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The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard A. Medeiros next Lesson 19

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Exponential Functions Exponential Functions © 2007 Herbert I. Gross next 2 x

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© 2007 Herbert I. Gross next Until now our discussion of exponents has been restricted to the cases in which the exponents were integers (that is, whole numbers and negative whole numbers). This was fine as far as compound interest problems were concerned. Namely, if your savings were earning 5% interest compounded annually, the bank would add the interest to your account at the end of the year. Introduction

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© 2007 Herbert I. Gross next On the other hand, suppose the cost of groceries increases at an annual rate of 5% a year and that a certain order of groceries costs $100 today. Then a year from today, the same order would cost $105 (i.e., 1.05 1 × $100). However, unlike the usual situation with bank interest, the cost of the groceries will not remain at $100 for 365 days and then suddenly jump.

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Note on Representation However, we usually write the new cost in the form 1.05 × $200. If the cost of an item increases by 5%, it means that what used to cost $1 will now cost $1.05. Since paying, say, $200 can be viewed as being equivalent to paying $1 two hundred times; if the item originally cost $200 it will now cost 200 × $1.05. next © 2007 Herbert I. Gross

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next Making the more realistic assumption that the cost of the groceries increases continuously, we now want to extend our concept of exponents to include such expressions as, say… (1.05) 1/2 × $100 …which would represent the cost of the groceries in 6 months (that is, half a year from now).

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© 2007 Herbert I. Gross next We do this in much the same way that we extended our definition of exponents from whole numbers to all integers. (1.05) 1/2 (1.05) 1/2 = next (1.05) 1/2 + 1/2 = (1.05) 1 = 1.05 Namely, we assume that our rules of arithmetic are to remain as they are. For example, to keep the rule for multiplying like bases in effect it would mean that… next

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© 2007 Herbert I. Gross next (1.05) 1/2 (1.05) 1/2 = (1.05) 1/2 + 1/2 = (1.05) 1 = 1.05 So if now we let x denote (1.05) 1/2, we see that… x(x) = 1.05 next x 2 = 1.05 And by the definition of the (positive) square root… x = √1.05 Since x = (1.05) 1/2, we see that… 1.05 1/2 = √1.05 next

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Technically speaking, if x 2 = 1.05, then © 2007 Herbert I. Gross next Note x = ± √1.05. However, based on the fact that the cost of the groceries is increasing by 5% per year, we know that the answer has to be + √1.05. To check whether 1.05 1/2 = √1.05, we may use the calculator to compute both… √1.05 and (1.05) 1/2. next

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To compute © 2007 Herbert I. Gross next Note √1.05, we enter 1.05 and then press the square root key to obtain… √1.05 = 1.024695… To compute (1.05) 1/2, we rewrite it as (1.05) 0.5 and then enter the key strokes… next and the display will also show 1.024695... 1.05xyxy.5=1.024695…

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© 2007 Herbert I. Gross next Again notice that the function f(x) = (1.05) x is not linear. f(x) = 1.05 x next f(0) = 1.05 0 = 1 …and we see that while 1 / 2 is midway between 0 and 1, f(1/2) ( that is, 1.024695…) is not midway between 1 ( i.e.,f(0) ) and 1.05 ( i.e.,f(1)). (It’s close, but to be linear it has to be exact.) next f( 1 / 2 ) = 1.05 1/2 = 1.024695… f(1) = 1.05 1 = 1.05 Note Namely… next

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© 2007 Herbert I. Gross next We could also have used the property (b m ) n = b mn with b = 1.05, m = 1 / 2 and n = 2 to obtain… [(1.05) 1/2 ] 2 = 1.05 1 = 1.05 next In fact if we want this property to remain true all the time, we see that for any positive integer p, the formula (b m ) n = b mn becomes… Note next (b m ) n = b mn1/pp (1/p) · p = b1= b1 = b= b In other words, b 1/p = √b. p

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© 2007 Herbert I. Gross next means. Note next Namely, it represents the number that has to be raised to the p th power to obtain b. Recall what Thus, for example, = 2 because 2 4 = 16. As a check, you may rewrite 1 / 4 as 0.25 and then enter the following sequence of key strokes on the calculator… 16xyxy.25= √b√b p √16 4 2 and the answer, 2, will appear in the display. next

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© 2007 Herbert I. Gross next Just as an anecdotal observation: even though b n does not mean the same thing as n b, in the special case where b = 2 and n = 4, it is true. That is, 4 2 = 2 4 ( = 16).

