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## Calc for BusLife Sciences

by: Alvena McDermott

11

0

73

# Calc for BusLife Sciences MATH 1314

Marketplace > University of Houston > Mathmatics > MATH 1314 > Calc for BusLife Sciences
Alvena McDermott
UH
GPA 3.69

Dianne Gross

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COURSE
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Dianne Gross
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KARMA
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This 73 page Class Notes was uploaded by Alvena McDermott on Saturday September 19, 2015. The Class Notes belongs to MATH 1314 at University of Houston taught by Dianne Gross in Fall. Since its upload, it has received 11 views. For similar materials see /class/208382/math-1314-university-of-houston in Mathmatics at University of Houston.

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Date Created: 09/19/15
M1314 Lessun 18 1 1314 Math Lesmn 18 Am and the In nite 1mm hegch cunfurms te semethmg fur wheh wehave afurmulafur geumetry Example 1 Suppuse m 5 and the areaunder the graph uf m hem x I U x 4 Appmdmzu39ngAm Under 2 Curve Nuw Suppuse the areaunda39 the curve is het semethmg whuse area eeh he eesuy eemputea wen heed te develup amethud fur ehmhg sueh an area Example 2 Here we39ll draw seme rectangles te zppmmmate the areaunda39 the curve under the curve h Eaehheetehgle has awxdth uf Ax Thehaghtxs aetehmmea hy Lhevalue af x The area is appmxlmated hy the sum efthe areas quhe rectangles M1314 LessmlE 2 Eurup 3 Nzxt we39llmcruse than an ufxecunges Whatyau shnuldsee 15 um um number afxecungesmcxuses the areawe campu39z usng 39nsme39hadbecamesmme annual Th An Undnlhz anhufz Fm n Ln be amngmve cman nmanan b Thzn39hearuuf39hexegmmdex m gaph uffxsgvznby A ygLX VM XnAX whzxe x xi xn are mum pm m the mural 11 5 Dream wxdlh Ax The sums arms ufxectmges are calledR39Imann sums undue named mg acmm mathzmmum M1314 Lesson 18 3 Example 5 Use left endpoints and 4 subdivisions of the interval to approximate the area under fx 2x2 1 on the interval 0 2 Example 6 Use right endpoints and 4 subdivisions of the interval to approximate the area under f x 2x2 1 on the interval 0 2 M1314 Lesson 18 Example 7 Use midpoints and 4 subdivisions of the interval to approximate the area under fx 2x2 1 on the interval 0 2 Example 8 Suppose f x 1 3x Approximate the area under the graph of f on the interval 0 12 using 6 subdivisions and left endpoints M1314 Lesson 18 5 The De nite Integral Letfbe de ned on 11 b If limfx1 fx2 fxH Axexists for all choices of b a representative points in the n subintervals of 11 b of equal width Ax then this limit is called the de nite integral of f from a to b The de nite integral is noted by rfxdx limfx1 fx2 fxn Ax The number a is called the lower limit of integration and the number b is called the upper limit of integration A function is said to be integrable on 11 b if it is continuous on the interval 11 b The de nite integral of a nonnegative function fxdx is equal to the geometric area u n m M1314 Lesson 18 6 The definite integral of a general function Page 976 in your book The part of the graph of f that goes below the xaxis is considered negative area Using definite integration we will look how to integrate this function form 0 to 5 ifxdx 7 ifxdx 0 2 From this section you should be able to Explain the procedure used to approximate area under a curve Use Riemann sums to approximate the area under a curve using right endpoinm left endpoints or midpoinm Explain what we mean by definite integral of a nonnegative function or a general function M1314 Lesson 16 1 Math 1314 Lesson 16 Antiderivatives So far in this course we have been interested in finding derivatives and in the applications of derivatives In this chapter we will look at the reverse process Here we will be given the answer and we ll have to find the problem This process is generally called integration We can use integration to solve a variety of problems Antiderivatives Definition A function Fis an antiderivative of fon interval Iif F39x fxfor all Xin I The process of finding an antiderivative is called antidifferentiation or finding an indefinite integral Example 1 Determine if Fis an antiderivative of fif Fxx3 x2 2x5 and fx x2 3x2 7 F39Cx3 gtlt 3x2 F Cgtlt31 Dlt3 J fhz ankqur lvao 106 ls va M1314 Lesson 16 2 Example 2 Suppose Hx x3 10 and Kx x3 27 If fx 3x2 show that each of H and Kis an antiderivative of 7 and draw a conclusion Cx33x K Cx353x HCKW and KCx dl er 0x constant In 2397 Notation We will use the integral sign I to indicate