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384 The common theme that links the functions of this chapter is that they occur as pairs of inverse functions. In particular, two of the most important functions that occur in mathematics and its applications are the exponential function and its inverse function, the logarithmic function. In this chapter we investigate their properties, compute their derivatives, and use them to describe exponential growth and decay in biology, physics, chemistry, and other sciences. We also study the inverses of trigonometric and hyperbolic functions. Finally, we look at a method (l’Hospital’s Rule) for computing difficult limits and apply it to sketching curves. There are two possible ways of defining the exponential and logarithmic functions and developing their properties and derivatives. One is to start with the exponential function (defined as in algebra or precalculus courses) and then define the logarithm as its inverse. That is the approach taken in Sections 7, 7, and 7 and is probably the most intuitive method. The other way is to start by defining the logarithm as an integral and then define the exponential function as its inverse. This approach is followed in Sections 7*, 7*, and 7* and, although it is less intuitive, many instructors prefer it because it is more rigorous and the properties follow more easily. You need only read one of these two approaches (whichever your instructor recommends). t!x"! loga x f !x"! a x The exponential and logarithmic functions are inverse functions of each other.
INVERSE FUNCTIONS:
EXPONENTIAL,
LOGARITHMIC, AND
INVERSE TRIGONOMETRIC
FUNCTIONS
7
INVERSE FUNCTIONS Table 1 gives data from an experiment in which a bacteria culture started with 100 bacte- ria in a limited nutrient medium; the size of the bacteria population was recorded at hourly intervals. The number of bacteria N is a function of the time t:. Suppose, however, that the biologist changes her point of view and becomes interested in the time required for the population to reach various levels. In other words, she is think- ing of t as a function of N. This function is called the inverse function of f, denoted by , and read “f inverse.” Thus is the time required for the population level to reach N. The values of can be found by reading Table 1 from right to left or by consulting Table 2. For instance, because Not all functions possess inverses. Let’s compare the functions and whose arrow diagrams are shown in Figure 1. Note that never takes on the same value twice (any two inputs in have different outputs), whereas does take on the same value twice (both 2 and 3 have the same output, 4). In symbols, but Functions that share this property with are called one-to-one functions. DEFINITION A function is called a one-to-one function if it never takes on the same value twice; that is, If a horizontal line intersects the graph of in more than one point, then we see from Figure 2 that there are numbers and such that. This means that is not one-to-one. Therefore we have the following geometric method for determining whether a function is one-to-one. HORIZONTAL LINE TEST A function is one-to-one if and only if no horizontal line intersects its graph more than once. x 1 x 2 f !x 1 "! f !x 2 " f f f !x 1 " " f !x 2 " whenever x 1 " x 2 1 f f f !x 1 " " f !x 2 " whenever x 1 " x 2 t! 2 "! t! 3 " A t f f t f! 1! 550 "! 6 f! 6 "! 550. f! 1 t! f! 1 !N" f! 1 N! f !t" 7. 385 TA B L E 2 t as a function of N N! time to reach N bacteria 100 0 168 1 259 2 358 3 445 4 509 5 550 6 573 7 586 8 t! f! 1 !N" TA B L E 1 N as a function of t t (hours)! population at time t 0 100 1 168 2 259 3 358 4 445 5 509 6 550 7 573 8 586 N! f !t" 4 3 2 1 10 4 2 A B g FIGURE 1 4 3 2 1 10 7 4 2 A B f f is one-to-one; g is not 0 fl ‡ y=ƒ FIGURE 2 This function is not one-to-one because f(⁄)=f(¤). y ⁄ ¤ x N In the language of inputs and outputs, this definition says that is one-to-one if each out- put corresponds to only one input. f
