows that the data is approximately linear. A higher degree polynomialfit, such as a cubic, may better model thedata.0300060(c) The-intercept represents the percentage of mice that developed tumors without any asbestos exposure.23.Ifis the original distance from the source, then the illumination is() =−2=2. Moving halfway to the lamp givesus an illumination of12=12−2=(2)2= 4(2), so the light is 4 times as bright.24.(a) Set= 90 inand solve for:90 = 30603952⇐⇒90306=03952⇐⇒=90306103952≈1533 lb(b) Set= 300 lband calculate:= 306 (300)03952≈2915 in(c) According to the model, a300 lbostrich needs a wingspan of292 intofly. Therefore, an ostrich with a72 inwingspancannot generate enough lift forflight.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 25 preview image24¤CHAPTER 1FUNCTIONS AND SEQUENCES25.(a) Using a computing device, we obtain a power function=, where≈31046and≈0308.(b) If= 291, then=≈178, so you would expect tofind 18 species of reptiles and amphibians on Dominica.26.(a)= 1000 431 2271499 528 750(b) The power model in part (a) is approximately=15. Squaring both sides gives us2=3, so the model matchesKepler’s Third Law,2=3.27.(a) Using a calculator to perform a 3rd-degree polynomial regressiongives= 001553−037252+ 39461+ 12108whereisage andis length. This polynomial is plotted along with ascatterplot of the data.01530AL(b) A 5-year old rock bass has a length of(5) = (00155)(5)3−(03725)(5)2+ (39461)(5) + 12108≈136 in(c) Using computer algebra software to solve forin the equation20 = 001553−037252+ 39461+ 12108gives≈1088years. Alternatively, the graph from part (a) can be used to estimate the age when= 20by drawing ahorizontal line at= 20to the curve and observing the age at this point.1.3New Functions from Old Functions1.(a) If the graph ofis shifted3units upward, its equation becomes=() + 3.(b) If the graph ofis shifted3units downward, its equation becomes=()−3.(c) If the graph ofis shifted3units to the right, its equation becomes=(−3).(d) If the graph ofis shifted3units to the left, its equation becomes=(+ 3).(e) If the graph ofis reflected about the-axis, its equation becomes=−().(f) If the graph ofis reflected about the-axis, its equation becomes=(−).(g) If the graph ofis stretched vertically by a factor of3, its equation becomes= 3().(h) If the graph ofis shrunk vertically by a factor of3, its equation becomes=13().2.(a) To obtain the graph of=() + 8from the graph of=(), shift the graph8units upward.(b) To obtain the graph of=(+ 8)from the graph of=(), shift the graph8units to the left.(c) To obtain the graph of= 8()from the graph of=(), stretch the graph vertically by a factor of8.(d) To obtain the graph of=(8)from the graph of=(), shrink the graph horizontally by a factor of8.(e) To obtain the graph of=−()−1from the graph of=(),first reflect the graph about the-axis, and then shift it1unit downward.(f) To obtain the graph of= 8(18)from the graph of=(), stretch the graph horizontally and vertically by a factorof8.3.(a) (graph 3) The graph ofis shifted4units to the right and has equation=(−4).(b) (graph 1) The graph ofis shifted3units upward and has equation=() + 3.(c) (graph 4) The graph ofis shrunk vertically by a factor of3and has equation=13().
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 26 preview imageSECTION 1.3NEW FUNCTIONS FROM OLD FUNCTIONS¤25(d) (graph 5) The graph ofis shifted4units to the left and reflected about the-axis. Its equation is=−(+ 4).(e) (graph 2) The graph ofis shifted6units to the left and stretched vertically by a factor of2. Its equation is= 2(+ 6).4.(a) To graph=()−2, we shift the graph of,2units downward.The point(12)on the graph ofcorresponds to the point(12−2) = (10).(b) To graph=(−2), we shift the graph of,2units to the right.The point(12)on the graph ofcorresponds to the point(1 + 22) = (32).(c) To graph=−2(), we reflect the graph about the-axis and stretch the graph vertically by a factor of2.The point(12)on the graph ofcorresponds to thepoint(1−2·2) = (1−4).(d) To graph=(13) + 1, we stretch the graphhorizontally by a factor of3and shift it1unit upward.The point(12)on the graph ofcorresponds to thepoint(1·32 + 1) = (33).5.(a) To graph=(2)we shrink the graph ofhorizontally by a factor of2.The point(4−1)on the graph ofcorresponds to thepoint12·4−1= (2−1).(b) To graph=12we stretch the graph ofhorizontally by a factor of2.The point(4−1)on the graph ofcorresponds to thepoint(2·4−1) = (8−1).(c) To graph=(−)we reflect the graph ofaboutthe-axis.The point(4−1)on the graph ofcorresponds to thepoint(−1·4−1) = (−4−1).(d) To graph=−(−)we reflect the graph ofaboutthe-axis, then about the-axis.The point(4−1)on the graph ofcorresponds to thepoint(−1·4−1·−1) = (−41).