integration antidifferentiation Problems will be written in the form jfx dx Fx C This indicates that the indefinite integral of fx with respect to the variable Xis FxC where Fx is an antiderivative of f Basic Rules Rule 1 The Indefinite Integral ofa Constant jkdxkxC Example3 dex 5XC M1314 Lesson 16 Rule 2 The Power Rule n1 Ixquot dx x n1 C n a 1 r4l 5 Example 4 1x4 dx 1 C 35 C 4 or gtltSC 12 Example5 j dx quot j X ampx 39 I X S 1 1 5 x4 Example6 j sdx quot39 d I 7 Z x2 M1314 Lesson 16 4 Rule 3 The Indefinite Integral of a Constant Multiple of a Function jcfxdx cffxdx Example 7 14x3dx 3 31 15de lt W q q 7L4 C1XCJ J Am zjxdx22ltqc Example 8 i4dx x 3 3 2 L C 2x C 3 3 Z 3 C 3x Rule 4 The Sum Difference Rule ifX i 9xdx ifXdxi19XdX ijidx 35de331dlx L 2353135 4 gt C Example 9 j2x2 5x 1dx M1314 Lesson 16 5 Rule 5 The Indefinite Integral of the Exponential Function jexdx eX C Example 10 5eX 4x3dx Sex q x SZK XqC Rule 6 The Indefinite Integral of the Function fx jldxlnxc x 0 x Xlll X0 d 4 A uh e me d dx 5gtlt fix tm o E i Hg 393gtogtlt Example 11 3x xx3dx K3ng 3JFZX Ii 3 T 321 S rX1 4 Zj h 39Ic 1 3 I m z gtltL5nlgtlt Zx C 2 z 3gtlt739Snx 4 C39 Z X M1314 Lesson 16 6 Applying the Rules 2 3 z 393 50 5 he M 3 t 3C3W szdx BX ggf39 93 C 3 3gtlt z a gXVC Exam le 13 x2 3 l ijdx Z 4quot 1 ix P I x2 x3 lt X X X 17x q nxC x1 7X q nx C Example 14 wx 3 7exdx 3 LX33Z39 76X kgvdx 5ltx 4na gtdx i75gtlt lzlhXC 34H C 4 76x 42 h gtlt 54le 2 nnxx c M1314 Lesson 16 7 Differential Equations A quotquot 39 39 I t39 is an t39 that involves the derivative or differential of some function So if we write f39 x 3x 5 we have a differential equation We will be interested in solving these A solution of a differential equation is any function that satisfies the differential equation So for the example above fx gxz 5x 3 is a solution of the differential equation since the derivative of f is 3x 5 The general solution of a differential equation is one which gives all of the solutions so the general solution for the example above will be fxx2 5xC If we are given a point that lies on the function we can find a particular solution that is we can find C If we know that f 2 1 we can substitute this information into our general solution and solve forC 27 itHal Qo d lt b10 Mm 32 13 567 K 10 67 L C5 A x1 5xs 51va 739 Porticm om SON310quot f 2 1 is called an initial condition M1314 Lesson 16 Initial Value Problems An initial value problem is a differential equation together with one or more initial conditions If we are given this information we can find the function f by first finding the general solution and then finding the value of Cthat satisfies the initial condition Example 15 Solve the initial value problem f39 x 2x 5 f2 3 Z 5Q2xr5393dx 5 ZgtltZ sx C X SgtltC 7 3 411 393 szi C23C 3 quotY CZ q 10 0 ac 43470 XL SXq Example 16 Solve the initial value problem f39x 3eX 2x f0 7 7 K 2 jCBCX39MBdX Be gt 232 xm 2 420quot Ha 36 1C7 C 7 3 C 4 510 3e gtlt1L M1314 Lesson 16 9 Example 17 A cable television provider estimates that the number of 3 its subscribers will grow at the rate of 100 210t4 new subscribers per month tmonths from the start date of the service Suppose 5000 subscribers signed up for the service before the start date How many subscribers will there be 16 months after the start date 1C Ce7IooZ t lo SJQOO 351 dk 4 Zrotkic L qo Z MQ gt quot loo j C 3000 39K 0000 1420 0 C03 I Q 000 7 90 om r lzof c39Socgtltgt 7 8 0 4002 001La3 ZQCIQD loo ZlQaO From this section you should be able to Explain what we mean by an antiderivalive inde nite integral a differential equation and an inilial value problem Determine if one function is an antiderivative of another function Use lhe basic rules to nd antiderivatives Simplify if necessary before applying the basic rules Solve inilial value problems M1314 lesson 22 1 Math 1314 Lesson 22 Functions of Several Variables So far we have looked at functions of a single variable In this section we will consider functions of more than one variable You are already familiar with some examples of these Pxy 2x 2y AP it P1 it These formulas are functions of several variables We have just never called them that before We will for the most part limit our discussion to functions of two variables Functions of Two Variables Definition A real valued function of two variables f consists of a set A of ordered pairs of real numbers x y called the domain of the function and a rule that associates with each ordered pair in the domain of f one and only one real numb r denoted by zfxv to F x z D orrvl You will need to learn several skills using functions of several