SOLUTION From the definition of we have The diagram in Figure 6 makes it clear how reverses the effect of in this case. M The letter is traditionally used as the independent variable, so when we concentrate on rather than on , we usually reverse the roles of and in Definition 2 and write By substituting for in Definition 2 and substituting for in (3), we get the following cancellation equations: The first cancellation equation says that if we start with , apply , and then apply we arrive back at , where we started (see the machine diagram in Figure 7). Thus undoes what does. The second equation says that undoes what does. For example, if , then and so the cancellation equations become These equations simply say that the cube function and the cube root function cancel each other when applied in succession. Now let’s see how to compute inverse functions. If we have a function and are able to solve this equation for in terms of , then according to Definition 2 we must have . If we want to call the independent variable x, we then interchange and and arrive at the equation y! f! 1 !x". x! f! 1 !y" x y x y y! f !x" f! f! 1 !x""! !x 1 % 3 " 3! x f! 1! f !x""! !x 3 " 1 % 3! x f !x"! x 3 f! 1 !x"! x 1 % 3 FIGURE 7 x f ƒ f –! x f f f! 1 x f! 1 x f f! 1 , f! f! 1 !x""! x for every x in B 4 f! 1! f !x""! x for every x in A y x 3 f! 1 !x"! y &? f !y"! x f! 1 f x y x FIGURE 6 The inverse function reverses inputs and outputs. B 5 7 _ f A 1 3 8 A 1 3 8 f –! B 5 7 _ f! 1 f f! 1 !! 10 "! 8 because f! 8 "!! 10 f! 1! 5 "! 1 because f! 1 "! 5 f! 1! 7 "! 3 because f! 3 "! 7 f! 1 SECTION 7 INVERSE FUNCTIONS | | | | 387
HOW TO FIND THE INVERSE FUNCTION OF A ONE-TO-ONE FUNCTION f STEP 1 Write. STEP 2 Solve this equation for in terms of (if possible). STEP 3 To express as a function of x, interchange and. The resulting equation is. EXAMPLE 4 Find the inverse function of. SOLUTION According to (5) we first write Then we solve this equation for : Finally, we interchange and : Therefore the inverse function is. M The principle of interchanging and to find the inverse function also gives us the method for obtaining the graph of from the graph of. Since if and only if , the point is on the graph of if and only if the point is on the graph of. But we get the point from by reflecting about the line. (See Figure 8.) Therefore, as illustrated by Figure 9: The graph of is obtained by reflecting the graph of about the line. EXAMPLE 5 Sketch the graphs of and its inverse function using the same coordinate axes. SOLUTION First we sketch the curve (the top half of the parabola , or ) and then we reflect about the line to get the graph of. (See Figure 10.) As a check on our graph, notice that the expression for is. So the graph of is the right half of the parabola y! !x 2! 1 and this seems reasonable from Figure 10. M f! 1 f! 1 !x"! !x 2! 1, x " 0 f! 1 f! 1 y 2!! 1! x x! !y 2! 1 y! x y! s! 1! x f !x"! s! 1! x f! 1 f y! x FIGURE 8 FIGURE 9 0 y x (b, a) (a, b) y=x 0 y x f –! y=x f f! 1 !b, a" !a, b" y! x f! 1 !b"! a !a, b" f !b, a" f! 1 f f !a"! b x y f! 1 !x"! s 3 x! 2 y! s 3 x! 2 x y x! s 3 y! 2 x 3! y! 2 x y! x 3 # 2 V f !x"! x 3 # 2 y! f! 1 !x" f! 1 x y x y y! f !x" 5 388 | | | | CHAPTER 7 INVERSE FUNCTIONS N In Example 4, notice how reverses the effect of. The function is the rule “Cube, then add 2”; is the rule “Subtract 2, then take the cube root.” f! 1 f f f! 1 0 y=x y=ƒ (0, _1) y=f –!(x) (_1, 0) FIGURE 10 y x
