Section 0.3 Power and exponential functions

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1 Section 0.3 Power and eponential functions (5/6/07) Overview: As we will see in later chapters, man mathematical models use power functions = n and eponential functions =. The definitions and asic properties of these functions, which are studied in precalculus courses, are reviewed in this section. We will discuss applications of these functions in later chapters, eginning with Chapter, where we stud their derivatives. We also descrie here how formulas for functions can e modified to translate, reflect, epand, and contract their graphs, The section closes with notes on the histor of analtic geometr. Topics: Power functions = n and their graphs Vertical and horizontal translation Reflection, magnification, and contraction Eponential functions = and their graphs Laws of eponents Historical notes Power functions A power function is a function of the form = n, where is the variale and n is a constant. If n is a positive integer, then n equals the product of n s, as in the formula 3 =. If n is a positive fraction, p/q, then n = p/q is the qth root of the pth power of, which also equals the pth power of the qth root of. In the case of n = 3 5, for eample, we have n = 5/3 = 3 5 = [ 3 ] 5. If n is a negative integer or fraction, so that n = m with m a positive integer or fraction, then n equals / m, as in the formulas 3 = 3 = 5/3 = 5/3 = 3 5 = [ 3 ] 5. all. To have formulas and identities involving n appl with zero eponents, 0 is defined to e for If n is not an integer or a fraction, it is irrational and has an infinite decimal epansion, which is used to define n for positive. The irrational numer, for eample, has the decimal epansion = , and if we want to define 0, we let n =. e the numer otained taking onl one digit after the decimal point in the epansion of, let n =. e the numer otained taking two digits after the decimal point, and so forth. This gives us an infinite string of rational numers n, n,n 3,... that approaches. We sa that is the limit of the numers n, n,n 3,... The numers 0 n, 0 n, 0 n 3,... are defined ecause the eponents n, n,n 3,... are rational. And, just as the numers n,n, n 3,... approach their limit, the numers 0 n,0 n,0 n 3,... approach their limit, which is defined to e 0. The first seven of the numers 0 n j are calculated in the net question. We will discuss limits in Chapter.

2 p. (5/6/07) Section 0.3, Power and eponential functions Question The following tale gives the first seven of numers n, n,n 3,... that approach = Use a calculator or computer to complete the second row of include approimate decimal values of 0 n, 0 n, 0 n 3,...,0 n 7. j 3 n j n j. = The same procedure is used to define n for an irrational n and positive. We let n j denote the rational numer otained taking j digits after the decimal point in the decimal epansion of n. Then the numers n j approach n and the numers n j approach n. The epression n is defined unless it involves dividing zero, taking an even root of a negative numer, or taking an irrational power of a negative numer. Consequentl, = n is defined for all with three eceptions: it is not defined at 0 if n is negative; it is not defined for negative if n is a fraction with an even denominator, and it is not defined for negative is n is irrational. The graphs of = n Figures through show four curves = n with odd integers n. The curve = with n = in Figure is a line through the origin. The shapes of the other curves are analzed in the following questions. = FIGURE Question (a) Wh does = 3 in Figure pass through the origin and wh is it aove the -ais for > 0 and elow the -ais for < 0? () Use the formula 3 = to eplain wh = 3 is much closer to the -ais than = for nonzero ver close to 0 and is much farther from the -ais than = for ver far from 0. = 3 = = /3 = = FIGURE FIGURE 3 FIGURE Notice that the curve = 3 in Figure gets steeper as moves awa from 0.