variables 1 Evaluating a function of several variables at a given point Example 1 Suppose fx y 3x2y 4xy6 Compute f00f2 1andf 1 3 x 0 Z 39Flt0039C 30ICO39qltOGgtDp 0097 2 l sczv CD qcnc map 41C 3 lz l Eg39HD ZZ CZl C hs 300163 C OGBEH C quot9 lz lS C 339l39 M1314 lesson 22 2 Example 2 The volume of a cylindrical tank with radius r and height h is given by the formula Find the volume of a tank with m and m feet liw rm m miczca zw 3 13 Lama39s it Example 3 Suppose you borrow money to buy a house You after 20 years of payments M1314 lesson 22 3 2 Find the domain of a function of several variables We ll have several kinds of situations that are similar to finding domain of functions of a single variable Problems arise when we have rational functions functions involving radicals and logarithmic functions Example 4 Find the domain of the function fx y 2x2 3y2 Polbn omlal a Domain is all Y Qonl humb evs 3x Example 5 Find the domain of the function fx y 2x 5y ampX53O 01x 5 E s X Tla Or Xig l lS dQl m gol gor all V20 hUxmers QXQQPJQ WMQV ngt oY X2713 lesson 22 2501 M1314 Example 6 Find the domain of the function fx y 116 x2 y2 lLo39XL22 O W 2 XZDZ Or NFL 326119 T141 domolh 04 ls c1 Chch oq radius 4 OEQ39EQOl ark OH in Cindx l cluxclsef all the Bdcses Graphing functions in space is quite difficult You will not need to do this Here are a couple of examples of graphs of functions of two variables 95 M I o 39W l I Wu W as O IIII wwwma Q 3 Hopefully you ll recognize how difficult it would be to graph these manually We usually graph function of two variables using a computer M1314 lesson 22 Here s an example of graphing just one point in space Example 7 Graph the point 2 3 5 E From this section you should be able to Evaluate a function of several variables Find the domain of a function of several variables M1314 Lesson 4 1 Math 1314 Lesson 4 Basic Rules of Differentiation We can use the limit de nition of the derivative to nd the derivative of every function but it isn t always convenient Fortunately there are some rules for nding derivatives which will make this easier First a bit of notation fx is a notation that means the derivative of f with respect to x evaluated at x Rule 1 The Derivative of a Constant c 0 where c is a constant Example 1 If fx l7 nd f39x Rule 2 The Power Rule xquot me for any real number 71 Example 2 If fx x5 nd f39x Example 3 If fx 6 nd f39x M1314 Lesson 4 Example 4 If fxi nd f39x 3 a x Rule 3 Derivative of a Constant Multiple of a Function d d 7 cfx 0 fx where c 15 any real number dx dx Example 5 If fx 3xquot nd f39x Example 6 If fx 5 nd f39x Rule 4 The SumDifference Rule immigmlfxigx Example 7 Find the derivative fx 4x3 2x2 4 x M 13 14 Lesson 4 Rule 5 The Derivative of the Exponential Function 61 x x 7 e e Example 8 Find the derivative fx J 4x3 2 66 Rule 6 The Derivative of an Exponential Function base is not 2 d 7 a In a r a I i l Example 9 Find the derivative f x 4x Rule 7 The Derivative of the Logaritlm c Function id 111 i x i 1 provided x 7 0 dx x Example 10 Find the derivative fx 5x 2 6 1nx M1314 Lesson 4 From this lesson you should be able to State the basic rules for nding derivatives Select the appropriate rule to use for a given problem Find the derivative of a function using the basic rules M1314 Lesson 20 1 Math 1314 Lesson 20 Evaluating Definite Integrals We will sometimes need these properties when computing definite integrals Properties of Definite Integrals Suppose fand gare integrable functions Then 1 f fxdx 0 2 y fxdx fxdx 3 f2cfxdx c fxdx 4 I2fx gx1dx J2fxdx 139xdx 5 f2fxdx I fxdx 13fxdx where a lt c lt b We will need to use substitution to evaluate some problems a 3 a g c Example 1 Evaluate f4xx2 3fdx 2 3 U olu t 2 in 0 13 Lquot 3 h 1 91 1 9 U x33 Xo u 013r 3 Z tp bj ONZXOX x3 qC3gt 3 0 isssz ZVL3 qu lxax 515 Lesson20 2 l 3 e 30 M1314 Example 2 Evaluate x2eX3dx 39 jlzudu 3 0 0 2X 7 Xo utCo3O V5 3quot Co luv 393de x UCamp31 gc Y3 du dex 91 2 3 3m x dx X450 75 0 X muqq 2 Example 3 Evaluate 3 13x 6 3 M3X 4 2m ucac sc 5 du0lx o x gtltz q3lt0ltn30 a dKL x1cgtlgtlt M1314 Lesson 20 3 Applications Example 4 A company purchases a new machine for which the rate of depreciation is given by 10000t 6 How much value is lost over the first three years that the machine is in use C D OJQOO QUCOOIOOQ 2 3 2 JO IoQQO E39CDOOOOgt039 LQLOooot10 2 5000 331 00300 C3 0 q51 WANG 893000 Example 5 Suppose you are driving a car and that your velocity can be approximated by vt 2tV25 t2 where tis measured in seconds and vis measured in feet per second How far will you travel in the 5 seconds from t 0 to t 5 O 0 y H 92 3 21 