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© 2007 Herbert I. Gross next For reasons such as this, in this course we will limit our study of exponential functions to cases in which the base is positive. While 2 4 = 16, it is also true that ( - 2) 4 = 16. However, when the base is negative “strange things” seem to happen. For example, if we look at the powers of - 1, we see that if the exponents are integers, the signs of the powers alternate between 1 and - 1. That is… ( - 1) 2 = 1, ( - 1) 3 = - 1, ( - 1) 4 = 1, ( - 1) 5 = - 1, etc.

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© 2007 Herbert I. Gross next Note next Once we know how to compute b 1/p we can compute b q/p for any common fraction q / p. Namely, q / p = 1 / p (q) and therefore, b q/p = (b 1/p ) q. For example, 16 3/4 = (16 1/4 ) 3 = 2 3 = 8 Notice that while 3 / 4 is 3 / 4 of the way between 0 and 1, 16 3/4 is only about halfway between 16 0 and 16 1.

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© 2007 Herbert I. Gross next Definition With the above discussion in mind, b n is now defined for any rational number n. ), they can be approximated by rational numbers to as great a degree of accuracy as we desire. √2√2 next As for irrational numbers (such as

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© 2007 Herbert I. Gross next = 1.414213562… √2√2 16 1.41 = 49.866533098…16 1.414 = 50.422648707…16 1.4142 = 50.450616714…16 1.41421 = 50.452015521…16 1.414213 = 50.452274481…16 1.4142135 = 50.452505113…16 1.41421356 = 50.452507911…16 1.414213562 = 50.452513507…16 1.414213563 = 50.452513925…16 = ? √2√2 next Thus, measured to 6 decimal place accuracy, 16 1.414213562, and 16 1.414213563 are all equal to 50.452514 16 √ 2 next On my calculator, was between 1.414213562 and 1.414213563; and the following results were obtained… √2√2 16 1.4 = 48.502930128…

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© 2007 Herbert I. Gross next Interest that is compounded annually comes with the disadvantage that the investor is “punished” if the money is withdrawn as little as one day before the end of the year. For example, suppose you invest $10,000 at an interest rate of 8% compounded annually. At year’s end, your investment will be worth $10,000(1.08) or $10,800. However, if you withdraw the money early (i.e., before the interest is payable), you will only get your original $10,000 back. Annual Compounding

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© 2007 Herbert I. Gross next Some banks compound interest semi-annually instead of annually. That is, the bank would pay you half (4%) of the 8% interest after half the year. In this way, after 6 months your $10,000 investment increases in value to $10,400. Semi-annual Compounding Thus, for the second half of that year, the other 4% (of your 8% yearly interest) is paid on $10,400. Since 4% of $10,400 is $416, the year ends with your having earned $816 in interest, which is $16 more than would be earned with annual compounding. next

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© 2007 Herbert I. Gross next As you see, the more often interest is compounded during the year, the more money your investment will earn. Compound Interest For example, suppose you again invest $10,000 at an interest rate of 8% but now it is being compounded quarterly, i.e., 4 times a year (every 3 months). Then at the end of 3 months, your investment earns 2% (that is 1/4 of 8%); the next 3 months the new amount of your investment increases by 2%, etc. next

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© 2007 Herbert I. Gross next The table below illustrates your $10,000 investment at 8% interest being compounded quarterly… Compound Interest next After 3 months(1.02) 1 × $10,000$10,200(1.02) 1 ($10,000)After 6 months(1.02) 1 × $10,200$10,404(1.02) 2 ($10,000)After 9 months(1.02) 1 × $10,404$10,612.08(1.02) 3 ($10,000)After 12 months(1.02) 1 × $10,612.08$10,824.16(1.02) 4 ($10,000)

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© 2007 Herbert I. Gross next The more often interest is compounded, the closer the investment will appear to be earning interest “continuously”. Compound Interest For example, suppose once again that we have invested $10,000 at an interest rate of 8%. To keep the arithmetic relatively simple, let’s look only at the powers of 10; that is, we will see what the investment looks like if it is compounded 10 times a year, or 100 times per year, or 1,000 times per year, etc. next

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© 2007 Herbert I. Gross next If the interest is compounded 10 times per year, it will increase the value of the investment by 8% ÷ 10, i.e., by 0.8% (the investment multiplied by 1.008), each of the 10 times during the year. Compound Interest That is: at the end of one year, the investment will be worth (1.008) 10 ($10,000); or to the nearest cent… next $10,829.42

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© 2007 Herbert I. Gross next If the interest is compounded 100 times per year, the value of the investment will increase by 8% ÷ 100 or 0.08% each time. That is, the investment is multiplied by 1.0008 each of the 100 times during the year. Compound Interest So after one year, the investment will be worth (1.008) 100 ($10,000); or to the nearest cent… next $10,832.52

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© 2007 Herbert I. Gross next If the interest is compounded 1,000 times per year: each time, it will increase the value of the investment by 8% ÷ 1000 or 0.008 (i.e., the investment is multiplied by 1.00008) each of the 1,000 times during the year. Compound Interest That is, the investment will be worth (1.00008) 1,000 ($10,000); or to the nearest cent… next $10,832.52 And if compounded 10,000 times a year, it will be worth (1.000008) 10,000 ($10,000); or, $10,832.87

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© 2007 Herbert I. Gross next Based on these calculations, we begin to realize that… Enrichment Note on Compound Interest Even though the investment is increasing in value each time interest is compounded, after a while the increase has become negligible. next For example, we just showed that the difference between compounding the interest 100 times a year and 1,000 times a year is only $0.35 per $10,000.