3 Section 0.3, Power and eponential functions p. 3 (5/6/07) Question 3 Question (a) Wh does = /3 in Figure 3 pass through the origin and wh is it aove the -ais for > 0 and elow the -ais for < 0? () Use the formula = /3 /3 to eplain wh = is closer to the -ais than = /3 for < < 0 and 0 < < and is farther from the -ais than = /3 for < and >. Notice that the curve = /3 in Figure 3 gets less steep as moves awa from 0. Eplain wh = in Figure (a) does not intersect the -ais and is aove it for > 0 and elow it for < 0, () is far from the -ais for small positive and small negative, and (c) is close to the -ais for large positive and large negative. The reasoning in Questions through can e used to show that, for an constant n not equal to 0 or, the portion of = n for positive is similar to the portion of one of the curves in Figures through for positive. For n > (Figure 5), the curve = n, like = 3 in Figure, curves up to the right from the origin. For 0 < n < (Figure 6), the curve = n, like = /3 in Figure 3, rises up from the origin ut gets less steep as moves to the right. For n < 0 (Figure 7), the curve = n, like = in Figure, comes down to the right of the -ais and approaches the -ais as moves to the right. = n (n > ) = n (0 < n < ) = n (n < 0) FIGURE 5 FIGURE 6 FIGURE 7 The nature of the graph = n for < 0 depends on the value of n. There are three possiilities: = n is either an odd function, is an even function, or is not defined for negative. We are using here the following definition. Definition A function = f() is even if f( ) = f() for all in its domain and is odd if f( ) = f() for all in its domain. The graphs of odd functions are smmetric aout the origin; the graphs of even functions are smmetric aout the -ais. The functions = 3, = /3, and = of Figures through are odd ecause ( ) 3 = and ( ) /3 = /3 for all and ( ) = for 0. Their graphs are smmetric aout the origin. The functions =, = /3, and = of Figures 8 through 0 are even ecause ( ) = and ( ) /3 = /3 for all and ( ) = for 0. Their graphs are smmetric aout the -ais. The functions = 5/, = /, and = / of Figures through are not defined for negative ecause the involve the square root / =. Consequentl, their graphs do not etend to the left of the -ais.

4 p. (5/6/07) Section 0.3, Power and eponential functions 3 = = /3 3 = FIGURE 8 FIGURE 9 FIGURE 0 = 5/ 3 = / = / 3 FIGURE FIGURE FIGURE 3 Eample Match the functions (a) =, () = 5, and (c) = / to their graphs in Figures through 6. 3 FIGURE FIGURE 5 FIGURE 6 Solution (a) Here the power n = is a negative numer. Consequentl, = is similar to = for positive. The function = is even, so its graph is smmetric aout the -ais and is in Figure 5. () The power 5 in this case is greater than, so = 5 is similar to = 3 for 0. The function = 5 is odd, so its graph is smmetric aout the origin and is in Figure 6. (c) Because n = is a fraction etween 0 and, = / is similar to = /3 for 0, ut since n = has an even denominator, the function is not defined for 0, and its graph is in Figure.

5 Section 0.3, Power and eponential functions p. 5 (5/6/07) Vertical and horizontal translation If we add a positive constant to a function = f(), we otain the function = f()+k, whose graph is otained from the graph of f raising it k units. Sutracting a positive constant k ields = f() k, whose graph is otained lowering the graph of = f() k units (Figure 7). The raising or lowering of a graph is called vertical translation. = f() + k k k = f() = f() k = f( + k) = f() = f( k) k k FIGURE 7 FIGURE 8 If, on the other hand, we add a positive constant k to the variale in the formula = f(), we otain = f( + k), whose value at 0 k is f (( 0 k) + k) = f( 0 ), which is the value of = f() at 0. Consequentl, = f( + k) is otained shifting the graph of f to the left k units, as shown in Figure 8. This action is called horizontal translation. Sutracting a positive constant k from the variale gives = f( k), whose value at 0 + k is f (( 0 + k) k) = f( 0 ), which is the value of = f() at 0. Hence, = f( k) is the curve = f() shifted k units to the right, as is also shown in Figure 8. Eample Solution Sketch the graph of the function = + 3 completing the square. We complete the square in the formula = + 3 adding and sutracting the square of half the coefficient of. This gives = ( +) +3 or = ( ) +. Its graph in Figure 9 is the curve = translated up units and unit to the right. 6 = ( ) + FIGURE 9 3 Reflection Multipling a function = f() gives the function = f() whose value at is the negative of the value of = f(). Its graph is the mirror image of = f() relative to the -ais. Multipling the variale gives the function = f( ) whose value at is the value of = f() at. Its graph is the mirror image of = f() relative to the -ais (Figure 0). Read this paragraph carefull: = f( + k) is = f() shifted k units to the left, not to the right.