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 27 preview image26¤CHAPTER 1FUNCTIONS AND SEQUENCES6.(a) The graph of= 2 sincan be obtained from the graphof= sinby stretching it vertically by a factor of2.(b) The graph of= 1 +√can be obtained fromthe graph of=√by shifting it upward1unit.7.=1+ 2: Start with the graph of the reciprocal function= 1and shift 2 units to the left.8.= (−1)3: Start with the graph of=3and shift 1 unit to the right.9.=−3√: Start with the graph of=3√and reflect about the-axis.10.=2+ 6+ 4 = (2+ 6+ 9)−5 = (+ 3)2−5: Start with the graph of=2, shift 3 units to the left, and then shift5 units downward.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 28 preview imageSECTION 1.3NEW FUNCTIONS FROM OLD FUNCTIONS¤2711.=√−2−1: Start with the graph of=√, shift 2 units to the right, and then shift 1 unit downward.12.= 4 sin 3: Start with the graph of= sin, compress horizontally by a factor of3, and then stretch vertically by afactor of4.13.= sin(2): Start with the graph of= sinand stretch horizontally by a factor of2.14.= 2−2: Start with the graph of= 1, stretch vertically by a factor of2, and then shift2units downward.15.=−3: Start with the graph of=3and reflect about the-axis. Note: Reflecting about the-axis gives the same resultsince substituting−forgives us= (−)3=−3.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 29 preview image28¤CHAPTER 1FUNCTIONS AND SEQUENCES16.= 1−2√+ 3: Start with the graph of=√, shift3units to the left, stretch vertically by a factor of2,reflect about the-axis, and then shift1unit upward.17.=12(1−cos): Start with the graph of= cos, reflect about the-axis, shift 1 unit upward, and then shrink vertically bya factor of 2.18.=||−2: Start with the graph of=||and shift2units downward.19.= 1−2−2=−(2+ 2) + 1 =−(2+ 2+ 1) + 2 =−(+ 1)2+ 2: Start with the graph of=2, reflect aboutthe-axis, shift 1 unit to the left, and then shift 2 units upward.
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 30 preview imageSECTION 1.3NEW FUNCTIONS FROM OLD FUNCTIONS¤2920.=14tan(−4): Start with the graph of= tan, shift4units to the right, and then compress vertically by a factor of4.21.This is just like the solution to Example 4 except the amplitude of the curve (the 30◦N curve in Figure 9 on June 21) is14−12 = 2. So the function is() = 12 + 2 sin2365(−80). March 31 is the90th day of the year, so the model gives(90)≈1234h. The daylight time (5:51AMto 6:18PM) is12hours and27minutes, or1245h. The model value differsfrom the actual value by1245−12341245≈0009, less than1%.22.Using a sine function to model the brightness of Delta Cephei as a function of time, we take its period to be54days, itsamplitude to be035(on the scale of magnitude), and its average magnitude to be40. If we take= 0at a time of averagebrightness, then the magnitude (brightness) as a function of timein days can be modeled by the formula() = 40 + 035 sin254.23.Let()be the water depth in meters athours after midnight. Apply the following transformations to the cosine function:•Vertical stretch by factor5since the amplitude needs to be12−22= 5 m•Horizontal stretch by factor122=6since the period needs to be12 h•Vertical shift7units upward since the function ranges between2and12which has a midpoint of12+22= 7 m•Horizontal shift675units to right to position the maximum at= 675 h(6:45AM)Combining these