ZS 1 at t 1 vi mid LEE Eylif z U 32539 u zsweZ Jco uZS cogtzs OM ztoH39 t S M3 2r s7 o L ON Z E ot z3 C CD36 Fax 1 y3 Lo 23 zgo 33333 ll ii I l M1314 Lesson 20 4 Example 6 The marginal daily profit function associated with production and sales of a video game is estimated to be P39x 0003x2 04x 17 where xis the number of units produced and sold daily and P39x is measured in dollars per unit Find the additional daily profit realizable if production and sales is increased from 200 units per day to 300 units per day 3 2 30 7 J 6000 3x w UH YBdX 333quot 4 39C yzx W 200 3 I m 3 7X13 0 T om1300302C30 gt Wang 1000M J ozx 20 2 oLCZOCD 7C39ZGOB 5 300 h a oool The Average Value of a Function We can use the definite integral to find the average value of a function Suppose fis an integrable function on the interval a b Then 1 a j fxdx This the average value of fover the interval is b is what average value represents FOO V k men13 l i I q 0 M1314 Lesson 20 Example 7 Find the average value of fx J over the interval 1 16 K a 30 de j Ha I V2 L lt 753 d 39 33 72 5 H 3 It a x2 rLIE zj 39 S 3 l qs 391 3 5 c 2 Casi D Z 0 22 T iii S 9 gt 93 is Example 8 Find the average value of fx x2 3x Son 2 5 rl l l g 7 S 1 1 x33x 5x K Elijox 393XSdX 63 quot2 L 3 21 291 scs as 2 sltzgt 33 2293 7A M1314 Lesson 20 6 Example 9 The sales of ABC Company in the first tyears of its operation is approximated by the function St tw02t2 4 where St is measured in millions of dollars What were the over itsmyears of 5M in firm company s 7 5 operation a J 5 0 Vb Jst CzeZl ly ok t 35 q M oi V 5 l o o 2 Sq Ul zolul Z 14 X20 u x20gt q q lf gtlt5 q 2cs1qq t CH duquot q L q oi fdt J AgZ1 7 dqampchE Z 2 4 ZZC3 lt4 393 3 V3 D 1 q 21 million 1 Lo 3 e qdqrobl qggt over the S w 5 DQQIS From this lesson you should be able to Use 1e C to compute de nite integrals including problems that require substitution De ne average value Use 1e FI39OC to nd average value Solve word problems using FI39OC M1314 Lesson 12 1 Math 1314 Lesson 12 Curve Sketching One of our objectives in this part of the course is to be able to graph functions In this lesson we ll add to some tools we already have to be able to sketch an accurate graph of each function From prerequisite material we can nd the domain y intercept and end behavior of the graph of a function and from the last two sections we can learn much about a function by analyzing the first and second derivatives We also know how to find the zeros of some functions We ll expand that group of function before we continue to curve sketching The Rational Zeros of a Polynomial Function The rational zeros of a function are the zeros of the function that can be written as a fraction such 1 as 2 or Somet1mes we can find the ratlonal zeros of a functlon by factor1ng Example 1 Find the rational zeros fx x3 16x Example 2 Find the rational zeros fx x4 11x2 18 M1314 Lesson 12 2 Note that zeros that are square roots are NOT rational roots Imaginary solutions to the equation fx 0 such as i 3139 are NOT rational roots Sometimes we won t be able to factor the function Then we ll need another method We ll use a theorem called the Rational Zeros Theorem First we ll nd all of the possible rational zeros of a given function using the Rational Zeros Theorem Then we can use a calculator or synthetic diVision to determine which 7 if any 7 of the possible rational zeros are actually zeros of the function Here s the theorem Rational Zeros Theorem Suppose fx anxquot a x 1 a0 where an 72 0 and a0 72 0 and all ofthe coefficients of the polynomial are integers If x E is a rational zero of the function where p and q have no common factors then p is a q factor of the constant term a0 and q is a factor of the leading coefficient an Example 3 Find the possible rational zeros of fx 2x3 3x2 8x 4 Example 4 Find all rational zeros of f x x3 6x2 3x 10 or state that there are none M1314 Lesson 12 3 Example 5 Find all rational zeros of f x x3 7x2 l6x 12 or state that there are none Example 6 Find all rational zeros of f x x4 4x3 3x2 4x 4or state that there are none M1314 Lesson 12 Example 7 Find all rational zeros of f x x3 3x2 l or state that there are none Example 8 Find all rational zeros of f x x4 5x3 4x2 or state that there are none M1314 Lesson 12 5 Example 9 Find all rational zeros of fx x3 4x2 4x 16 or state that there are none Curve Sketching Now we ll turn our attention to graphing functions You will need to be able to use the following guide to sketch the graphs of functions A Guide to Curve Sketching 1 Determine the domain off 2 Find the rational x intercepts and y intercept of