6 p. 6 (5/6/07) Section 0.3, Power and eponential functions = f( ) = f() = f() FIGURE 0 Question 5 Draw the graph of = f( ), where f is the function whose graph is in Figure 0. Eample 3 Solution Sketch the graphs of (a) = and () =. (a) The graph of = in Figure is the mirror image relative to the -ais of the curve = in Figure. () The graph of = in Figure is the mirror image relative to the -ais of =. = = 3 3 FIGURE FIGURE Magnification and contraction If we multipl a function = f() a constant k >, we otain the function = kf(), whose graph is otained from = f() multipling the -coordinate of ever point on it k. This magnifies the curve verticall, as shown the middle and upper curves in Figure 3. Similarl, if we divide the function k >, we otain = f(), whose graph is otained from = f() dividing the -coordinate of k each point k. This contracts the curve verticall, as shown the lower curve in Figure 3. (k > ) = kf() (k > ) = f(k) = f() = f() = k f() = f(/k) FIGURE 3 FIGURE

7 Section 0.3, Power and eponential functions p. 7 (5/6/07) Multipling the variale a constant k > contracts the graph horizontall since = f(k) for positive k has the same value f (k( 0 /k)) = f( 0 ) at 0 /k as = f() has at 0. Similarl, dividing the variale k > magnifies the graph horizontall (Figure ). Eample The curve drawn with a heav line in Figure 5 is the graph of = G(). Is the other curve the graph of = G(), = G( ), = G(), or = G()? = G() FIGURE Solution The second curve in Figure 5 has the equation = G() ecause it is otained magnifing = G() a factor of verticall and contracting it a factor of horizontall. Eponential functions An eponential function is a function of the form =, where the eponent is the variale and is a positive constant, called the ase. All eponential functions are defined and positive for all, and their graphs pass through the point (0, ) since 0 = for an positive. If =, then is the constant function = (Figure 6). If is greater than, then the graph approaches the -ais on the left and curves up on the right (Figure 7). If 0 < <, then the graph approaches the -ais on the right and curves up on the left (Figure 8). 6 = ( = ) = ( > ) 6 6 = (0 < < ) FIGURE 6 FIGURE 7 FIGURE 8 Notice that the function = with positive is neither even or odd; its graph is not smmetric aout the -ais nor aout the origin. Read this paragraph carefull: for k > 0, = f(k) is = f() contracted a factor of k, not epanded a factor of k.

8 p. 8 (5/6/07) Section 0.3, Power and eponential functions Eample 5 Draw the curve = 5 + 3( ). Solution The curve = 5 + 3( ) is = magnified verticall a factor of 3 and then translated up 5 units. It is drawn in Figure 9, where the values (0) = 5 + 3( 0 ) = = 8 and () = 5 + 3( ) = 5 + = 7 on it have een plotted. The curve has = 5 as a horizontal asmptote. = 5 + 3( ) = 5 FIGURE 9 Question 6 Draw the curve = 5 3( ). Eample 6 A sample of radioactive radium-66 has mass M(t) = 6 ( ) t/60 grams at time t (ears). (a) What is the mass of the sample at t = 0, t = 60, and t = 30 ears and how are these numers related? () Draw the graph of M = M(t) in a tm-plane. Solution (a) The formula M(t) = 6 ( ) t/60 gives M(0) = 6 ( ) 0 = 6, M(60) = 6 ( ) = 8, and M(30) = 6 ( ) =. Consequentl, M(60) is half of M(0) and M(30) is half of M(60). (These calculations illustrate the fact that the halflife of radium is 60 ears.) () The graph of M = M(t) in Figure 30 is otained plotting the values from part (a). 6 8 M (grams) M = 6 ( ) t/60 FIGURE t (ears)

