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Solution Manual for Biocalculus: Calculus, Probability, and Statistics for the Life Sciences, 1st Edition

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DIAGNOSTIC 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)uplicible1

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2¤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.post, in

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TEST 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.or ducessi

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4¤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.post, in

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TEST 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,or ducessi

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6¤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.post, in

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TEST 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.or ducessi

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8¤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.post, in

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1FUNCTIONS 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]uplicible9

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10¤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.post, in

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SECTION 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.or ducessi

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12¤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−.post, in

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SECTION 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].or ducessi

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Solution Manual for Biocalculus: Calculus, Probability, and Statistics for the Life Sciences, 1st Edition - Page 15 preview image

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14¤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(−∞∞).post, in

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Solution Manual for Biocalculus: Calculus, Probability, and Statistics for the Life Sciences, 1st Edition - Page 16 preview image

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SECTION 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.or ducessi
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Biocalculus: Calculus, Probability,...

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