Replacing by the general number in the formula of Theorem 7, we get If we write , then , so Equation 8, when expressed in Leibniz notation, becomes If it is known in advance that is differentiable, then its derivative can be computed more easily than in the proof of Theorem 7 by using implicit differentiation. If , then. Differentiating the equation implicitly with respect to , remembering that is a function of , and using the Chain Rule, we get Therefore EXAMPLE 6 Although the function , , is not one-to-one and therefore does not have an inverse function, we can turn it into a one-to-one function by restricting its domain. For instance, the function , , is one-to-one (by the Horizon- tal Line Test) and has domain and range. (See Figure 12.) Thus has an inverse function with domain and range. Without computing a formula for we can still calculate. Since , we have. Also. So by Theorem 7 we have In this case it is easy to find explicitly. In fact, ,. [In general, we could use the method given by (5).] Then , so , which agrees with the preceding computation. The functions and are graphed in Figure 13. M EXAMPLE 7 If , find. SOLUTION Notice that is one-to-one because and so is increasing. To use Theorem 7 we need to know and we can find it by inspection: Therefore! f! 1 "$! 1 "! M 1 f $! f! 1! 1 "" ! 1 f $! 0 " ! 1 2! sin 0 ! 1 2 f! 0 "! 1? f! 1! 1 "! 0 f f! 1! 1 " f $!x"! 2! sin x ( 0 f V f !x"! 2 x # cos x! f! 1 "$! 1 " f f! 1
! f! 1 "$!x"! 1 %( 2 sx )! f! 1 "$! 1 "! 12
f! 1 f! 1 !x"! sx 0 ) x ) 4 ! f! 1 "$! 1 "! 1 f $! f! 1! 1 "" ! 1 f $! 1 " ! 1 2 f! 1 "! 1 f! 1! 1 "! 1 f $!x"! 2 x ! f! 1 "$! f! 1 "$! 1 " f! 1 #0, 4$ #0, 2$ #0, 2$ #0, 4$ f f !x"! x 20 ) x ) 2 y! x 2 x!! dy dx ! 1 f $!y" ! 1 dx dy f $!y" dy dx ! 1 x y x y! f! 1 !x" f !y"! x f !y"! x NOTE 2 f! 1 dy dx ! 1 dx dy y! f! 1 !x" f !y"! x ! f! 1 "$!x"! 1 f $! f! 1 !x"" 8 NOTE 1 a x 390 | | | | CHAPTER 7 INVERSE FUNCTIONS FIGURE 12 0 y (a) y=≈, x! R 0 2 x y (2, 4) (b) ƒ=≈, 0 ̄x ̄ 0 y x (2, 4) (4, 2) (1, 1) f FIGURE 13 f –!
SECTION 7 INVERSE FUNCTIONS | | | | 391 (c) What is the value of? (d) Estimate the value of. The formula , where , expresses the Celsius temperature C as a function of the Fahrenheit tem- perature F. Find a formula for the inverse function and interpret it. What is the domain of the inverse function? 22. In the theory of relativity, the mass of a particle with speed is where is the rest mass of the particle and is the speed of light in a vacuum. Find the inverse function of and explain its meaning. 23–28 Find a formula for the inverse of the function. 23. 25. 26. 27. 28. , ; 29–30 Find an explicit formula for and use it to graph , and the line on the same screen. To check your work, see whether the graphs of and are reflections about this line. 29. , 30. , 31–32 Use the given graph of to sketch the graph of. 31. 32. 33– (a) Show that is one-to-one. (b) Use Theorem 7 to find. (c) Calculate f! 1 !x"and state the domain and range of f! 1. ! f! 1 "$!a" f y 0 2 x 1 y 0 1 x 1 f f! 1 f !x"! x 4 # 1 x " 0 f !x" ! sx 2 # 2 x x ( 0 f f! 1 y! x f! 1 f! 1 ,f y! f! x "! 2 x 2! 8 x x " 2 1! sx 1 # sx f !x"! s 10! 3 x y! 2 x 3 # 3 f !x"! 4 x! 1 2 x # 3 f! x "! 3! 2 x 24. f m 0 c m! f !v"! m 0 s 1! v 2 %c 2 v 21. C! 59 !F! 32 " F " !459. y 0 1 x 1 f! 1! 0 "
- (a) What is a one-to-one function? f! 1! 2 " (b) How can you tell from the graph of a function whether it is one-to-one?