9 Section 0.3, Power and eponential functions p. 9 (5/6/07) The natural eponential function As we will see later, the most useful eponential function in calculus is the natural eponential function = e, whose ase is an irrational numer e = that will e defined in Chapter. The graph of = e is shown in Figure 3. = e FIGURE 3 Laws of eponents The following rules for working with eponents are valid for an numers and if is positive. If is negative, the hold for all values of and such that all epressions involved are defined. = + () ( ) = () (c) = c (3) = () The advantage of using eponential notation and these rules is illustrated in the net eample. Eample 7 (a) Simplif the formula = 3 without using fractional or negative eponents taking the sith power of oth sides, simplifing, and taking the sith root. () Simplif = 3 using fractional and negative eponents and rules () through (). Solution (a) Taking sith powers of oth sides of = ( ) 6 6 = ( 3 ) 6 = 3 gives ( )( )( ) ( )( ) = =. Then taking sith roots ields = 6 since > 0. () Fractional eponents enale us to make a more direct calculation: = 3 = / /3 = / /3 = /6 = 6.

10 p. 0 (5/6/07) Section 0.3, Power and eponential functions Historical notes General algeraic equations were first studied in the siteenth centur, ut without the modern convention of using a single letter for the unknown. The Italian phsician and algeraist Gerolamo Cardano (50 57), for eample, used the Latin sentence, Cuus p 6 reus aequalis 0, for the equation that we would write = 0. In Cardano s sentence, the word cuus denotes the cue of the unknown, p stands for plus, reus denotes the unknown, and aequalis means equals. The use of a single letter for the unknown and of eponents for positive integer powers was popularized a treatise La Géométrie, written in 637 the French philosopher Réné Descartes ( ) as an appendi to a work on the philosoph of science. Réné Descartes Pierre Fermat ( ) (60 665) Descartes and a French lawer Pierre Fermat (60 665) are considered the inventors of analtic geometr as a tool for giving geometric meaning to aspects of algera and calculus. Greek mathematicians, including Euclid (ca. 300 BC), Archimedes (87 BC), and Apollonius (ca. 5 BC), used the equivalent of coordinate sstems with the theor of proportions for studing geometric figures, and algera was emploed in the siteenth centur to solve geometric prolems. It was Descartes and Fermat, however, who first studied curves defined equations as well as their geometric properties and who made etensive use of the association etween the algera of equations and the geometr of curves. Fermat s work was not pulished until after his death, more than fort ears after the pulication of Descartes La Géométrie, so Descartes often receives more credit for creating what now is known as analtic or cartesian geometr. Fermat and Decartes generall used onl positive coordinates. Negative coordinates were first used sstematicall Isaac Newton (6 77) in his Enumeration of Curves of Third Degree (676). Fermat s name has een in the news in recent ears ecause his famous last theorem, that n + n = z n has no nonzero integer solutions,, and z for integers n >, has finall een proved. Responses 0.3 Response The tale is completed elow. (Notice that all of the values of 0 n j egin with 5, the last five egin with 5.95, and the last two egin with This illustrates that the numers approach the infinite decimal 5.95 = 0.) j j j. =

11 Section 0.3, Power and eponential functions p. (5/6/07) Response (a) = 3 passes through the origin ecause 0 3 = 0, is aove the -ais for > 0 ecause 3 is positive for > 0, and is elow the -ais for < 0 ecause 3 is negative for < 0. () = 3 is much closer to the -ais than = for ver small nonzero and is much farther from the -ais than = for large positive or negative ecause 3 = equals multiplied a ver small positive numer for small nonzero and multiplied a large positive numer for large positive or negative. Response 3 Response Response 5 (a) = /3 passes through the origin ecause 0 /3 = 0 and is aove the -ais for positive and elow the -ais for negative ecause /3 is positive for positive and negative for negative. () = is closer to the -ais than = /3 for < < 0 and 0 < < and is farther from the -ais than = /3 for < and > ecause = /3 /3 equals /3 multiplied a positive numer less than if < < 0 or 0 < < and equals /3 multiplied a numer greater than if < or >. (a) = does not intersect the -ais ecause = / is not defined at = 0. () = is far from the -ais for ver small close to zero ecause = / is ver large when is ver small. (c) = is close to the -ais if is a large positive or negative numer ecause then = / is ver small. = f( ) is = f() reflected aout the -ais and aout the -ais. Figure R5 = f( ) 5 = 5 3( ) = 5 5 Figure R5 Figure R6 Response 6 = 5 3( ) is = 5 + 3( ) (Figure 9) reflected aout the - and -aes. Figure R6

