Exponentiation does not distribute over addition or subtraction.Enough parentheses must be used when entering an exponential into your calculator or computer so that the typed expression
Trang 1Exponentiation does not distribute over addition or subtraction.
Enough parentheses must be used when entering an exponential into your calculator or computer so that the typed expression is interpreted as you intended
These points are illustrated below
bc2= bcc = (bc)2 e.g., for x = 0, −x2is negative while (−x)2is positive (ab)n= anbn e.g., (−2x)2= (−2)2· x2= 4x2 = −2x2= −(2x2)
a b
n
=a
n
3= 4 3
1/2
=4
1/2
31/2 = 2
31/2 =√2
3 But
(a + b)n = an+ bn e.g., (2 + 3)2= 52= 25 but 22+ 32= 4 + 9 = 13 (a − b)n = an− bn e.g.,√
25 − 9 = (25 − 9)1/2= 161/2= 4 but 251/2− 91/2= 5 − 3 = 2 Entering
in a calculator gives 2x5+√2 not 2x5+√2or (2x)5+√2 To enter (2x)5+√2you can type
Different calculators may have slightly different conventions Check yours out either
by using the instruction manual or by playing around with numbers you know Rounding off the bases of exponentials can cause substantial inaccuracy Let your calculator work for you; do not round off early
EXERCISE 9.2 Use a calculator or computer to evaluate the following to two decimal places
(a) π3(1 +23)50/3 (b) 100(1 +0.0512 )12·60 (c) (3.001)0.0013.001−33
Answer
(a) 154500.10 (b) 1996.07 (c) 56.73
EXERCISE 9.3 Each statement that follows is incorrect Find the incorrect step and correct it.
Trang 2i) 3−8x−2=√3−83x−2=1
2x
−2/3
ii)
B3C−1 2B
−3
=
B2C−1 2
−3
=B
−6C3
2−3 =2
−3C3
B6 = C
3
8B6
iii) 4x2+ 16y2= 4(x2+ 4y2) =√4 x2+ 4y2= 2(x + 2y) = 2x + 4y
n
Bn +C
n
Dn =A
nDn+ CnBn
BnDn =(AD)
n
+ (CB)n (BD)n = AD + CB
BD
n
v) 1 2
−1 1
x +1 y
−1
= 2 1
x +1 y
−1
= 2(x + y) = 2x + 2y
Answers are provided at the end of the chapter.
i)
√
2√
x3
(2x)−1/2 ii) 4(9y−x)1/2 iii)b
x+w− bx
3
√ 64x3)1/2
1 2
3R+ QR+1
Q2R
Answers are provided at the end of the chapter.
The Exponential Function The function f is called an exponential function if it can be written in the form f (t) = Cbt, where C is a constant and b is a positive constant This function is called exponential because the variable, t, is in the exponent The domain of the exponential function is all real numbers,
and the range is all positive real numbers If a quantity Q grows or decays exponentially
with time t, then Q(t) = Cbtfor some constants C and b There are two unknown constants,
so two data points are necessary in order to find them At t = 0, Q(0) = Cb0so C = Q(0) Sometimes the quantity Q(0) is denoted by Q0, indicating the initial quantity; we write
Q(t ) = Q0bt
In Figure 9.3(a) on the following page are the graphs of f (t) = btfor b = 1, 2, 3, and 10
In Figure 9.3(b) are graphs of f (t) = b−tfor b = 2, 3, and 10 or, equivalently, of f (t) = bt
for b =12,13, and101 The key notion is that numbers greater than 1 increase when squared, cubed, etc
22= 4; 23= 8; 24= 16; 25= 32; while numbers between 0 and 1 decrease when squared, cubed, etc
1 2
2
=1
4;
1 2
3
=1
8;
1 2
4
= 1
16;
1 2
5
= 1
32. Recall that x−k=x1k Therefore,12t=12tt =21t = 2−t; the graph of (12)t= 2−t can
be obtained by reflecting the graph of 2t across the y-axis Similarly, 13t= 3−t and
1 10
t
= 10−t Therefore, the functions graphed in Figure 9.3(b) can also be written as
f (t ) =12
t
, f (t) =13t, and f (t) =101t, respectively
Trang 31 2 3
4 6 8 10 12
2
f=2t
f=3t
f=10t
t
f
f=1t
–1 –2
2 4 6 8 10 12
t
f f=3-t f=10-t
f=2-t
Figure 9.3
Observe that btis always positive
Suppose b > 1
i The graph of bt is increasing and concave up bt increases at an ever-increasing rate
ii The more negative t becomes the smaller btbecomes In fact, by taking t negative enough, bt can be made arbitrarily close to zero limt →−∞bt= 0
Suppose 0 < b < 1
i The graph of btis decreasing and concave up
ii As t increases without bound bt gets arbitrarily close to 0 We can write this as
t → ∞, bt
→ 0; equivalently, limt →∞bt
= 0
The graph of bt has a horizontal asymptote at y = 0
EXERCISE 9.5
(a) Using your graphing calculator or computer, graph f (t) = C · 2tfor different values of
C (Be sure to include both positive and negative values of C.) How does the value of C affect the graph? Does it alter the y-intercept? The horizontal asymptote? If something changes, explain how it changes Do your answers make sense to you?