transformations gives the water depth function() = 5 cos6(−675)+ 724.Let()be the total volume of air in mL afterseconds. Because the respiratory cycle is periodic, a sine function can be usedas a model by applying the following transformations:•Vertical stretch by factor250since the amplitude needs to be5002= 250mL•Horizontal stretch by factor42=2since the period needs to be4 s•Vertical shift2250units upward since the function ranges between2000and2500which has a midpoint of2000+25002= 2250mLCombining these transformations gives the volume function() = 250 sin2+ 225025.Let()be the gene frequency afteryears. The gene frequency dynamics can be modeled using a sine function with thefollowing transformations:•Vertical stretch by factor30since the amplitude needs to be80−202= 30%•Horizontal stretch by factor32since the period needs to be3years•Vertical shift50units upward since the function ranges between80and20which has a midpoint of80+202= 50Combining these transformations gives the gene frequency function() = 30 sin23+ 5026.Let()be the density of neutrophils in cells/L afterdays. The density is periodic and can be modeled using a cosinefunction with the following transformations:•Vertical stretch by factor1000since the amplitude needs to be2000−02= 1000•Horizontal stretch by factor212since the period needs to be21days (or3weeks)•Vertical shift1000units upward since the function ranges between0and2000which has a midpoint of2000+02= 1000Combining these transformations gives the density function() = 1000 cos2+ 1000
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Test Bank for Biocalculus: Calculus for Life Sciences, 1st Edition - Page 31 preview image30¤CHAPTER 1FUNCTIONS AND SEQUENCES27.() =3+ 22;() = 32−1.=Rfor bothand.(a)(+)() = (3+ 22) + (32−1) =3+ 52−1,=R.(b)(−)() = (3+ 22)−(32−1) =3−2+ 1,=R.(c)( )() = (3+ 22)(32−1) = 35+ 64−3−22,=R.(d)() =3+ 2232−1,=|6=±1√3since32−16= 0.28.() =√3−,= (−∞3];() =√2−1,= (−∞−1]∪[1∞).(a)(+)() =√3−+√2−1,= (−∞−1]∪[13], which is the intersection of the domains ofand.(b)(−)() =√3−− √2−1,= (−∞−1]∪[13].(c)( )() =√3−·√2−1,= (−∞−1]∪[13].(d)() =√3−√2−1,= (−∞−1)∪(13]. We must exclude=±1since these values would makeundefined.29.() =2−1,=R;() = 2+ 1,=R.(a)(◦)() =(()) =(2+ 1) = (2+ 1)2−1 = (42+ 4+ 1)−1 = 42+ 4,=R.(b)(◦)() =(()) =(2−1) = 2(2−1) + 1 = (22−2) + 1 = 22−1,=R.(c)(◦)() =(()) =(2−1) = (2−1)2−1 = (4−22+ 1)−1 =4−22,=R.(d)(◦)() =(()) =(2+ 1) = 2(2+ 1) + 1 = (4+ 2) + 1 = 4+ 3,=R.30.() =−2;() =2+ 3+ 4.=Rfor bothand, and hence for their composites.(a)(◦)() =(()) =(2+ 3+ 4) = (2+ 3+ 4)−2 =2+ 3+ 2.(b)(◦)() =(()) =(−2) = (−2)2+ 3(−2) + 4 =2−4+ 4 + 3−6 + 4 =2−+ 2.(c)(◦)() =(()) =(−2) = (−2)−2 =−4.(d)(◦)() =(()) =(2+ 3+ 4) = (2+ 3+ 4)2+ 3(2+ 3+ 4) + 4= (4+ 92+ 16 + 63+ 82+ 24) + 32+ 9+ 12 + 4=4+ 63+ 202+ 33+ 3231.() = 1−3;() = cos.=Rfor bothand, and hence for their composites.(a)(◦)() =(()) =(cos) = 1−3 cos.(b)(◦)() =(()) =(1−3) = cos(1−3).(c)(◦)() =(()) =(1−3) = 1−3(1−3) = 1−3 + 9= 9−2.(d)(◦)() =(()) =(cos) = cos(cos)[Note that this isnotcos·cos.]32.() =√,= [0∞);() =3√1−,=R.(a)(◦)() =(()) =3√1−=3√1−=6√1−.The domain of◦is{|3√1−≥0}={|1−≥0}={|≤1}= (−∞1].(b)(◦)() =(()) =(√) =31− √.The domain of◦is{|is in the domain ofand()is in the domain of}. This is the domain of,that is,[0∞).(c)(◦)() =(()) =(√) =√=4√. The domain of◦is{|≥0and√≥0}= [0∞).
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