12 p. (5/6/07) Section 0.3, Power and eponential functions Interactive Eamples 0.3 Interactive solutions are on the we page http// ashenk/.. Solve the equations (a) 3/ = 8 and () /3 = 6 for.. The curve in Figure 3 has the equation = a with constants a and. What are those constants? 8 6 FIGURE Do the two curves in Figure 33 have the equations = 5 + e, = 5 e, = 5 + e, or = 5 e? 5 0 FIGURE 33. Solve the equation 93 9 = 3 for using the fact that if = with a positive, then =. 5. (a) Determine the general shape of the curve = 3 / without generating it on a calculator or computer. () Sketch the curve plotting at least one point on it. 6. (a) Determine the general shape of the curve = + without generating it on a calculator or computer. () Sketch the curve plotting at least one point on it. In the pulished tet the interactive solutions of these eamples will e on an accompaning CD disk which can e run an computer rowser without using an internet connection.

13 Section 0.3, Power and eponential functions p. 3 (5/6/07) 7. Figure 3 shows the graph of a function = P() and the curves = P() and = P() +. Which curve is which? 3 FIGURE The curve drawn with a fine line in Figure 35 is the graph of = H(). (a) Is the other curve aove the -ais the graph of = H() or of = H(/)? () Give equations in terms of H for the two curves elow the -ais. = H() FIGURE 35 Eercises 0.3 A Answer provided. O Outline of solution provided. C Graphing calculator or computer required. CONCEPTS: C. Generate the curves = and = together in the window.5.5, and cop them on our paper. Then use the equation = ( ) to eplain wh = is elow = for some values of and aove it for others. C. Generate the curves = and = together in the window.5.5, and cop them on our paper. Then use the equation = ( ) to eplain wh = is elow = for some values of and aove it for others. 3. Derive the identit = + for a positive constant in the case of = and = 3 writing = and 3 =.. Derive the identit ( ) = for a positive constant in the case of = and = 3 writing ( ) 3 as ( ) ( ) ( ) and then writing for. 5. How can the function z = e either a power function or an eponential function?

14 p. (5/6/07) Section 0.3, Power and eponential functions BASICS: 6. Find all real solutions of (a) = k +, () = k 3, and (c) = k + 0. Here k is a positive constant. 7. Solve the following equations for recognizing powers of and 0 and using the fact that if = with positive, then =. (a) = 8 () = 8 (c) 0 = 0.00 (d) 0 0 = 00 (e) (0 ) = 00 (f) 6 = 6 8. The middle curve in Figure 36 is = 5 =. Which is which? + (/) 5 +. The other curves are = 5 + () and 6 FIGURE 36 In Eercises 9 through solve the equations for. 9. O 3 = 5 0. A 3. 5 = 6 = 0. ( ) 3 =. In Eercises 3 through 9 solve the equations for using the fact that if = with positive, then =. /3 3. O 3 = O 7 = (7 ) 0. 5 = 8 7. A 3 = 9(3 ) ( 5. A ) = 8. ( ) = = 7 What are the domains of the functions in Eeercises 0 through 3? 0. O = /3. A = / 3. = 3/. =