- (a) Suppose is a one-to-one function with domain and range. How is the inverse function defined? What is the domain of? What is the range of? (b) If you are given a formula for , how do you find a formula for? (c) If you are given the graph of , how do you find the graph of? 3–16 A function is given by a table of values, a graph, a formula, or a verbal description. Determine whether it is one-to-one.
- , is the height of a football t seconds after kickoff.
- is your height at age t.
- If is a one-to-one function such that , what is?
- If , find. If , find.
- The graph of is given. (a) Why is one-to-one? (b) What are the domain and range of f! 1? f f
- h!x"! x # sx h! 1! 6 " f !x"! x # cos x f! 1! 1 " f! 1! 9 " f f! 2 "! 9 f !t"
- f !t" h!x"! 1 # cos x 0 ) x ) & h!x"! 1 # cos x
t!x"! 1 %x t!x"! ( x (
f !x"! x 2! 2 x f !x"! 10! 3 x y x 8. x y y x x y 3. f! 1 f f! 1 f f! 1 f! 1 B f! 1 f A 7 E X E R C I S E S x 1 2 3 4 5 6 f !x" 1 2 3 5 2 2. x 1 2 3 4 5 6 f !x" 1 2 4 8 16 32
There are holes in the graph in Figure 1 corresponding to irrational values of x. We want to fill in the holes by defining , where , so that is an increasing continu- ous function. In particular, since the irrational number satisfies we must have and we know what and mean because 1 and 1 are rational numbers. Similarly, if we use better approximations for , we obtain better approximations for : .... .... .... It can be shown that there is exactly one number that is greater than all of the numbers ... and less than all of the numbers ... We define to be this number. Using the preceding approximation process we can com- pute it correct to six decimal places: Similarly, we can define (or , if ) where x is any irrational number. Figure 2 shows how all the holes in Figure 1 have been filled to complete the graph of the function . In general, if is any positive number, we define This definition makes sense because any irrational number can be approximated as closely as we like by a rational number. For instance, because has the decimal repre- sentation... , Definition 1 says that is the limit of the sequence of numbers Similarly, is the limit of the sequence of numbers It can be shown that Definition 1 uniquely specifies and makes the function continuous. ax f !x"! a x 5 3, 5 3, 5 3, 5 3, 5 3, 5 3, 5 3,... 5 & 2 1, 2 1, 2 1, 2 1, 2 1, 2 1, 2 1,... s 3! 1 2 s 3 s 3 ax ! lim r lx 1 ar r rational a f !x"! 2 x, x!! 2 x ax a ( 0 2 s 3 ) 3. 2 s 3 2 1, 2 1, 2 1, 2 1, 2 1, 2 1, 2 1, 2 1, 2 1, 2 1, 1 + s 3 + 1? 2 1 + 2 s 3 + 2 1. 1 + s 3 + 1? 2 1 + 2 s 3 + 2 1. 1 + s 3 + 1? 2 1 + 2 s 3 + 2 1. 1 + s 3 + 1? 2 1 + 2 s 3 + 2 1. s 3 2 s 3 2 1 2 1. 2 1 + 2 s 3 + 2 1. 1 + s 3 + 1. s 3 f !x"! 2 x x!! f SECTION 7 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES | | | | 393 0 1 x y 1 FIGURE 2 y=2®, x real N A proof of this fact is given in J. Marsden and A. Weinstein, Calculus Unlimited (Menlo Park, CA: Benjamin/Cummings, 1981). For an online version, see cds.caltech/~marsden/ volume/cu/CU