the inction If there are no rational x intercepts sa so 3 Determine the end behavior of the function 4 For an exponential inction determine any horizontal asymptotes 5 Determine where the inction is increasing and where it is decreasing 6 Find the x andy coordinates of any relative extrema 7 Determine where the inction is concave upward and where it is concave downward 8 Find the x andy coordinates of any points of in ection 9 If necessary plot a few additional points to determine the shape ofthe graph 10 Sketch the function Recall the generalizations about end behavior of a polynomial inction from College Algebra Evendegree polynomials look like Odddegree polynomials look like yix2 yix3 15 775 yx3 POLH y7x3 NOHL M1314 Lesson 12 Example 10 Use the guide to curve sketching to sketch f x x4 4x3 M1314 Lesson 12 7 Sometimes a function has some zeros that are not rational We may occasionally give you the approximate zeros of the function and ask you to complete the rest of the guide to curve sketching Example 11 Use the guide to curve sketching to sketch fx x3 6x2 15x 3 Note the approximate zeros of the function are 022 172 and 7 94 M1314 Lesson 12 8 Example 12 Use the guide to curve sketching to sketch fx x3 7x2 16x 12 Note we found the rational zeros in example 5 M1314 Lesson 12 Example 13 Use the guide to curve sketching to sketch fx x3 8x2 19x 12 M1314 Lesson 12 Example 14 Use the guide to curve sketching to sketch f x xe M1314 Lesson 12 11 For the next two problems you are given all of the information listed in the guide to curve sketching You just need to use it to graph the function Example 15 Sketch the function if you are given the following information X concave concave M1314 Lesson 12 Example 16 Sketch the function if you are given the following information X concave concave M1314 Lesson 12 13 Example 17 Here is the graph of a polynomial function Which of the statements below isare true The function has three zeros The graph of the function is increasing on one interval and decreasing on two intervals The graph of the function has one relative maximum and one relative minimum The graph of the function has two in ection points The function could be a quartic function 4 11 degree with a positive leading coefficient V eP Nf From this section you should be able to Find any rational zeros of a 3ml or 4th degree polynomial Use the guide to curve sketching to sketch the graph of a polynomial or exponential Sketch a graph of a function given all of the information from the guide to curve sketching Answer questions about the graph of a function given the graph of the function M1314 Lesson 1 1 Math 1314 Lesson 1 Limits What is calculus The body of mathematics that we call calculus resulted from the investigation of two basic questions by mathematicians in the 18th century 1 How can we find the line tangent to a curve at a given point on the curve 6 1305 vbva Slope b 2 nesatiuq Slope arber in L3 no slope line 7 W I 3 e fh 1 M1314 Lesson 1 2 2 How can we find the area of a region bounded by an arbitrary curve The investigation of each of these questions relies on the process of finding a limit so we39ll start by informally defining a limit and follow that by learning techniques for finding limits Limits Finding a limit amounts to answering the following question What is happening to the ywalue of a function as the x value approaches a specific target number If the ywalue is approaching a specific number then we can state the limit of the function as x gets close to the target number Rx l 1 O L x 2 1 225 W 29369 M1314 Lesson 1 3 Example 1 907 FGO hm J PO 3 O jgj X 57 S 23921 31M 4 gt0 1 4 32 x 1 2 f2 Umf5026 39 4984f54 431091 2 3 2 g 539 7 9w X o gm FltX7lt l Xe It does not matter whether or not the xvalue every reaches the target number It might or it might not Example 2 hm 390 0 XAI Z quotma 3872 X o 1 m 365 0 X 7 z 300 M1314 Lesson 1 When can a limit fail to exist We will look at two cases where a limit fails to exist note there are more but some are beyond the scope of this course Case 1 The yvalue approaches one number from numbers smaller than the target number and it approaches a second number from numbers larger than the target number PgtO 39 5 l l l q i linoFIDO does no C t i m emb DME X7 Case 2 At the target number for the xvalue the graph of the function has a vertical asymptote i iim 35070449 I X wL l l 79 72 77 76 is 74 7 72 71 4557291 390 M1314 Lesson 1 5 For either of these two cases we would say that the limit as xapproaches the target number does not existquot De nition We say that a function fhas limit L as