(b) Using your graphing calculator or computer, graph f (t) = 2−t+ D How does the value of D affect the graph? Does it alter the y-intercept? The horizontal asymptote?
If something changes, explain how it changes Do your answers make sense to you? (c) Given your answers to parts (a) and (b), take a guess at what the graph of f (x) = 3 − 2x
looks like (Note: This function is the same as g(t) = −2t
+ 3.)6Check your answer using your graphing calculator Then try h(t) = −3 − 2−t
(d) Using your graphing calculator, graph f (t) = 2kt for different values of k (Be sure
to include both positive and negative values of k.) How does the value of k affect the
6 Did changing the order in which the terms were written make the problem easier for you? If so, you need to practice fiddling around with the form in which expressions are written so you can help yourself by transforming an expression that looks unfamiliar
Trang 4graph? Does it alter the y-intercept? The horizontal asymptote? If something changes, explain how it changes Rewriting f (t) as 2k t
helps in interpreting the results
Manipulating Exponents: Taking Control
The three exponent laws given at the beginning of this section are identities; they can be read from left to right or from right to left Flexibility and a sense of your own goals helps
in applying them when working with exponential functions
problem An observer hypothesizes that in a certain region of the Sinai the number of plastic bags littering the desert is increasing exponentially Two months after he arrived he estimates there are 100 plastic bags littering the region, and 7 months after he arrived he estimates there are 120 plastic bags He sets the date of his arrival as a benchmark time of t = 0 Using the two data points and the observer’s hypothesis of exponential increase, find an equation for B(t), the number of bags in this region t months after the observer arrived
Approach 1
The hypothesis of exponential growth implies
B(t ) = Cbt for constants C and b
We know that
when t = 2, B = 100 and when t = 7, B = 120
Therefore,
100 = Cb2 and
120 = Cb7
We have a pair of simultaneous equations with two unknowns Our goal is to eliminate a variable One strategy is to divide one equation by the other, eliminating C Equivalently,
we could solve each equation for C and set them equal We’ll do this
C =100
b2 and
C =120
b7 So
100
b2 =120b7 Multiplying both sides by b7/100 leaves b’s on one side
b7
b2=120 100
b5= 1.2 Raise both sides of the equation to 1/5 to solve for b
(b5)1/5= (1.2)1/5
b = (1.2)1/5
Trang 5We now have
B(t ) = C[(1.2)1/5]t B(t ) = C(1.2)t /5
We can use either of the data points to solve for C
100 = C(1.2)2/5 100
(1.2)2/5= C
(6/5)0.4 = 100(5/6)0.4 Therefore,
B(t ) = 100(5/6)0.4(1.2)t /5
Approach 2
We have two data points: when t = 2, B = 100 and when t = 7, B = 120 In 5 months the number of bags has increased by 20 Because we’re hypothesizing exponential growth, the
percentincrease is the important bit of information, not the actual increase itself Twenty
is 20% of 100, so every 5 months the number of bags increases by 20%
B(t ) = C(1.2)t /5 Now find C as done above
Note that if we were willing to adjust our benchmark time of t = 0 from the observer’s day of arrival in the Sinai to the day he made the first bag count we would have
B(t ) = 100(1.2)t /5, where t = 0 corresponds to 2 months after his arrival
the interval [x, x + 3] is given by
final value − initial value initial value =f (x + 3) − f (x)f (x) , or Ca
2(x+3)− Ca2x
Simplify this expression
SOLUTION
C[a2(x+3)− a2x]
Ca2x Factor out the C
=a
2x+6− a2x
a2x We want to factor out a2x, so we write
a2x+6as a2x· a6
=a
2x
· a6− a2x
a2x Factor out a2x
=a
2x[a6− 1]
a2x
= a6− 1
Trang 6Notice that the percent change in f over any interval 3 units in length is a constant The
percent change does not depend upon the particular endpoints of the interval This is a defining characteristic of exponential functions
P R O B L E M S F O R S E C T I O N 9 2
For Problems 1 through 9, simplify the following expressions.
1 x−1+ z−1 (z + x)x−2
2 (xy)
−3
xy−3
3
√ 2x3+12y3x4
6y2x5
−1
zx−1+ z−1
5 z
0x−1y−2
z−2x−1y2
6 xn+1y
2n
x y
n
7 (ab
x)−2 (ab)−x
8 (ab)
−x
a−x+ b−y
9 (a−x+1b)
3
(a2b3)x
For Problems 10 through 15, factor bxout of the following expressions Check your answer by multiplying out.
10 bx+y+ bx
11 b2x+ bx+1
12 3b2x+1− 4b2x−1
13 (ab2)x+ ab
−x
Trang 714 b3x− (2b)−1+2x
15.√
abx− a√b3x
For Problems 16 through 21, if the statement is always true, write “True;” if the statement is not always true, produce a counterexample In these problems, a and b are positive constants.