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© 2007 Herbert I. Gross next If interest is compounded 1,000,000 times per year, each compounding time will increase the value of the investment by 8% ÷ 1,000,000 or 0.000008%. That is: the value of the investment is multiplied by 1.00000008 each of the 1,000,000 times during the year. Enrichment Note on Compound Interest Thus at year’s end, the $10,000 investment will be worth (1.00000008) 1,000,000 ($10,000) or, to the nearest cent, $10,832.87; which to the nearest cent is the same amount we got when the interest was compounded 10,000 times. next

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© 2007 Herbert I. Gross next Enrichment Note on Compound Interest next The following chart summarizes our discussion. It shows the value of a $10,000 investment at the end of one year if the interest rate is 8% and the interest is compounded n times per year.

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© 2007 Herbert I. Gross next Value in Dollars to the Nearest Cent n 1 10,800 4 10,824.32 10 10,829.42 100 10,832.52 1,000 10,832.84 10,000 10,832.87 100,000 10,832.87 1,000,000 10,832.87 continuously 10,832.87 next Enrichment Note on Compound Interest * As n increases, the value of the investment increases; but eventually, since rounding is to the nearest cent, the amount of that increase becomes unnoticeable. *

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© 2007 Herbert I. Gross next Notice that as the number of times per year interest is compounded increases, the effect on the value of the investment is less and less; and eventually becomes negligible. It is in this context that we may say that the interest is being compounded continuously Enrichment Note on Compound Interest next We are “making a fuss” about this effect because continuous compounding is a very important phenomenon in nature.

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Continuous Compounding There is a number, denoted by e (which to 9 decimal place accuracy is 2.718281828) such that if the $10,000 is invested at an interest rate of 8% compounded continuously, at the end of the year the value of the investment will be e 0.08 ($10,000). As a check, to the nearest cent… (2.718281828) 0.08 ($10,000) = $10,832.87 next

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© 2007 Herbert I. Gross next e 0.08 ($10,000) Enrichment Note on Compound Interest next If we replace 0.08 by r and $10,000 by P in e 0.08 ($10,000) = $10,832.87, we obtain the formula… A = Pe r …where A is the value of the investment at the end of one year. Generalization:

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© 2007 Herbert I. Gross next Enrichment Note on Compound Interest next At the end of the first year, the value of the investment (A) is Pe r. It is this amount (Pe r ) that is multiplied by e r to determine the value of the investment at the end of the year. Hence, at the end of two years, the value of investment A is, (Pe r )e r or Pe 2r A = Pe r Generalization

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© 2007 Herbert I. Gross next Enrichment Note on Compound Interest next Since at the end of two years, the value of the investment A is Pe 2r, it is this amount that is multiplied by e r to determine the value of the investment at the end of the third year. Hence, at the end of three years, the value of investment A is (Pe 2r )e r or Pe 3r And continuing in this way, we see that at the end of t years the value is given by… A = Pe tr = Pe rt

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© 2007 Herbert I. Gross next If $8,000 is invested at an interest rate of 7% compounded continuously for 6 years, the value of the investment (A) at that time will be given by… “Interest”ing Example next A = $8,000e 0.07(6) = $8,000e 0.42 …and if we round e off to 2.718281828, the formula above becomes… A = $8,000(2.718281828) 0.42 …and with the aid of a calculator we see that to the nearest cent, the value of the investment is $12,175.69. next A = Pe rt

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© 2007 Herbert I. Gross next As a “reality” check, we could assume that the interest rate was 7% compounded 1,000 times a year for 6 years; which means that in 6 years the interest would have been compounded 6,000 times. “Interest”ing Example In that case, the value of the investment would have been $8,000(1.00007) 6,000 or $12,175.51. So this indicates that our answer, $12,175.69, is indeed reasonable. next

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© 2007 Herbert I. Gross next Discrete Variables Sometimes the domain (input) of a function has to be an integer (often a whole number). For example, if toys cost $4 each, the cost (C) of x toys would be given by… C = f(x) = 4x In this case, the domain of f is the set of whole numbers (because you can’t buy either a fractional part of a toy or a negative number of toys). That is, we can buy no toys, 1 toy, 2 toys, etc. In this case we refer to x as being a discrete variable. next