xapproaches the target number a written hm 5 gt0 Z X quot3 q if the value x can be made as close to the number L as we please by taking xsufficiently close to but not equal to a lim fx L Note that L is a single real number Evaluating Limits There are several methods for evaluating limits We will discuss these three 1 substituting 2 factoring and reducing 3 nding limits at in nity To use the first two of these methods we will need to apply several properties of limits M1314 Lesson 1 Properties of limits Suppose lim fx L and lim gx M Then x gta x a DWNll 5 lIm linfxr ling fxr Lr for any real number r cfx c fx cL for any real number c linfx i gx fx i gx L i M Ef glfx9x fXi39X V39 x I f x a d dM o xalxia pl OVIe We39ll use these properties to evaluate limits Substitution Example 3 Evaluate lim3x2 4x 5 x gt2 3 21 Cz7 s 1 3 5 15 M1314 Lesson 1 7 Example 4 Evaluate lim 2x 3 x gt0 x 1 dwCQZB o S UnderstomA w 3 Example 5 Evaluate lim3xE x 4 3 147 3 C83 2 1 What do you do when substitution gives you a value in the form a where kis any nonzero real number 5 11 Example 6 Evaluate ling quot 2 lt33 x x 3 3 3 Linwt does noL Zyis t M1314 Lesson 1 8 Indeterminate Forms What do you do when substitution gives you the value 0 a I l is Ixaba This is called an indeterminate form It means that you are not done with the problem You must try another method for evaluating the limit See if you can factor the function If you can you may be able to reduce the fraction and then substitute 2 7 Example 7 Evaluate linlw liq LS x gt 9 X2 1 0quot O m um Lg P xi gt 2 1 Example 8 Evaluate ling wig x g x 2 Z 0 MM g aZCX Q Hm gt93 Z 3 5572 xix xvaz M1314 Lesson 1 9 So far we have looked at problems where the target number is a specific real number Sometimes we are interested in finding out what happens to our function as xincreases or decreases without bound Limits at Infinity 2x2 x2 1 I happens to x as we let the value of xget larger and larger Example 9 Consider the function fx What x 10 50 100 1000 10000 100000 1000000 10000000 l l 3l lq l l l mm N max A Z Z We say that a function x has the limit L as x increases without bound or as xapproaches infinity written lim fx Lif x can be made arbitrarily close X93 to L by taking xlarge enough We say that a function x has the limit L as x decreases without bound or as xapproaches negative infinity written lim fx L if x can be made X oo arbitrarily close to L by taking x to be negative and suf ciently large in absolute value We can also find a limit at in nity by looking at the graph of a function M1314 Lessonl 10 Example 10 Evaluate lim 2x 7 2 X gtw x waximam SmmqmuemN We can also find limits at infinity algebraically or by recognizing the end behavior of a polynomial function Example 11 Evaluate lim4x3 7x5 00 a X n 9 oo 1 qua397x igvt39oo L Xeoo M1314 Lessonl 11 Limits at infinity problems often involve rational expressions fractions The technique we can use to evaluate limits at infinity is to divide every term in the numerator and the denominator of the rational expression by xquot where n is the highest power of x present in the denominator of the expression Then we can apply this theorem Theorem Suppose n gt 0 Then min 0 and X gtw x lim in 0 provided inis de ned x p oo x x After applying this limit we can determine what the answer should be YOU MUST KNOW THIS PROCEDURE 2x2 5x1 m Example 12 Evaluate Ii 2 Xm 3x 2x 7 ampLSX 37 TT yja 20410 Z I X7 XLM fi l 30 3 Xquot XL X2 DixEde 93 highPSL VOPICKBR pomp m the dqnominator M1314 Lessonl 12 Often students prefer to just learn some rules for finding limits at infinity The highest power of the variable in a polynomial is called the degree of the polynomial We can compare the degree of the numerator with the degree of the denominator and come up with some generalizations o If the degree of the numerator is smaller than the degree of the denominator then limm 0 3L l 0 H 90 x 7 x3 o If the degree of the numerator is the same as the degree of the denominator then you can find lim m by making a fraction from the leading H 9x coefficients of the numerator and denomigator and then reducing to lowest terms M OM 5 2 1 x eoo le 3Jr 1 F Z If the degree of the numerator is larger than the degree of the denominator then it39s best to work the problem by dividing each term by the highest power of xin the denominator and simplifying You can then decide if the function approaches no or uo depending on the relaative powers and the coefficients m le 5 quot klx 4 AL 3o ZXL M x 2K 2 The notation lim fx no indicates that as the value f Z X 1 xincreases the value of the function increases