The graphs of members of the family of functions are shown in Figure 3 for var- ious values of the base a. Notice that all of these graphs pass through the same point because for. Notice also that as the base a gets larger, the exponential func- tion grows more rapidly (for ). Figure 4 shows how the exponential function compares with the power function . The graphs intersect three times, but ultimately the exponential curve grows far more rapidly than the parabola. (See also Figure 5.) You can see from Figure 3 that there are basically three kinds of exponential functions . If , the exponential function decreases; if , it is a constant; and if , it increases. These three cases are illustrated in Figure 6. Since , the graph of is just the reflection of the graph of about the -axis. The properties of the exponential function are summarized in the following theorem. THEOREM If and , then is a continuous function with domain and range. In particular, for all. If is a decreasing function; if , is an increasing function. If , and , , then
- ax!y ! 3. !ax "y ! axy 4. !ab"x ! axbx ax ay ax"y ! axay x y!! a # 1 f ab # 0 ! !0, %" ax # 0 x 0 $ a $ 1, f !x"! ax 2 a # 0 a " 1 f !x"! ax 1 (0, 1) (a) y=a®, 0<a<1 (b) y=1® (c) y=a®, a> (0, 1) FIGURE 6 0 x y 0 x y 0 x y y ! 1 #a"x ! 1 #ax ! a!x y!! 1 #a"x y! ax a # 1 y! ax 0 $ a $ 1 a! 1 y! x 2 y! x 2 y! 2 x y! 2 x FIGURE 3 0 x y 2 y=2® 4 y=≈ 100 200 6 FIGURE 5 0 x y 2 4 10 y=2® y=≈ FIGURE 4 0 x y 1 1® 1® ” 12 ’® ” 14 ’® 10® 4® 2® x # 0 a 0! 1 a " 0 !0, 1" y! ax 394 | | | | CHAPTER 7 INVERSE FUNCTIONS Members of the family of exponential functions
, so it exhibits the rapid growth that we observed in Figures 2 and 5. Under ideal conditions (unlimited space and nutrition and freedom from disease), this exponential growth is typical of what actually occurs in nature. What about the human population? Table 1 shows data for the population of the world in the 20th century and Figure 8 shows the corresponding scatter plot. The pattern of the data points in Figure 8 suggests exponential growth, so we use a graphing calculator with exponential regression capability to apply the method of least squares and obtain the exponential model Figure 9 shows the graph of this exponential function together with the original data points. We see that the exponential curve fits the data reasonably well. The period of rela- tively slow population growth is explained by the two world wars and the Great Depres- sion of the 1930s. DERIVATIVES OF EXPONENTIAL FUNCTIONS Let’s try to compute the derivative of the exponential function using the defini- tion of a derivative: ! lim h l 0 axah ! ax h ! lim h l 0 ax!ah ! 1 " h f &!x"! lim h l 0 f !x " h"! f !x" h ! lim h l 0 ax"h ! ax h f !x"! ax FIGURE 9 Exponential model for population growth 1900 6 x10' P 1920 1940 1960 1980 2000 t P! !0" " !1"t FIGURE 8 Scatter plot for world population growth 1900 6 x10' P 1920 1940 1960 1980 2000 t y! 2 t 396 | | | | CHAPTER 7 INVERSE FUNCTIONS TA B L E 1 Population Year (millions) 1900 1650 1910 1750 1920 1860 1930 2070 1940 2300 1950 2560 1960 3040 1970 3710 1980 4450 1990 5280 2000 6080
The factor doesn’t depend on h, so we can take it in front of the limit: Notice that the limit is the value of the derivative of at , that is, Therefore we have shown that if the exponential function is differentiable at 0, then it is differentiable everywhere and This equation says that the rate of change of any exponential function is proportional to the function itself. (The slope is proportional to the height.) Numerical evidence for the existence of is given in the table at the left for the cases and. (Values are stated correct to four decimal places.) It appears that the limits exist and In fact, it can be proved that these limits exist