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© 2007 Herbert I. Gross next Continuous Variables At other times, the domain is a continuous set of numbers. For example, if the speed of an object is 4 miles per minute, at the end of x minutes it has traveled m miles, where… m = f(x) = 4x In this case, the domain of f includes every instant that the object was traveling. In this case, we refer to x as being a continuous variable. next

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© 2007 Herbert I. Gross next Note on Notation So while f seems to be the same in both C = f(x) = 4x and m = f(x) = 4x, they have different domains; which means that they are different functions. C = f(n) = 4n To distinguish between a discrete and a continuous variable, we usually use the letter “n” to denote a discrete variable and the letter “x” to denote a continuous variable. So in the case of buying toys, we would rewrite the formula C = f(x) = 4x as… next

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© 2007 Herbert I. Gross next Graphing C = f(n) = 4n The graph of this function is the set of discrete points (0,0), (1,4), (2,8), (3,12) and in general (n,4n) for any whole number, n. next (0,0) (1,4) (2,8) (3,12) (n,4n) Geometrically… n c

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© 2007 Herbert I. Gross next Graphing m = f(x) = 4x In this case the graph would become the straight line y = 4x; that is, the set of all points in the plane of the form (x,4x). In essence it converts the set of discrete points in the previous graph into the graph shown on the right. next (n,4n) (0,0) (1,4) (2,8) (3,12) (x,4x) y = 4x n c x m

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© 2007 Herbert I. Gross next Graphing Exponential Functions Recall that in earlier lessons we learned how to graph any function of the form f(n) = b n ; where n is any integer. next ( - 3, 1 / 8 ), ( - 2, 1 / 4 ), ( - 1, 1 / 2 ), (0,2), (1,2), (2,4), (3,8), …(n,2 n ) As a specific example: suppose we let b = 2. In that case, the graph y = 2 n would consist of the discrete points in the plane: ( - 3,2 -3 ), ( - 2,2 -2 ), ( - 1,2 -1 ), (0,2 0 ), (1,2 1 ), (2,2 2 ), (3,2 3 )… (n,2 n ). That is…

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© 2007 Herbert I. Gross ( - 1, 1 / 2 ) (1,2) (3,8) (0,1) (-2,1/4)(-2,1/4) ( - 3, 1 / 8 ) (4,16) next Thus, the geometric graph would include the set of points indicated below.

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Now that we have learned to handle non-integral exponents, the graph can be converted into the continuous curve y = 2 x that is shown below… © 2007 Herbert I. Gross No matter whether x is positive or negative, 2 x is always positive. Therefore, the curve y = 2 x never touches or goes below the x-axis. next

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© 2007 Herbert I. Gross next Graphing Exponential Functions There is an interesting relationship between the graphs of f(x) = 2 x and g(x) = ( 1 / 2 ) x. Namely, since 1 / 2 = 2 -1 we may rewrite ( 1 / 2 ) x as (2 -1 ) x, which means that g(x) = (2 -1 ) x = 2 -x = f( - x). The fact that g(x) = f( - x) means that its graph is the reflection of the graph of f(x) with respect to the y-axis. More specifically…

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© 2007 Herbert I. Gross next y= 2 X y= 2 -X Note that the two curves intersect at the point (0,1). More generally, points on the y-axis remain stationary when a curve is reflected about the y-axis.

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© 2007 Herbert I. Gross next Graphing Exponential Functions Our present illustration was a special case of f(x) = b x for which b = 2. Replacing 2 by any number b leads to a graph that is similar to the previous one. All the graphs will contain the point (0,b 0 ); that is, the point (0,1); and for each choice of b, the graph will pass through the point (1,b) next

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© 2007 Herbert I. Gross next Building Vocabulary If x is positive, we refer to f(x) = b x as exponential growth (because as x increases, so does f(x)). On the other hand, if x is negative, we refer to f(x) = b x, as exponential decay (because as x increases, f(x) decreases). next

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© 2007 Herbert I. Gross next Radioactive Decay In science, one often talks about the half-life of a radioactive substance. It is the amount of time that it takes for the substance to “decay” to half of its original weight. For example, if the half-life of a particular substance is 1,000 years, it means that a thousand years from now, it will be half of its present weight. next

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© 2007 Herbert I. Gross next Radioactive Decay So if the original weight is 64 grams, 1,000 years from now its weight will be 32 grams; 1,000 years later (that is 2,000 years from now), its weight will be 16 grams; and 3,000 years from now, its weight will be 8 grams, etc… The half-life of radioactive carbon is often used as an aid for determining the age of an ancient artifact. next

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