without a bound This limit does not exist but the co notation is more descrIptIve so we WIII use It M L X ea x aso M1314 Lessonl 13 4 Example 13 Evaluate limw q X m x x1 LXQ L 53 W M 1xzzoo 2 5 1 0 X gtoo 323 x2 X1 I O 5x23x4 Example 14 Evaluate m Ii i X m 4x2 2x 8 K im 4x5 0 Example 15 Evaluate I 2 X m x 9x 9 From this lesson you should be able to State the two basic problems of the calculus De ne limit indeterminate form Find a limit as xapproaches a target number from a graph of f State when a limit fails to exist Evaluate limits where substitution gives k at 0 Evaluate limits by substitution or by factoring Evaluate limits at infinity M1314 lesson 3 1 Math 1314 Lesson 3 The Derivative The Limit De nition of the Derivative We now address the rst of the two questions of calculus the tangent line question We are interested in nding the slope of the tangent line at a speci c point W I I We could attempt to answer this question by graphing the function and its tangent line at the point of interest However with many functions we d get an approximation at best We need a way to nd the slope of the tangent line analytically for every problem that will be exact every time We can draw a secant line across the curve then take the coordinates of the two points on the curve P and Q and use the slope formula to approximate the slope of the tangent line Consider this function lesson 3 M1314 ow suppose we move point Q closer to point P When we do this we ll get a better approximation of the slope of the tangent 1ne Suppose we move point Q even closer to point P We get an even better approximation We are letting the distance betweenP and Q get smaller and smaller What does this sound like M1314 lesson 3 3 Now let s give these two points names We ll express them as ordered pairs Am Now we ll apply the slope formula to these two points This expression is called a difference quotient The last thing that we want to do is to let the distance betweenP and Q get arbitrarily small so we ll take a limit This gives us the definition of the slope of the tangent line Definition The slope of the tangent line to the graph of fat the point Px f x is given by hm fx hgt7 fx hgt0 1 provided the limit exists The difference quotient gives us the average rate of change We find the instantaneous rate of change when we take the limit of the difference quotient The derivative of f with respect to x is the function f 39 read fprime defined by f39x The domain of f39x is the set of all x for which the limit exists M1314 lesson 3 4 We can use the derivative of a function to solve many types of problems Once we have completed our study of the derivative we will be able to solve problems such as these Core Problems Core Problem 2 A ball is thrown straight up from the top of a building that is 64 feet high with an initial velocity of 96 feet per second so that its height in feet after t seconds is given by st l6t2 96t 64 1 Find the average velocity of the ball over the time intervals 2 3 2 25 2 21 2 Find the instantaneous velocity of the ball when t 2 3 Compare your results for questions 1 and 2 4 When does the ball hit the ground What is its velocity at the moment when its hits the ground Core Problem 3 The oxygen content of a pond t days after organic waste has been dumped into t2 4t 4 W graph of the function and interpret your results the pond is given by f t 100 J t gt 0 percent of the normal level Sketch the Core Problem 4 A tennis racket manufacturer finds that the total daily cost of manufacturing x rackets per day is given by the function C x 400 4x 00001x2 The rackets can be sold at a price of p dollars where p 10 00004x Assume that all rackets that are manufactured can be sold Find the daily level of production that will yield a maximum profit The Four Step Process for Finding the Derivative Now that we know what the derivative is we need to be able to find the derivative of a function We ll use an algebraic process to do so We ll use a FourStep Process to find the derivative The steps are as follows 1 Find fxh 2 Find fx h fx 3 Form the difference quotient fx h fx h 4 Find the limit of the difference quotient as h gets close to 0 lim hgt0 fxhfx h v fxh fx Sofx lg11 h M1314 lesson 3 Example 1 Find the rule for the slope of the tangent line for the function f x 3x 2 Step 1 Step 2 Step 3 Step 4 Example 2 Use the fourstep process to nd the derivative of f x x2 Step 1 Step 2 Step 3 Step 4 M1314 lesson 3 Example 3 Find the derivative of f x x Example 4 Find the derivative if f x 4x2 2x 3 M1314 lesson 3 7 Example 5 Suppose f x x2 2x 5 Find the average rate of change