and, correct to six decimal places, the val- ues are Thus, from Equation 4, we have Of all possible choices for the base in Equation 4, the simplest differentiation formula occurs when. In view of the estimates of for and , it seems rea- sonable that there is a number between 2 and 3 for which. It is traditional to denote this value by the letter. Thus we have the following definition. DEFINITION OF THE NUMBER e Geometrically this means that of all the possible exponential functions , the function f !x"! exis the one whose tangent line at ( 0, 1"has a slope f &! 0 "that is exactly 1. y! ax lim h l 0 eh ! 1 h e is the number such that! 1 7 e a f &! 0 "! 1 f &! 0 "! 1 f &! 0 " a! 2 a! 3 a d dx ! 3 x " $ !1" 3 x d dx 6! 2 x " $ !0" 2 x d dx
! 3 x " %
x! 0 $ 1. d dx
! 2 x " %
x! 0 5 $ 0. f &! 0 "! lim h l 0 3 h ! 1 h for a! 3, $ 1. f &! 0 "! lim h l 0 2 h ! 1 h for a! 2, $ 0. a! 2 a! 3 f &! 0 " 4 f &!x"! f &! 0 "ax f !x"! ax lim h l 0 ah ! 1 h ! f &! 0 " f 0 f &!x"! ax lim h l 0 ah ! 1 h ax SECTION 7 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES | | | | 397 h 0 0 1. 0 0 1. 0 0 1. 0 0 1. 3 h ! 1 h 2 h ! 1 h
. Thus, by the Chain Rule, Putting , we have , so and It can be shown that the approximate value to 20 decimal places is The decimal expansion of is nonrepeating because is an irrational number. EXAMPLE 4 In Example 6 in Section 3 we considered a population of bacteria cells in a homogeneous nutrient medium. We showed that if the population doubles every hour, then the population after hours is where is the initial population. Now we can use (4) and (5) to compute the growth rate: For instance, if the initial population is cells, then the growth rate after two hours is M EXAMPLE 5 Find the absolute maximum value of the function. SOLUTION We differentiate to find any critical numbers: Since exponential functions are always positive, we see that when , that is, when. Similarly, when. By the First Derivative Test for Absolute Extreme Values, has an absolute maximum value when and the value is M EXPONENTIAL GRAPHS The exponential function is one of the most frequently occurring functions in calculus and its applications, so it is important to be familiar with its graph (Figure 12) and properties. We summarize these properties as follows, using the fact that this function is just a special case of the exponential functions considered in Theorem 2 but with base a! e # 1. f !x"! ex f! 1 "!! 1 "e! 1! 1 e $ 0. f x! 1 x $ 1 f &!x" $ 0 x # 1 f &!x" # 0 1! x # 0 f &!x"! xe!x!! 1 " " e!x! 1 "! e!x! 1! x" V f !x"! xe!x !! 4000 "!0" $ 2773 cells#h dn
dt %t! 2
$! 1000 "!0" 2 t &t! 2
n 0! 1000 dn dt $ n 0 !0" 2 t n 0 n! n 02 t t e e e $ 2. e! 21 #k $ 21 #0 $ 2. x! 0 1! ck c! 1 #k ex ! d dx !ex "! d dx ! 2 cx "! k 2 cx d dx !cx"! ck 2 cx f &! 0 " $ 0. SECTION 7 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES | | | | 399 N The rate of growth is proportional to the size of the population. y= ́ 0 x 1 y 1 FIGURE 12 The natural exponential function