15 Section 0.3, Power and eponential functions p. 5 (5/6/07) Determine the general shapes of the curves in Eercises through 3 without generating them on our calculator or computer analzing their equations. Then sketch them plotting at least one point on each.. A = 5/ 5. = 0 + 5/ 6. O = 5 /( + ) 8. = + ( ) 3 9. O = 30. A = 0 e 7. A = + 3. = 0e /0 3. The lower curve in Figure 37 is = P() and the upper curve is = A + P(B) with constants A and B. What are A and B? 3 = A + P(B) = P() FIGURE Figure 38 shows the curves = L(+k) and = L( k) for a function = L() and a positive constant k. (a) Which is which and what is the value of k? () Draw the graph of = L(). FIGURE A According to Newton s law of gravit, an oject that weighs one pound on the surface of the earth weighs w = 6r pounds when it is r thousand miles from the center of the earth. (The radius of the earth is thousand miles.) Sketch the portion of w = 6r for r in an rw-plane. 35. A A o throws a all straight up in the air at time t = (seconds) and catches it at t =. Because there is no air resistance, the all is h = 6 6t feet aove his hand for t. Sketch the portion of h = 6 6t for t in a th-plane 36. A It costs a factor 5 dollars to manufacture each pint of a chemical, plus an overhead of 00 on each atch produced. The cost for a atch of pints is therefore dollars, and the average cost of a atch of pints is A = = dollars per pint. Draw the portion of A = for > 0 in an A-plane. 37. A culture of acteria contains 500 acteria initialll and, ecause the numer doules ever 3 das, the culture contains N = 500( t/3 ) acteria t das later. Draw the graph of this function in a tn-plane.

16 p. 6 (5/6/07) Section 0.3, Power and eponential functions 38. When air pressure is measured in atmospheres, the air pressure at the surface of the earth is atmosphere. At an altitude h < 80 kilometers aove the surface of the earth, the air pressure is P = ( ) h/5.8. Draw the graph of this function in an hp-plane. EXPLORATION: 0. O (5 )(5 ) = 5 3 for.. A Solve (5 )(5 ) = 5 for.. A The curve = () can e otained from = contracting it horizontall or magnifing it verticall. Eplain. 3. The curve = e + can e otained from = e horizonta translation or vertical magnification. Eplain.. A Which of the curves in Figures 39 through is the graph of = a + / with a > 0? Is positive or negative? Give our reasoning. FIGURE 39 FIGURE 0 FIGURE FIGURE FIGURE 3 FIGURE 5. Which of the curves in Figures 39 through is the graph of = a + 3 for constants a and? Is a positive or negative? Is positive or negative? Give our reasoning. 6. Which of the curves in Figures 39 through is the graph of = a + / for some constants a and? Is a positive or negative? Is positive or negative? Give our reasoning. 7. Which of the curves in Figure 5 has the equation = + a (a) with 0 < a <, () with + 0 < < a, and (c) with a < 0 and > 0? Give our reasoning.

17 Section 0.3, Power and eponential functions p. 7 (5/6/07) I II III FIGURE 5 C 8. Generate = + a on our calculator or computer in the window, 5, first for a = 0, and, and then for a = 0,, and. Eplain how changing a changes the graph and wh. C 9. How does changing change the curve = and wh? (Generate the curve for sample values of in the window, with -scale = 0.) 50. The curve = M() is shown in Figure 6. Draw the curves (a) = M(/), () = M(), (c) = M(), (d) = M()/, (e) = M()+, (f) = M( ), and (g) = M(/). 6 = M() FIGURE Figure 7 shows the graphs of =.5, = e, and = 6. (a) Which is the upper curve, which is the middle curve, and which is the lower curve for > 0? () Which is the upper curve, which is the middle curve, and which is the lower curve for < 0? 3 FIGURE 7 5. Find constants and C such that E() = C has the values in the following tale: 0 3 E()

18 p. 8 (5/6/07) Section 0.3, Power and eponential functions 53. Figure 8 shows the curve = with constant > and the mirror image of this curve aout the origin. What is the equation of the second curve? FIGURE 8 C 5. Generate the curves = and = in the window.5 6,.5 5 to see that the equation = has two positive solutions and one negative solution. Find the positive solutions trial and error and use a calculator or computer to find the approimate value of the negative solution. 55. Give a formula for the surface area A = A(V ) (square meters) of a cue as a function of its volume V (cuic meters) and draw the graph of this function. (End of Section 0.3)

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