of f over the interval 2 5 We can now answer Core Problem 2 Problem 2 A ball is thrown straight up from the top of a building that is 64 feet high with an initial velocity of 96 feet per second so that its height in feet after tseconds is given by st l6t2 96t 64 1 Find the average velocity of the ball over the time intervals 2 3 2 25 2 21 2 Find the instantaneous velocity of the ball when t 2 3 Compare your results for questions 1 and 2 4 When does the ball hit the ground What is its velocity at the moment when its hits the ground M1314 lesson 3 From this lesson you should be able to Explain what a derivative is State the limit de nition of the derivative Use the fourstep process to nd the derivative of a polynomial function Find the average rate of change M1314 Lesson 3 1 Math 1314 Lesson 3 The Derivative The Limit Definition of the Derivative We now address the first of the two questions of calculus the tangent line question We are interested in finding the slope of the tangent line at a specific point 1 to quotL ling Tt tanger quot 79 quotT1 POSI39E VV slop V T1 95W 5 L Tl We need a way to find the slope of the tangent line analytically for every problem that will be exact every time We can draw a secant line across the curve then take the coordinates of the two points on the curve Pand Q and use the slope formula to approximate the slope of the tangent line M1314 Lesson 3 2 Consider this function Sac 1 Id i 1 i c Q Now suppose we move point choser to point P When we do this we ll get a better approximation of the slope of the tangent line reenaner Sine M1314 Lesson 3 3 Suppose we move point Qeven closer to point P We get an even better approximation We are letting the distance between Pand Qget smaller and smaller What does thIs sound lIke Pand QH 3 dog 0 Une a PVOXHMGJCIOH P QI 0 04 the Q1 Don 201 SW P5 3 Now let s give these two points names We ll express them as ordered pairs M1314 Lesson 3 4 Now we39ll apply the slope formula to these two points 43xh7 43 x7 Cgtlt HO 39 00 Kirk gtlt h This expression is called a difference quotient The last thing that we want to do is to let the distance between Pand Qget arbitrarily small so we39ll take a limit This gives us the de nition of the slope of the tangent line Definition The slope of the tangent line to the graph of fat the point Px fxis given by m fx h fx h gt0 h provided the limit exists The difference quotient gives us the average rate of change We find the instantaneous rate of change when we take the limit of the difference quotient The derivative of fwith respect to xis the function f39 read fprimequot defined by f39xll1ingw The domain of f39 x is the set of all xfor which the limit exists We can use the derivative of a function to solve many types of problems But first we need a method for finding the derivative M1314 Lesson 3 5 The FourStep Process for Finding the Derivative Now that we know what the derivative is we need to be able to find the derivative of a function We39ll use an algebraic process to do so We39ll use a FourStep Process to find the derivative The steps are as follows 1 Find fxh 2 Find fx h fx 3 Form the difference quotient 4 Find the limit of the difference quotient as h gets close to 0 MM h gt0 h So f39x1imw Example 1 Find the rule for the slope of the tangent line for the function fx 3x 2 Step1 FCgtHh7 2 3gtltgtQ2 3x3k2 Step 2 le HD 4 Ex 37013 Vi M Z 3h Step3 ltX4 D CX 2 lb 3 h h HKD X7 Step4 hm 3 ha M1314 Lesson 3 6 Example 2 Use the fourstep process to find the derivative of fx x2 Step 1 xw CxmZ CxHOCxHQ XLi xldiX lo hz gtltZ th hz HH i in IH l a Rslth79ltgtlt33 Step2 th 16 h M 7 MC h Step3 Qxmltgx3 gtlt4Hn 2x h Axln Step4 m JUN h 391 oZX i O 2X h o M1314 Lesson 3 7 Example 3 Find the derivative of fx x J DJFCXHD Xh J L A hj Fgtlth 00 XH X quot XCXHQ gtltCgtlt10 h X quot Q h xCxHD in FCX4V5 FquotO aw L T W KQJm W 175 X GD lim wk 4 12 gtltgtlt X h a M1314 Lesson 3 8 Example 4 Find the derivative if fx 4x2 2x 3 Q quotg 3910 q OHS0739 2Cxh 3 5 Li XZ iZXYN rh27 1X Zh 3 H xhzxv MW ZxZh 3 QFgtlth3 FxgtM3xhK MZ z42hB WM axmcwizh 9ltgtlth7 ux 8xh ihZ zh 3x Lm Z h h 9 H m 464147 450 hm 8x LHn Z 8 x 9o 2 h o h M50 lt3 x2 o M1314 Lesson 3 Example 5 Suppose fx x2 2x5 Find them of fover the interval 2 5 mzafm Jech Rz 2 Si 5 gt77 5 2 39 3 2 l 435 53125345 zs Ioszo 03 27K 2C27S 439 st From this lesson you should be able to Explain what a derivative is State the limit definition of the derivative Use the fourstep process to find the derivative of a polynomial function Find the average rate of change

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