PROPERTIES OF THE NATURAL EXPONENTIAL FUNCTION The exponential func- tion is an increasing continuous function with domain and range . Thus for all. Also So the -axis is a horizontal asymptote of. EXAMPLE 6 Find. SOLUTION We divide numerator and denominator by : We have used the fact that as and so M EXAMPLE 7 Use the first and second derivatives of , together with asymp- totes, to sketch its graph. SOLUTION Notice that the domain of is , so we check for vertical asymptotes by computing the left and right limits as. As , we know that , so and this shows that is a vertical asymptote. As , we have , so As , we have and so This shows that is a horizontal asymptote. Now let’s compute the derivative. The Chain Rule gives Since and for all , we have for all. Thus is decreasing on and on. There is no critical number, so the function has no maximum or minimum. The second derivative is f '!x"!! x 2 e 1 #x!! 1 #x 2 "! e 1 #x! 2 x" x 4 ! e 1 #x! 2 x " 1 " x 4 !!%, 0" !0, %" e 1 #x # 0 x 2 # 0 x " 0 f &!x" $ 0 x " 0 f f &!x"!! e 1 #x x 2 y! 1 lim x l(% e 1 #x ! e 0! 1 x l (% 1 #x l 0 lim x l 0! e 1 #x ! lim t l !% et ! 0 x! 0 x l 0! t! 1 #x l !% lim x l 0 " e 1 #x ! lim t l % et ! % x l 0 x l 0 " t! 1 #x l %
f 'x & x " 0 (
f !x"! e 1 #x lim x l% e! 2 x ! lim t l!% et ! 0 t!! 2 x l !% x l % ! 1 1 " 0 ! 1 lim x l % e 2 x e 2 x " 1 ! lim x l % 1 1 " e! 2 x ! 1 1 " lim x l % e! 2 x e 2 x lim x l% e 2 x e 2 x " 1 x f !x"! ex lim x l % lim ex ! % x l!% ex ! 0 !0, %" ex # 0 x f !x"! ex! 10 400 | | | | CHAPTER 7 INVERSE FUNCTIONS
402 | | | | CHAPTER 7 INVERSE FUNCTIONS 17–18 Find the exponential function whose graph is given. 18. 19. Suppose the graphs of and are drawn on a coordinate grid where the unit of measurement is 1 inch. Show that, at a distance 2 ft to the right of the origin, the height of the graph of is 48 ft but the height of the graph of is about 265 mi. ; 20. Compare the rates of growth of the functions and by graphing both functions in several viewing rect- angles. Find all points of intersection of the graphs correct to one decimal place. ; Compare the functions and by graphing both and in several viewing rectangles. When does the graph of finally surpass the graph of? ; 22. Use a graph to estimate the values of such that . 23–30 Find the limit. 23. 24. 25. 26. 27. 28. 29. x lim l % !e! 2 x cos x" 30. x llim! )# 2 "" etan x lim x l 2! lim e 3 #! 2 !x" x l 2 " e 3 #! 2 !x" x lim l % x lim l % e!x 2 e 3 x ! e! 3 x e 3 x " e! 3 x x lim l % !1"x x liml !% !1"x e x # 1,000,000, x t f f t 21. f !x"! x 10 t!x"! e x t!x"! 5 x f !x"! x 5 t f f !x"! x 2 t!x"! 2 x ”2, 29 ’ 0 2 y x 0 (1, 6) (3, 24) y x 17.
- (a) Write an equation that defines the exponential function f !x"! Ca x with base. (b) What is the domain of this function? (c) If , what is the range of this function? (d) Sketch the general shape of the graph of the exponential function for each of the following cases. (i) (ii) (iii)
- (a) How is the number defined? (b) What is an approximate value for? (c) What is the natural exponential function? ; 3–6 Graph the given functions on a common screen. How are these graphs related? 3. , , , 4. , , , , , , 6. , , , 7–12 Make a rough sketch of the graph of the function. Do not use a calculator. Just use the graphs given in Figures 3 and 12 and, if necessary, the transformations of Section 1.
- Starting with the graph of , write the equation of the graph that results from (a) shifting 2 units downward (b) shifting 2 units to the right (c) reflecting about the x-axis (d) reflecting about the y-axis (e) reflecting about the x-axis and then about the y-axis
- Starting with the graph of , find the equation of the graph that results from (a) reflecting about the line (b) reflecting about the line 15–16 Find the domain of each function.