16.√
a2x= ax
17 (ax
+ bx)1 = a + b
18
a2x
b−2x = (ab)x
19 a−x+ ax= 1
20 axbcx= (axb)(ac)x
21 32x+ 23x= 9x+ 8x
For Problems 22 through 24, simplify as much as possible.
22 a
x+y− a2x
ax
23 a
2x− b4x
ax+ bx
24 A4+p− A5p
A2+p− A3p
25 The graphs below are graphs of functions of the form f (t) = Cat For each graph, determine the sign of C and whether a ∈ (0, 1) or a > 1
f f
Trang 8For Problems 26 through 29, find a function of the form f (x) = Cax+ D to fit the
graph given.
26
x
f
(1, 5)
3
27
x
y = 2
f
(–1, 3) 3 4
2 1
28
x
y = –1
f
(1, –4) –2
–1
–3 – 4
29
x
f
(2, 5) 1.5
30 True or False: If the statement is always true, write “True;” if the statement is not always true, produce a counterexample
Trang 9(a) (a2+ b2)1/2= a + b (b) (a + b)−1=a+b1 , a, b = 0
(c) (a + b)−1=a1+1b, a, b = 0 (d) R−1/2= −√1
R, R > 0 (e) xz
+ xz= 2xz (f ) xzxz= x(z2)
(g) xzxz= x2z
31 Mix and Match: Below are functions To each function match the graph that best fits the function Since there are no units on the graphs, you may match one graph to several
different functions Note: First do this problem without using a graphing calculator.
You can then check your answers with the calculator If you made a mismatch, figure out what your mistake was and learn from it
(a) f (x) = 3(32)x (b) f (x) = −2(0.4)x
(c) f (x) = 2 − 3−x (d) f (x) = 4(23)x
(e) f (x) = 4−x (f ) f (x) = 1 + 2x
x
f
(vii)
32 Simplify as much as possible:
(a) x
2y
+ xy+2
√ x
x −1/2 y − 1
y − xy2 (c) AB+4− A3B
AB(A2− AB) (d) y
3w
− yw+4
yw(yw+ y2)
33 Let f (x) = ax We know that replacing x by x + k shifts the graph of f left k units For exponential functions this is equivalent to vertical stretching or shrinking Explain
In Problems 34 through 37, evaluate the limits.
34 (a) lim
x→−∞−2(1.1)x
35 (a) lim
x→∞(0.89x
x→−∞(0.89x
− 1)
36 (a) lim
x→∞(35)−x (b) lim
x→−∞(35)−x
Trang 1037 (a) lim
x→∞−5(23)x+ 7 (b) lim
x→−∞−5(23)x+ 7
38 Factor bxout of each of the following expressions
(a) 3bx
− b2x (b) (3b)x
− bx+2 (c) b3x/2− b2x−1
9.3 APPLICATIONS OF THE EXPONENTIAL FUNCTION
In Section 9.2 we looked at exponential growth in the context of population growth under ideal conditions In this section we’ll look at the growth of money in an interest-earning account and the decay of radioactive materials
The Growth of Money in a Bank Account
Exponential functions arise in questions dealing with money earning interest in a bank account and in calculations of loan paybacks If money is deposited in a bank and left there (with no additional deposits), then the growth of the amount of money in the account is due to the interest earned The amount of interest earned during each period is directly proportional to the amount of money in the account Let’s look at an example
(a) Write a function that gives the amount of money in the account after t years
(b) How long does it take for the money to double?
SOLUTION
(a) At the end of the first year, the interest added to the $5000 we already had is 4% of
$5000, or $200 (= 0.04 · $5000) We start the next year with $5200, and at the end of that year get interest amounting to 4% of $5200, or $208, giving a total of $5408, and
so on Each year the amount in the account is 104% of the previous year’s amount Recall that if a population grows at a rate of 4% per year, then P (t) = P0 (1.04)t, where P0is the population at t = 0 Similarly, if money in a bank account grows at a rate of 4% per year, then M(t) = M0 (1.04)t, where M0is the amount of money in the account at t = 0 and t is measured in years Our function is
M(t ) = 5000 (1.04)t (b) To find out how long it takes for the money to double we want to solve 10,000 =
5000 (1.04)t, or, equivalently, 2 = (1.04)t Observe that doubling time is independent
of the amount of money we have in the account Suppose the amount in the bank initially
is denoted by B0 The doubling time is the solution to the equation 2B0= B0 (1.04)t Notice that we are back to the equation 2 = (1.04)t
This equation can be solved for t analytically using logarithms to obtain an exact answer.7The solution can be approximated using a graphing calculator Solutions to an equation of the form f (x) = k, where k is a constant, can be approximated graphically
in two ways.8
7 We will take this up in Chapter 13.
8 You might also have an “equation solver” on your calculator that allows you to approximate the solution.