- (a) (b)
- (a) t!t"! sin!e!t " (b) t!t"! s 1! 2 t f !x"! 1 1! e x f !x"! 1 1 " e x x! 2 y! 4 y! e x
- y! e x 11! 1! 12 e!x y! 2! 1! e x "
- y!! 2 !x y! 1 " 2 e x y! 4 x ! 3 y! 4 x! 3 y! 0 x y! 0 y! 0 y! 0
####### 5. y! 3 x y! 10 x y! ( 13 )x y! ( 101 )x
y! e x y! e !x y! 8 x y! 8 !x y! 2 x y! e x y! 5 x y! 20 x e e a # 1 a! 1 0 $ a $ 1 a " 1 a # 0 7 E X E R C I S E S
rumor at time and and are positive constants. [In Sec- tion 10 we will see that this is a reasonable model for .] (a) Find. (b) Find the rate of spread of the rumor. ; (c) Graph for the case , with measured in hours. Use the graph to estimate how long it will take for 80% of the population to hear the rumor. ; 60. An object is attached to the end of a vibrating spring and its displacement from its equilibrium position is , where is measured in seconds and is measured in centimeters. (a) Graph the displacement function together with the func- tions and. How are these graphs related? Can you explain why? (b) Use the graph to estimate the maximum value of the dis- placement. Does it occur when the graph touches the graph of? (c) What is the velocity of the object when it first returns to its equilibrium position? (d) Use the graph to estimate the time after which the displacement is no more than 2 cm from equilibrium. 61. Find the absolute maximum value of the function . 62. Find the absolute minimum value of the function ,. 63–64 Find the absolute maximum and absolute minimum values of on the given interval. 63. , 64. , 65–66 Find (a) the intervals of increase or decrease, (b) the inter- vals of concavity, and (c) the points of inflection. 65. 66. 67–68 Discuss the curve using the guidelines of Section 4. 68. ; 69. A drug response curve describes the level of medication in the bloodstream after a drug is administered. A surge function is often used to model the response curve, reflecting an initial surge in the drug level and then a more gradual decline. If, for a particular drug, , and is measured in minutes, estimate the times corresponding to the inflection points and explain their significance. If you have a graphing device, use it to graph the drug response curve. p! 4, k! 0 t A! 0, S!t"! At pe!kt 67! e! 1 #!x" 1 " y! e!x sin x f !x"! e x x 2 f !x"!! 1! x"e!x f !x"! xe!x )!1, 4* f !x"! x 2 e!x# 2 )!1, 6* 2 # 8 f t!x"! e x#xx # 0 f !x"! x! e x y! 8 e!t# 2 y! 8 e!t# 2 y!! 8 e!t# 2 y! 8 e!t# 2 sin 4t t y p a! 10 k! 0 t lim t l % p!t" p!t" 31– 46 Differentiate the function. t a k 31. 32. 34. 35. 36. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47– 48 Find an equation of the tangent line to the curve at the given point. 47. , Find if. 50. Find an equation of the tangent line to the curve at the point. 51. Show that the function satisfies the differen- tial equation. 52. Show that the function satisfies the differ- ential equation. 53. For what values of does the function satisfy the equation? 54. Find the values of for which satisfies the equation . If , find a formula for. 56. Find the thousandth derivative of. 57. (a) Use the Intermediate Value Theorem to show that there is a root of the equation. (b) Use Newton’s method to find the root of the equation in part (a) correct to six decimal places. ; 58. Use a graph to find an initial approximation (to one decimal place) to the root of the equation. Then use Newton’s method to find the root correct to eight decimal places. 59. Under certain circumstances a rumor spreads according to the equation where p!t"is the proportion of the population that knows the p!t"! 1 1 " ae!kt 4 e!x 2 sin x! x 2! x " 1 e x " x! 0 f !x"! xe!x 55. f !x"! e 2 x f !n"!x" y " y&! y'
- y! e *x y' " 6 y& " 8 y! 0 r y! erx y' " 2 y& " y! 0 y! Ae!x " Bxe!x 2 y'! y&! y! 0 y! e x " e! x / 2 xe y " ye x ! 1 !0, 1" e x 2 y
- y&! x " y y! !1, e" e x x y! e 2 x cos )x, !0, 1" 48.
y! cos+ f !t"! sin 2 !esin 2 t "
1! e 2 x
1 " e 2 x ,
y! y! s 1 " xe! 2 x ae x " b ce x " d y! e u ! e!u e u " e!u y! ee x y! s 1 " 2 e 3 x y! ek tan sx 37. F!t"! et sin 2t f !t"! sin!et " " esin t f !u"! e 1 #u t!x"! sx e x 33! e ax 3 y! e u !cos u " cu" y! e x 1 " x f !x"! !x 3 " 2 x"e x SECTION 7 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES | | | | 403
Inverse Function 6 Questions
Course: Mathematics
University: Sikkim Manipal University
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